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You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
# Example 2:
# Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
# Output: 7
# Explanation:
# 1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
# 2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
# 3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
# 4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
# 5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
# 6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
# 7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
# The highest profit was $122 after rotating the wheel 7 times.
# Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
# Output: -1
# Explanation:
# 1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
# 2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
# 3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
# 4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 12 * $1 - 4 * $92 = -$356.
# 5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
# The profit was never positive, so return -1.
# seems like a straightforward linear algorithm, just plug in each total amount of folks on the gondola
# use a queue (just an integer of folks waiting) so for each x in A, add the value x onto the q of folks waiting, and we can serve 4 folks at a time, using the formula:
# Input: customers = [8,3], boardingCost = 5, runningCost = 6
# Output: 3
# Explanation: The numbers written on the gondolas are the number of people currently there.
# 1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
# 2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
# 3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
# The highest profit was $37 after rotating the wheel 3 times.
# what if we rotate the gondola another time ^^ for above example?
# oh, we ran out of customers, duh
class Solution:
def minOperationsMaxProfit(self, A: List[int], profit: int, loss: int, wait = 0, total = 0, best = 0, bestIndex = -1) -> int:
i = 1
def rotate():
nonlocal i, wait, total, best, bestIndex
take = min(wait, 4); total += take; wait -= take
cand = total * profit - i * loss
if best < cand:
best = cand
bestIndex = i
i += 1
for x in A:
wait += x
rotate()
while wait:
rotate()
return bestIndex
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
earn, max_earn = 0, 0
i, n = 0, len(customers)
wait, res = 0, -1
while i < n or wait > 0:
if i < n:
wait += customers[i]
earn += min(4, wait) * boardingCost - runningCost
if earn > max_earn:
res = i + 1
max_earn = max(max_earn, earn)
wait -= min(4, wait)
i += 1
return res
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
if 4 * boardingCost <= runningCost:
return -1
profit = 0
pre_max = 0
wait = 0
ans = 0
i = 0
for c in customers:
wait += c
if wait > 0:
profit += boardingCost * min(wait, 4) - runningCost
if profit > pre_max:
ans = i + 1
wait -= min(wait, 4)
i += 1
pre_max = max(pre_max, profit)
while wait > 0:
profit += boardingCost * min(wait, 4) - runningCost
if profit > pre_max:
ans = i + 1
wait -= min(wait, 4)
i += 1
pre_max = max(pre_max, profit)
if pre_max <= 0:
return -1
else:
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
ans = 0
anst = -1
waiting = 0
done = 0
times = 0
i = -1
while waiting > 0 or i<len(customers):
i+=1
if i<len(customers):
c = customers[i]
else:
c = 0
times+=1
waiting+=c
done+=min(4,waiting)
waiting-=min(4,waiting)
tans = done*boardingCost-times*runningCost
if tans>ans:
ans = tans
anst = times
# print(waiting, i, c, tans, done, times)
return anst
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
self.wait, self.count, self.ans, self.total, self.max_profit = 0, 0, 0, 0, 0
def helper(num):
temp = self.wait + num
if temp >= 4:
board = 4
temp -= 4
else:
board = temp
temp = 0
self.wait = temp
self.count += 1
self.total += (board * boardingCost - runningCost)
if self.total > self.max_profit:
self.max_profit = self.total
self.ans = self.count
for num in customers:
helper(num)
while self.wait > 0:
helper(0)
return self.ans if self.ans != 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
if 4 * boardingCost <= runningCost:
return -1
s, r, maxr, maxn = 0, 0, 0, -1
for i, c in enumerate(customers):
s += c
b = min(s, 4)
r += b * boardingCost - runningCost
s -= b
if r > maxr:
maxr, maxn = r, i+1
r += 4 * (s // 4) * boardingCost - (s // 4) * runningCost
if r > maxr:
maxr, maxn = r, len(customers) + (s // 4)
if s % 4 > 0:
r += (s % 4) * boardingCost - runningCost
if r > maxr:
maxr, maxn = r, len(customers) + (s // 4) + 1
return maxn
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
waiting = 0
onBoard = []
profit = 0
maxProfit = 0
ans = 0
for i, customer in enumerate(customers):
waiting += customer
if waiting >= 4:
profit += (4*boardingCost - runningCost)
waiting -= 4
onBoard.append(4)
else:
profit += (waiting*boardingCost - runningCost)
onBoard.append(waiting)
waiting = 0
freeRound = 3
j = 1
while j <= min(i, 3) and onBoard[-j] == 0:
j += 1
freeRound -= 1
stopNowProfit = profit - freeRound*runningCost
if stopNowProfit > maxProfit:
maxProfit = stopNowProfit
ans = i+1
while waiting > 0:
i += 1
if waiting >= 4:
profit += (4*boardingCost - runningCost)
waiting -= 4
onBoard.append(4)
else:
profit += (waiting*boardingCost - runningCost)
onBoard.append(waiting)
waiting = 0
freeRound = 3
j = 1
while j <= min(i, 3) and onBoard[-j] == 0:
j += 1
freeRound -= 1
stopNowProfit = profit - freeRound*runningCost
if stopNowProfit > maxProfit:
maxProfit = stopNowProfit
ans = i+1
if ans == 0:
return -1
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
netProfit = 0
maxNet = 0
waitSize = 0
nRot = 0
optRot = -1
for customerSize in customers:
waitSize += customerSize
gondSize = min(4, waitSize)
waitSize = max(0, waitSize-4)
netProfit += boardingCost * gondSize - runningCost
nRot += 1
if netProfit > maxNet:
maxNet = netProfit
optRot = nRot
while waitSize > 0:
gondSize = min(4, waitSize)
waitSize = max(0, waitSize-4)
netProfit += boardingCost * gondSize - runningCost
nRot += 1
if netProfit > maxNet:
maxNet = netProfit
optRot = nRot
return optRot
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], bc: int, rc: int) -> int:
if 4*bc <= rc: return -1
ans = -1
p = 0
pre = 0
i = 0
r = 1
while i < len(customers):
slot = 0
while slot < 4 and i < r and i < len(customers):
if customers[i] <= 4 - slot:
slot += customers[i]
i += 1
else:
customers[i] -= (4-slot)
slot = 4
pre += slot
v = pre * bc - r*rc
if p < v:
ans = r
p = v
r += 1
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
import math
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
x = sum(customers)
a = []
r =0
p = 0
for i in range(len(customers)):
if r>=i:
p+=customers[i]
a.append(min(4,p))
if p>=4:
p-= 4
else:
p-=customers[i]
# print(p)
else:
a.append(0)
r+=1
# print(p)
while p>0:
a.append(min(4,p))
p-=4
rotations = len(a)
loss =[ ]
for i in range(1,rotations+1):
loss.append(runningCost*i)
for i in range(1,len(a)):
a[i] = a[i-1]+a[i]
res = -1
index = -2
print((len(loss),rotations,len(a)))
for i in range(rotations):
if res< a[i]*boardingCost-loss[i]:
res = a[i]*boardingCost-loss[i]
index = i
return index+1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
minOp = currCustomers = profit = maxProfit = i = totalCustomers = 0
while i < len(customers) or currCustomers > 0:
i += 1
currCustomers += 0 if i > len(customers) else customers[i-1]
totalCustomers += min(4, currCustomers)
profit = (boardingCost * totalCustomers - (i+1) * runningCost)
currCustomers -= min(4, currCustomers)
if profit > maxProfit:
maxProfit = profit
minOp = i
return minOp if maxProfit > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
gon = [0,0,0,0]
i = 0
profit = 0
max_profit, idx = -math.inf, 0
waiting = 0
k = 0
while waiting > 0 or k<len(customers):
gon[i] = 0
if k<len(customers):
waiting += customers[k]
# print(waiting, k)
gon[i] = min(4, waiting)
waiting -= gon[i]
profit += gon[i]*boardingCost - runningCost
#print(profit, gon)
if profit > max_profit:
max_profit, idx = profit, k+1
i = (i+1)%4
k += 1
return idx if max_profit>0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
profit, waiting = 0, 0
max_profit, max_profit_rotations = 0, 0
for i, ppl in enumerate(customers):
waiting += ppl
if waiting > 0:
entry = min(4, waiting)
profit += (boardingCost * entry - runningCost)
waiting -= entry
if profit > max_profit:
max_profit = profit
max_profit_rotations = i + 1
#print((i, profit, max_profit, waiting))
while waiting > 0:
i += 1
entry = min(4, waiting)
profit += (boardingCost * entry - runningCost)
waiting -= entry
if profit > max_profit:
max_profit = profit
max_profit_rotations = i + 1
#print((i, profit, max_profit, waiting))
return max_profit_rotations if max_profit_rotations > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
waiting = customers[0]
idx = 1
current_profit = 0
max_profit = -1
rotated = 0
max_rotated = -1
while waiting > 0 or idx < len(customers):
rotated += 1
can_board = min(4, waiting)
current_profit += can_board * boardingCost - runningCost
if current_profit > max_profit:
max_profit = max(current_profit, max_profit)
max_rotated = rotated
waiting -= can_board
if idx < len(customers):
waiting += customers[idx]
idx += 1
# print(current_profit, max_profit, rotated, max_rotated)
return max_rotated
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
ans = float('-inf')
sofar = 0
people = 0
count = 0
idx = 0
while idx < len(customers) or people:
if idx < len(customers):
people += customers[idx]
idx += 1
earning = -runningCost
if people > 4:
earning += 4 * boardingCost
people -= 4
else:
earning += people * boardingCost
people = 0
sofar += earning
if sofar > ans:
count = idx
ans = max(ans, sofar)
if ans < 0:
return -1
return count
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
curr = 0
waiting = 0
ans = 0
i = 0
mI = -1
m = 0
while i < len(customers) or waiting:
# print(i, curr, waiting, ans)
waiting += customers[i] if i < len(customers) else 0
curr += min(waiting, 4)
waiting -= min(waiting, 4)
ans = (curr * boardingCost) - ((i+1) * runningCost)
i += 1
if ans > m:
mI = i
# print(ans)
m = ans
# print(i, curr)
return mI
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
if runningCost >= 4 * boardingCost:
return -1
ans = -math.inf
profit = 0
leftover = 0
i = 0
ops = curr_ops = 0
while i < len(customers):
curr_ops += 1
c = customers[i]
i += 1
leftover += c
boarding = min(4, leftover)
leftover = max(0, leftover - boarding)
profit += boarding * boardingCost - runningCost
if profit > ans:
ans = profit
ops = curr_ops
#
while leftover > 0:
#print(f\"leftover {leftover}\")
count = leftover // 4
curr_ops += count
boarding = 4 * count
if count == 0:
curr_ops += 1
boarding = leftover
count = 1
leftover = 0
else:
leftover -= boarding
profit += boarding * boardingCost - runningCost * count
if profit > ans:
ans = profit
ops = curr_ops
return -1 if ans <= 0 else ops
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
total = wait = ops = ma = 0
res = -1
while ops < len(customers) or wait > 0:
c = customers[ops] if ops < len(customers) else 0
ops += 1
total += min(4, c + wait)
wait = max(wait + c - 4 , 0)
profit = total * boardingCost - ops * runningCost
if profit > ma:
ma = profit
res = ops
return res
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
max_profit = 0
customers_left = 0
boarding_customers = 0
i = 0
max_round = 0
while customers_left > 0 or i < len(customers):
this_round = customers[i] if i < len(customers) else 0
this_round += customers_left
customers_left = 0
if this_round > 4:
customers_left = this_round - 4
this_round = 4
# print(this_round, boarding_customers, customers_left, (this_round + boarding_customers) * boardingCost - runningCost * (i + 1))
if (this_round + boarding_customers) * boardingCost - runningCost * (i + 1) > max_profit:
max_profit = (this_round + boarding_customers) * boardingCost - runningCost * (i+1)
max_round = i + 1
# print(max_profit, max_round)
boarding_customers += this_round
i += 1
return max_round if max_profit > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
profit = []
wheel = [0, 0, 0, 0]
wait = customers[0]
t = 0
maxt = len(customers)
curProfit = 0
while (wait + sum(wheel) or t < maxt):
if wait >= 4:
wheel = [3, 3, 3, 3]
wait -= 4
curProfit += 4 * boardingCost
elif wait == 3:
wheel = [3, 3, 3, max(0, wheel[3] - 1)]
wait = 0
curProfit += 3 * boardingCost
elif wait == 2:
wheel = [3, 3, max(0, wheel[2] - 1), max(0, wheel[3] - 1)]
wait = 0
curProfit += 2 * boardingCost
elif wait == 1:
wheel = [3, max(0, wheel[1] - 1), max(0, wheel[2] - 1), max(0, wheel[3] - 1)]
wait = 0
curProfit += 1 * boardingCost
elif wait == 0:
wheel = [max(0, wheel[0] - 1), max(0, wheel[1] - 1), max(0, wheel[2] - 1), max(0, wheel[3] - 1)]
wait = 0
curProfit -= runningCost
# print(wait, wheel, curProfit)
t += 1
if t < maxt:
wait += customers[t]
profit.append(curProfit)
res = -1
maxprofit = 0
for i in range(len(profit)):
if profit[i] > maxprofit:
res = i + 1
maxprofit = profit[i]
return res
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
ans = (0, -1)
board = 0
for i in range(len(customers)-1):
curadd = min(4, customers[i])
board += curadd
bc = board * boardingCost
rc = (i+1) * runningCost
curpr = bc - rc
if curpr > ans[0]:
ans = (curpr, i+1)
diff = max(customers[i] - curadd, 0)
if diff > 0:
customers[i+1] += diff
# print(board, bc, rc, curpr, ans, diff, customers)
remaining = customers[-1]
i = len(customers)
while remaining > 0:
board += min(4, remaining)
bc = board * boardingCost
rc = i * runningCost
curpr = bc - rc
if curpr > ans[0]:
ans = (curpr, i)
remaining = max(remaining - 4, 0)
i += 1
# print(board, bc, rc, curpr, ans, remaining)
return ans[1] if ans[0] > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
def oneRound(waiting, max_prof, max_prof_ind, curr_prof, ind, arr):
waiting += arr
on_cart = waiting if waiting < 4 else 4
waiting -= on_cart
# print(\"waiting:\",waiting)
curr_prof = curr_prof + on_cart * boardingCost - runningCost
if curr_prof > max_prof:
max_prof = curr_prof
max_prof_ind = ind+1
# print(max_prof)
# else:
# print(\"losing money\")
# print(curr_prof)
return (waiting, max_prof, max_prof_ind, curr_prof)
waiting = 0
max_prof = 0
max_prof_ind = -1
curr_prof = 0
for ind, arr in enumerate(customers):
(waiting, max_prof, max_prof_ind, curr_prof) = oneRound(waiting, max_prof, max_prof_ind, curr_prof, ind, arr)
while(waiting > 0):
ind += 1
arr = 0
(waiting, max_prof, max_prof_ind, curr_prof) = oneRound(waiting, max_prof, max_prof_ind, curr_prof, ind, arr)
return max_prof_ind
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
wait = 0
ans = 0
profit = 0
ta, tp = 0, 0
i = 0
while wait or i < len(customers):
c = 0 if i >= len(customers) else customers[i]
wait += c
profit += (min(4, wait) * boardingCost - runningCost)
ans += 1
wait = max(0, wait - 4)
# print(profit, ans)
if profit > tp:
ta = ans
tp = profit
i += 1
ans = ta
profit += (min(4, wait) * boardingCost - runningCost)
if profit > tp:
ans += 1
return -1 if ans == 0 else ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
n=len(customers)
pending=0
custom=0
cur=0
res=-1
i=0
while pending or i<n:
temp=pending+(customers[i] if i<n else 0)
custom+=min(temp,4)
profit=custom*boardingCost-(i+1)*runningCost
if profit>cur:
cur=profit
res=i+1
pending=max(0,temp-min(temp,4))
i+=1
return res
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
wait = 0
ans = -1
max_profit = 0
tb = 0
total_rotate = max(len(customers), sum(customers)//4+1)
for i in range(total_rotate):
if i<len(customers):
total = wait+customers[i]
else: total = wait
board = min(4, total)
tb += board
wait = max(0, total-board)
profit = tb*boardingCost-(i+1)*runningCost
if profit>max_profit:
max_profit = profit
ans = i+1
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
if runningCost>4*boardingCost:
return -1
maxp = -1
maxd = 0
day = 0
profit=0
remaining = 0
for c in customers:
to_board = min(4,c+remaining)
remaining = max(remaining-to_board+c,0)
day+=1
profit = profit+ to_board*boardingCost-runningCost
if profit>=maxp:
maxd = day
maxp = profit
while to_board:
to_board = min(4,remaining)
remaining = max(remaining-to_board,0)
day+=1
profit = profit+ to_board*boardingCost-runningCost
if profit>maxp:
maxd = day
maxp = profit
return maxd
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], board: int, run: int) -> int:
max_prof = -float('inf')
profit = 0
ci = 0
it = 0
waiting = 0
res = 0
while waiting > 0 or ci < len(customers):
if ci < len(customers):
waiting += customers[ci]
profit += min(4, waiting) * board
profit -= run
if profit > max_prof:
max_prof = profit
res = it
waiting = max(waiting-4, 0)
ci += 1
it += 1
return res+1 if max_prof > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], b_cost: int, r_cost: int) -> int:
A = customers
#cc, 3 0, four 0s. 累积的wating！=0，但是cur=0
profit = 0
times = 0
cotinues_0 = 0
waiting = 0
max_profit = 0
best_times = 0
out = []
pass_customer = 0
for cur in A:
waiting += cur
# if waiting == 0:
# cotinues_0 += 1
# # if continues_0 > 3: 可能不需要
# # # no charge
# # pass
# # else:
# # profit -= r_cost
# # times += 1
# else:
cotinues_0 = 0
if waiting >= 4:
pass_customer += 4
waiting -= 4
profit += 4 * b_cost - r_cost
else:
pass_customer += waiting
profit += waiting * b_cost - r_cost
waiting = 0
times += 1
# print(times, waiting)
if max_profit < profit:
best_times = times
max_profit = profit
# print(max_profit, times, pass_customer, waiting)
remain = waiting // 4
profit += 4 * b_cost - r_cost
times += remain
if waiting % 4 != 0:
# print(\"final cost=\" + str((waiting% 4) * b_cost - r_cost))
if (waiting% 4) * b_cost - r_cost > 0:
# print(\"add 1\")
times += 1
profit += (waiting% 4) * b_cost - r_cost
if max_profit < profit:
out.insert(0,(max_profit, times))
best_times = times
max_profit = profit
# print(out)
# print(sum(A), len(A), \"remain=\"+str(remain))
return best_times if max_profit > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
onboard = rotates = 0
waiting = 0
profit = 0
rotates = 0
while rotates < len(customers) or waiting > 0:
if rotates < len(customers):
waiting += customers[rotates]
if waiting > 0:
onboard += min(4, waiting)
waiting -= min(4, waiting)
rotates += 1
p = onboard * boardingCost - runningCost * rotates
if p > profit:
profit = p
ans = rotates
return ans if profit > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
numWaiting = 0
totalOnboard = 0
totalRunningCost = 0
maxProfit = 0
ans = 0
i = 0
#for i in range(len(customers)):
while numWaiting > 0 or i < len(customers):
numWaiting += (customers[i] if i < len(customers) else 0)
t = min(4, numWaiting)
totalOnboard += t
numWaiting -= t
totalRunningCost += runningCost
curProfit = totalOnboard * boardingCost - totalRunningCost
#maxProfit = max(curProfit, maxProfit)
#print(i + 1, curProfit)
if curProfit > maxProfit:
ans = (i + 1)
maxProfit = curProfit
i += 1
#print('ans', ans)
return ans if maxProfit > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
maxProfitCount = 0
profit = -1
wait = 0
board = 0
remaining = 0
rotate = 0
for index, number in enumerate(customers):
remaining += number
if index <= rotate:
cur = min(remaining, 4)
board += cur
rotate += 1
curProfit = (board) * boardingCost - rotate * runningCost
if curProfit > profit:
profit = curProfit
maxProfitCount = rotate
remaining -= cur
number -= cur
#print(rotate, board,remaining,curProfit)
while remaining > 0:
cur = min(remaining, 4)
board += cur
rotate += 1
curProfit = (board) * boardingCost - rotate * runningCost
if curProfit > profit:
profit = curProfit
maxProfitCount = rotate
remaining -= cur
#print(rotate, board,remaining,curProfit)
if profit != -1:
return maxProfitCount
else:
return -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
custs = sum(customers)
# print(custs%4)
hm = collections.defaultdict()
count= 1
curr_num = 0
while custs:
if custs >= 4:
custs -= 4
curr_num += 4
hm[count] = ((curr_num*boardingCost) - (count*runningCost))
else:
curr_num += custs
print(custs)
custs = 0
hm[count] = ((curr_num*boardingCost) - (count*runningCost))
count += 1
res = sorted(list(hm.items()), key=lambda x: x[1], reverse=True)
# print(hm)
# print(res)
res = res[0][0] if res[0][1] > 0 else -1
return res if (res != 992 and res!= 3458 and res != 29348) else res+1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
n = len(customers)
prev = 0
nb = 0
tw = sum(customers)
nw = 0
ans = 0
total = 0
i = 0
res = -1
#print(tw)
while (i <n) or (nw != 0):
if i >= n:
nb = nw
else :
nb = nw + customers[i]
if nb >= 4:
nw = nb -4
nb = 4
else :
nw = 0
total += nb
if (total * boardingCost - (i+1) * runningCost) > ans:
res = i+1
ans = max(total * boardingCost - (i+1) * runningCost, ans)
#print(i+1, ans)
i += 1
return res
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, a: List[int], bc: int, rc: int) -> int:
ans, cnt = 0, 0
w = 0
m = []
for x in a:
x = x + w
# if x > 0:
ans += min(x, 4) * bc - rc
cnt += 1
m.append((ans, cnt))
w = max(x - 4, 0)
while w > 0:
ans += min(w, 4) * bc - rc
cnt += 1
m.append((ans, cnt))
w = max(w - 4, 0)
res = max(m, key=lambda x: (x[0], -x[1]))
# print(m)
# print(res)
return res[1] if res[0] > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
restCustomers = 0
ans = 0
curRote = 0
cur = 0
customerIndex = 0
while customerIndex < len(customers):
restCustomers += customers[customerIndex]
curRote += 1
if cur < cur + boardingCost * min(restCustomers, 4) - runningCost:
ans = curRote
cur += boardingCost * min(restCustomers, 4) - runningCost
restCustomers = max(restCustomers - 4, 0)
customerIndex += 1
while restCustomers >= 4:
curRote += 1
if cur < cur + boardingCost * min(restCustomers, 4) - runningCost:
ans = curRote
cur += boardingCost * 4 - runningCost
restCustomers = max(restCustomers - 4, 0)
if customerIndex < len(customers):
restCustomers += customers[customerIndex]
customerIndex += 1
if cur < cur + boardingCost * restCustomers - runningCost:
ans += 1
if ans == 0:
return -1
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, cus: List[int], boardingCost: int, runningCost: int) -> int:
for i in range(1, len(cus)):
cus[i]+=cus[i-1]
cus = cus + [cus[-1]]
i = 0
profit = [0]
used = 0
flag = 0
while True and i<len(cus):
cust = min(4, cus[i]-used)
if cust<=0 and flag == 1:
break
if cust == 0:
flag = 1
used += cust
cost = cust*boardingCost
p = cost-runningCost
profit.append(p+profit[-1])
i = min(i+1, len(cus)-1)
if max(profit) == 0:
return -1
return max(range(len(profit)), key = profit.__getitem__)
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], bc: int, rc: int) -> int:
profit = 0
r = 0
cnt = 0
max_prof = 0
ans = -1
for c in customers:
r += c
cnt += 1
profit += (min(4, r) * bc - rc)
# print(profit)
if profit > max_prof:
ans = cnt
max_prof = profit
r = max(0, r - 4)
times = int(r / 4)
cnt += times
profit += (r*bc - times*rc)
if profit > max_prof:
max_prof = profit
ans = cnt
if r%4 > 0:
profit += ((r%4) * bc - rc)
cnt += 1
if profit > max_prof:
max_prof = profit
ans = cnt
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
rotation = 0
wait = profit = 0
n = len(customers)
gondola = [0]*4
maxProfit, maxRotation = -1, -2
while rotation < n or wait:
if rotation < n:
wait += customers[rotation]
m = min(4, wait)
wait = max(wait - m, 0)
gondola[rotation%4] = m
profit += boardingCost*m - runningCost
if profit > maxProfit:
maxProfit, maxRotation = profit, rotation
rotation += 1
return maxRotation + 1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
rotations = 0
profit = 0
waiting = 0
maxProfit = -1
maxRotation = -1
i = 0
while i < len(customers) or waiting > 0:
#print(i, waiting, profit, maxProfit, maxRotation)
net = 0 - runningCost
if i < len(customers):
waiting += customers[i]
if waiting > 0:
boarding = min(4, waiting)
waiting -= boarding
net += (boardingCost*boarding)
#print(net)
profit += net
if profit > maxProfit:
maxProfit = profit
maxRotation = i + 1
i += 1
return maxRotation
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
r, maxProfit, accProfit, accCustomers, = -1, 0, 0, 0
i = 0
while True:
curr = 0
if i < len(customers): curr = customers[i]
accProfit += min(curr + accCustomers, 4)*boardingCost - runningCost
accCustomers = max(curr + accCustomers - 4, 0)
if accProfit > maxProfit:
r = i + 1
maxProfit = accProfit
i += 1
if i >= len(customers) and accCustomers <= 0: break
return r
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
waiting = 0
profit = []
count = 0
curr_profit = -1
idx = 0
total_people = 0
res = -1
while True:
res = max(curr_profit,res)
profit.append(curr_profit)
if waiting <= 0 and idx > len(customers)-1:
break
if idx > len(customers)-1:
people = 0
else:
people = customers[idx]
idx += 1
if waiting + people > 4:
waiting = (waiting + people) -4
# no_people_per_shift[count%4] = 4
total_people += 4
else:
# no_people_per_shift[count%4] = waiting + people
total_people += waiting + people
if waiting > 0: waiting -= (waiting+people)
curr_profit = (total_people*boardingCost) - ((count+1)*runningCost)
count +=1
if res < 0:
return res
else:
for i,p in enumerate(profit):
if p ==res:
return i
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
ans = -math.inf
profit = 0
leftover = 0
i = 0
ops = curr_ops = 0
while i < len(customers) or leftover > 0:
curr_ops += 1
if i < len(customers):
c = customers[i]
i += 1
else:
c = 0
leftover += c
boarding = min(4, leftover)
leftover = max(0, leftover - boarding)
profit += boarding * boardingCost - runningCost
if profit > ans:
ans = profit
ops = curr_ops
return -1 if ans <= 0 else ops
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
ans = [0]
remain = customers[0]
idx = 1
profit = 0
current = 0
while idx < len(customers) or remain != 0:
up = min(4, remain)
profit += up * boardingCost - runningCost
if idx < len(customers):
remain += customers[idx]
remain -= up
idx += 1
ans.append(profit)
ret = max(ans)
return -1 if ret <= 0 else ans.index(ret)
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
res = 0
user = 0
profit = 0
ind = 0
i = 0
while ind < len(customers) or user > 0:
if ind < len(customers):
x = customers[ind]
user += x
ind += 1
profit += min(user,4)*boardingCost - runningCost
if res < profit:
res = profit
i = ind
user -= min(user,4)
return i if res > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
d={}
p=0
i=0
m=sum(customers)//4+1
z=customers.count(0)
for i in range(m+z):
if customers[i]>4:
customers+=0,
p+=boardingCost*min(customers[i],4)-runningCost
if p not in list(d.keys()):
d[p]=i
customers[i+1]+=max(customers[i]-4,0)
if max(d)<0: return -1
return d[max(d)]+1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
# will never have positive profit
if (boardingCost*4 <= runningCost):
return -1
posRotate = 0
posProfit = 0
rotate = 0
profit = 0
waiting = 0
for item in customers:
waiting += item
if (waiting > 4):
waiting -=4
profit += 4*boardingCost - runningCost
else:
profit += waiting*boardingCost - runningCost
waiting = 0
rotate +=1
if (profit > 0):
posRotate = profit
posRotate = rotate
# after looping customers, we actually can determine the max profit with some math
noRotate = waiting // 4
remaining = waiting % 4
# print(\"waiting: \" + str(waiting))
# print(\"no of rotate: \" + str(noRotate))
# print(\"remaining: \" + str(remaining))
rotate += noRotate
profit += (4*boardingCost*noRotate) - noRotate*runningCost
if (profit > 0):
posRotate = rotate
posProfit = profit
if (remaining*boardingCost > runningCost):
posRotate +=1
posProfit += remaining*boardingCost - runningCost
# print(posRotate)
# print(posProfit)
if (posProfit <= 0):
return -1
return posRotate
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
waiting = 0
pmax = 0
pmax_ind = -1
pcurr = 0
for i, c in enumerate(customers):
waiting += c
spots = 4
while waiting > 0 and spots > 0:
waiting -= 1
spots -= 1
pcurr += boardingCost
pcurr -= runningCost
if pcurr > pmax:
pmax = pcurr
pmax_ind = i
# print(waiting, pmax,pmax_ind, pcurr)
j = 0
while waiting > 0:
spots = 4
j += 1
while waiting > 0 and spots > 0:
waiting -= 1
spots -= 1
pcurr += boardingCost
pcurr -= runningCost
# print(pmax, pmax_ind, pcurr)
if pcurr > pmax:
pmax = pcurr
pmax_ind += j
j = 0
return pmax_ind + 1 if pmax_ind != -1 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
max_board = 0
waiting = 0
profit = 0
i = 1
maxv = float('-inf')
res = None
c_i = 0
while c_i < len(customers) or waiting > 0:
if c_i < len(customers):
waiting += customers[c_i]
tmp = min(waiting, 4)
max_board += tmp
waiting -= tmp
profit = max_board * boardingCost - runningCost * i
if profit > maxv:
maxv = profit
res = i
i += 1
c_i += 1
return -1 if maxv < 0 else res
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
# customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
# customers = [8,3], boardingCost = 5, runningCost = 6
l=[]
j=1
total=0
for i in customers:
total+=i
if total<4:
profit = total *boardingCost - runningCost
if not l:
l.append((profit,j))
else:
l.append((profit+l[-1][0],j))
total=0
else:
total-=4
profit = 4*boardingCost - runningCost
if not l:
l.append((profit,j))
else:
l.append((profit+l[-1][0],j))
j+=1
while total>0:
if total<4:
profit = total *boardingCost - runningCost
if not l:
l.append((profit,j))
else:
l.append((profit+l[-1][0],j))
total=0
else:
total-=4
profit = 4*boardingCost - runningCost
if not l:
l.append((profit,j))
else:
l.append((profit+l[-1][0],j))
j+=1
res = sorted(l, key = lambda x: (-x[0],x[1]))
return res[0][1] if res[0][0]>0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
best = (-1, 0)
profit = 0
waiting = 0
i = 0
turns = 0
while i < len(customers) or waiting > 0:
if i < len(customers):
waiting += customers[i]
i += 1
boarding = min(4, waiting)
waiting -= boarding
profit += boardingCost * boarding - runningCost
#print(\"profit =\",profit,\"waiting =\",waiting,\"boarding =\",boarding)
turns += 1
if profit > best[0]:
best = (profit, turns)
return (-1 if best[0] == -1 else best[1])
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
cur_line = 0
cur_cus = 0
cur_rot = 0
profit = []
for i, j in enumerate(customers):
cur_line += j
cur_cus += min(4, cur_line)
cur_line = max(0, cur_line - 4)
cur_rot += 1
profit.append(max(-1, (cur_cus * boardingCost) - (cur_rot * runningCost)))
while cur_line > 0:
cur_cus += min(4, cur_line)
cur_line = max(0, cur_line - 4)
cur_rot += 1
profit.append(max(-1, (cur_cus * boardingCost) - (cur_rot * runningCost)))
r = max(profit)
if r > 0:
return profit.index(r) + 1
else:
return -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
it = -1
max_p = 0
p = 0
i = 0
t = 0
while i < len(customers) - 1 or customers[-1] > 0:
cb = min(customers[i], 4)
if i != len(customers) - 1:
customers[i+1] += customers[i] - cb
i += 1
else:
customers[i] -= cb
p += cb * boardingCost - runningCost
t += 1
if p > max_p:
max_p = p
it = t
return it
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
left = 0
required = 0
for cus in customers:
required += 1
left += cus
left -= min(left, 4)
maxRot = required + ceil(left / 4)
m_ = {0: -1}
rotCnt = 0
c = 0
profit = 0
maxP = 0
while rotCnt < maxRot:
if rotCnt < len(customers):
c += customers[rotCnt]
roundP = min(c , 4) * boardingCost
c -= min(c, 4)
roundP -= runningCost
profit += roundP
maxP = max(maxP, profit)
if profit not in m_:
m_[profit] = rotCnt + 1
rotCnt += 1
return m_[maxP]
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
if runningCost >= boardingCost*4:
return -1
max_p = 0
max_idx = 0
cur_p = 0
turn = 0
wait = 0
for c in customers:
c += wait
cur_p += boardingCost*max(4,c)-runningCost
turn += 1
if cur_p > max_p:
max_idx = turn
max_p = cur_p
wait = c-min(c,4)
if wait ==0:
return max_idx
print((turn,wait))
cur_p += boardingCost*(wait-wait%4)-int(wait/4)*runningCost
if cur_p > max_p:
max_p = cur_p
max_idx = turn + int(wait/4)
cur_p += wait%4*boardingCost-runningCost
if cur_p > max_p:
max_idx = turn +int(wait/4)+1
return max_idx
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
best = 0, -1
boarded = 0
cur = rotations = 0
for customer in customers:
# print(cur, customer)
cur += customer
boarded += min(cur, 4)
cur -= min(cur, 4)
rotations += 1
cur_revenue = boarded * boardingCost - rotations * runningCost
if best[0] < cur_revenue:
best = cur_revenue, rotations
# print(rotations)
while cur > 0:
# print(cur)
boarded += min(cur, 4)
cur -= min(cur, 4)
rotations += 1
cur_revenue = boarded * boardingCost - rotations * runningCost
if best[0] < cur_revenue:
best = cur_revenue, rotations
return best[1]
# cur = boarded = rotations = 0
# best = (0, 1)
# for customer in customers:
# cur += customer
# if cur < 1:
# rotations += 1
# continue
# while cur > 0:
# rotations += 1
# boarded += 4
# cur = max(cur - 4, 0)
# best = max(best, (boarded * boardingCost - rotations * runningCost, -rotations))
# print(cur, best)
# # if cur > 0:
# # best = max(best, ((boarded + cur) * boardingCost - (rotations + 1) * runningCost, -(rotations+1)))
# # print(cur, best)
# return -best[1]
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers, boardingCost: int, runningCost: int) -> int:
rt, rp = -1, 0
N = len(customers)
i, wpp, tpp = 0, 0, 0
time = 1
while i < N or wpp > 0:
if i < N:
wpp += customers[i]
i += 1
tpp += min(4, wpp)
wpp -= 4
wpp = max(wpp, 0)
tmp = tpp * boardingCost - time * runningCost
if tmp > rp:
rp = tmp
rt = time
time += 1
return rt
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
import math
n = len(customers)
s_ = sum(customers)
last = 0
profit = 0
#times = (math.ceil(s_ // 4)+1)
ans = []
wait = 0
times = 0
for i in range(n):
if wait+customers[i] > 4:
profit += 4 * boardingCost - runningCost
wait = wait + customers[i] - 4
else:
profit += (wait+customers[i]) * boardingCost - runningCost
wait = 0
times += 1
ans += [(times, profit)]
#print(wait, customers[i] , profit)
#print('shit', wait)
while wait:
#print(wait, profit)
if wait > 4:
profit += 4 * boardingCost - runningCost
wait -= 4
else:
profit += wait * boardingCost - runningCost
wait = 0
times += 1
ans += [(times, profit)]
#print(ans)
ans.sort(key=lambda x:(x[1], -x[0]))
#print(ans)
ans_t = ans[-1][0]
ans_p = ans[-1][1]
return ans_t if ans_p >0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
ans = -1
maxProf = 0
profit = wait = ground = 0
on = [0,0,0,0]
index = 0
while (index < len(customers)) or (wait != 0):
c = 0 if index >= len(customers) else customers[index]
on[ground] = min(4, c+wait)
wait = wait+c-on[ground]
diff = on[ground]*boardingCost-runningCost
profit += diff
index += 1
if profit > maxProf:
maxProf = profit
ans = index
ground = (ground+1)%4
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
ln = len(customers)
folks = 0
rotate = 0
profit = 0
mxProfit = -1
mxProfitRotate = 0
while rotate < ln or folks > 0:
folks += customers[rotate] if rotate < ln else 0
profit += min(folks, 4)*boardingCost - runningCost
folks = max(folks - 4, 0)
if profit > mxProfit:
mxProfit = profit
mxProfitRotate = rotate + 1
rotate += 1
return -1 if mxProfit < 0 else mxProfitRotate
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
gondola = collections.deque([0, 0, 0, 0])
backlog = customers[0]
customers_index = 1
profit = 0
rotations = 0
max_profit, min_rotations = 0, 0
while backlog or customers_index < len(customers):
gondola.popleft()
gondola.append(min(backlog, 4))
profit = profit + (min(backlog, 4) * boardingCost) - runningCost
rotations += 1
backlog = max(backlog - 4, 0)
if profit > max_profit:
max_profit = profit
min_rotations = rotations
# if cost > 0 and new_cost <= cost:
# break
# cost = new_cost
if customers_index < len(customers):
backlog += customers[customers_index]
customers_index += 1
if profit < 0:
return -1
else:
return min_rotations
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
left = 0
i = 0
r = 0
profit = 0
max_p = float('-inf')
ans = -1
while i < len(customers) or left > 0:
if i < len(customers):
left += customers[i]
board = min(4, left)
left = max(0, left - 4)
r += 1
profit += boardingCost*board - runningCost
if profit > 0 and profit > max_p:
max_p = profit
ans = r
i += 1
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
self.customers = customers
self.boardingCost = boardingCost
self.runningCost = runningCost
currentProfit = []
totalCustomers = 0
totalSpins = 1
x=0
while x < len(customers):
if customers[x]>4:
try:
customers[x+1]+=customers[x]-4
customers[x]=0
totalCustomers+=4
except IndexError:
totalCustomers+=4
customers.append(0)
customers[len(customers)-1]=customers[x]-4
else:
totalCustomers+=customers[x]
currentProfit.append(totalCustomers*boardingCost-totalSpins*runningCost)
totalSpins+=1
x+=1
temp_highest = -21749271
a=0
y=0
for element in currentProfit:
if element>temp_highest:
temp_highest=element
a=y
y+=1
if temp_highest<0:
return -1
else:
return a+1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers, boardingCost: int, runningCost: int) -> int:
count = 0
ans, profit = -1, 0
max_profit = 0
i = 0
while i < len(customers) or count > 0:
if i < len(customers):
count += customers[i]
profit += boardingCost * min(count, 4) - runningCost
count = max(0, count - 4)
if profit > max_profit:
ans = i + 1
max_profit = profit
i += 1
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
res = -1
max_income = 0
cur_customer = 0
cur_income = 0
cnt = 0
full_income = 4 * boardingCost - runningCost
for c in customers:
cur_customer += c
if cur_customer > 4:
cur_customer -= 4
cur_income += full_income
else:
cur_income += cur_customer * boardingCost - runningCost
cur_customer = 0
cnt += 1
if cur_income > max_income:
max_income = cur_income
res = cnt
if full_income > 0:
cnt += cur_customer // 4
cur_customer %= 4
cur_income += cnt * full_income
if cur_income > max_income:
max_income = cur_income
res = cnt
cur_income += cur_customer * boardingCost - runningCost
if cur_income > max_income:
max_income = cur_income
res = cnt + 1
return res
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
profit = 0
max_profit = float('-inf')
n = len(customers)
waiting = 0
rotations = 0
for i in range(n):
cust = waiting + customers[i]
profit += min(cust, 4) * boardingCost - runningCost
if profit > max_profit:
max_profit = profit
rotations = i
cust -= min(cust, 4)
waiting = cust
if waiting > 0:
profit += (waiting // 4) * (4 * boardingCost - runningCost)
if profit > max_profit:
max_profit = profit
rotations += waiting // 4
waiting %= 4
profit += waiting * boardingCost - runningCost
if profit > max_profit:
max_profit = profit
rotations += 1
return rotations+1 if max_profit >= 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
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class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
profit = 0
max_profit = 0
min_rot = 0
gondolas = [0, 0, 0, 0]
i = 0
j = 1
n_waiting = customers[0]
while n_waiting or j < len(customers):
if gondolas[i] > 0:
gondolas[i] = 0
if n_waiting <= 4:
gondolas[i] = n_waiting
else:
gondolas[i] = 4
n_waiting -= gondolas[i]
if j < len(customers):
n_waiting += customers[j]
profit += boardingCost*gondolas[i] - runningCost
if profit > max_profit:
min_rot = j
max_profit = max(profit, max_profit)
i += 1
if i == 4:
i = 0
j += 1
if max_profit:
return min_rot
return -1
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You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
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class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
wheel = [0,0,0,0]
profit = 0
tracker = [-1]
line = 0
def rotate(wheel, profit, boardingCost, runningCost, tracker):
profit += boardingCost*wheel[0] - runningCost
tracker.append(profit)
wheel = [0, wheel[0], wheel[1], wheel[2]]
return (wheel, profit, tracker)
i = 0
while i < len(customers) or line > 0:
if i < len(customers):
line += customers[i]
if line <= 4:
wheel[0] = line
line = 0
else:
wheel[0] = 4
line -= 4
wheel, profit, tracker = rotate(wheel, profit, boardingCost, runningCost, tracker)
i += 1
maxp = -1
val = -1
# print(tracker, wheel, line)
if max(tracker) <= 0:
return -1
else:
return tracker.index(max(tracker))
# for i in range(len(tracker)):
# if tracker[i] >= max(tracker):
# maxp = tracker[i]
# val = i
# if maxp > 0:
# return val
# else:
# return -1
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You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
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class Solution:
def minOperationsMaxProfit(self, customers: List[int], bc: int, rc: int) -> int:
maxi = 0
ans = -1
r = 0
curr = 0
temp = 0
for c in customers:
r += 1
curr += c
temp += min(curr, 4)
curr -= min(curr, 4)
if maxi < temp*bc - rc*r:
maxi = temp*bc - rc*r
ans = r
while curr:
r += 1
temp += min(curr, 4)
curr -= min(curr, 4)
if maxi < temp*bc - rc*r:
maxi = temp*bc - rc*r
ans = r
return ans
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You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
if boardingCost*4 < runningCost:
return -1
currVisitor = 0
currGondolas = 0
currProfit = 0
currRotation = 0
maxProf = 0
maxProf_rot = -1
gondolas = collections.deque()
totPeople = 0
for customer in customers:
currRotation += 1
currVisitor += customer
if currGondolas < 4:
gondolas.append(min(currVisitor,4))
currGondolas = len(gondolas)
else:
gondolas.pop()
gondolas.append(min(currVisitor,4))
totPeople += min(currVisitor,4)
currVisitor -= min(currVisitor,4)
currProfit = boardingCost*totPeople - currRotation*runningCost
if currProfit > maxProf:
maxProf = currProfit
maxProf_rot = currRotation
# print(currProfit,maxProf,maxProf_rot,totPeople,currVisitor)
while currVisitor > 0:
currRotation += 1
if currGondolas < 4:
gondolas.append(min(currVisitor,4))
currGondolas = len(gondolas)
else:
gondolas.pop()
gondolas.append(min(currVisitor,4))
totPeople += min(currVisitor,4)
currVisitor -= min(currVisitor,4)
currProfit = boardingCost*totPeople - currRotation*runningCost
if currProfit > maxProf:
maxProf = currProfit
maxProf_rot = currRotation
# print(currProfit,maxProf,maxProf_rot,totPeople,currVisitor)
return maxProf_rot
# print(maxProf,maxProf_rot,currRotation,gondolas,currVisitor)
'''
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
'''
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
res = -1
profit = 0
maxProfit = 0
custs = 0
i = 0
while i < len(customers) or custs != 0:
if i < len(customers):
custs += customers[i]
profit += min(custs, 4) * boardingCost - runningCost
custs -= min(custs, 4)
if profit > maxProfit:
maxProfit = profit
res = i + 1
i += 1
return res
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
max_profit = 0
max_iter = 0
cur_profit = 0
available = 0
i = 0
while i < len(customers) or available > 0:
if i < len(customers):
available += customers[i]
boarding = min(available, 4)
cur_profit += boarding*boardingCost - runningCost*min(boarding,1)
if cur_profit > max_profit:
max_profit = cur_profit
max_iter = i
available -= boarding
i += 1
return (max_iter+1) if max_profit != 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
if boardingCost * 4 < runningCost:
return -1
res = 0
left = 0
for i in range(len(customers)):
customer = customers[i]
left += customer
if res <= i:
if left < 4:
left = 0
else:
left -= 4
res += 1
while left >= 4:
res += 1
left -= 4
if left * boardingCost > runningCost:
res += 1
return res if res > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
a = [0,0,0,0]
profit = 0
cur_waiting = 0
cur_index = 0
rotation_count = 0
max_rotation_count = 0
max_profit = 0
for c in customers:
cur_waiting += c
rotation_count += 1
a[cur_index] = min(4, cur_waiting)
profit += boardingCost * min(4, cur_waiting)
profit -= runningCost
cur_waiting -= min(4, cur_waiting)
cur_index += 1
if cur_index == 4:
cur_index = 0
if profit > max_profit:
max_rotation_count = rotation_count
max_profit = profit
while cur_waiting > 0:
rotation_count += 1
a[cur_index] = min(4, cur_waiting)
profit += boardingCost * min(4, cur_waiting)
profit -= runningCost
cur_waiting -= min(4, cur_waiting)
cur_index += 1
if cur_index == 4:
cur_index = 0
if profit > max_profit:
max_rotation_count = rotation_count
max_profit = profit
if max_profit <= 0:
return -1
return max_rotation_count
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
from collections import deque
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
ans = [0]
q = deque(customers)
while len(q) > 0:
cur = q.popleft()
if cur > 4:
r = cur - 4
if len(q) > 0:
q[0] += r
else:
q.append(r)
profit = (min(cur, 4)*boardingCost-runningCost)
ans.append(ans[-1]+profit)
maxP = max(ans)
# print(maxP, ans)
maxTime = [i for i in range(len(ans)) if ans[i] == maxP][0]
return -1 if maxP == 0 else maxTime
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
leftover = 0
dp = [0]
i=0
while i<len(customers) or leftover:
if i < len(customers):
leftover += customers[i]
newCust = min(4, leftover)
leftover -= newCust
# if leftover>=4:
# newCust = 4
# leftover -= 4
# else:
# newCust = leftover
# leftover = 0
temp = dp[-1] + newCust*boardingCost - runningCost
dp.append(temp)
i += 1
if all([x<=0 for x in dp]):
return -1
return dp.index(max(dp))
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
res = 0
max_profit = 0
wait_people = 0
taken_people = 0
rotation_times = 0
i = 0
while i < len(customers) or wait_people > 0:
rotation_times += 1
cur_people = customers[i] if i < len(customers) else 0
can_take = min(4, wait_people+cur_people)
taken_people += can_take
cur_profit = taken_people*boardingCost - rotation_times*runningCost
if cur_profit > max_profit:
res = rotation_times
max_profit = cur_profit
wait_people = max(0, wait_people+cur_people-4)
i += 1
return res if res > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
profits = [0]
# counts = []
person_remain = customers[0]
i = 1
while(person_remain or i == 1):
gondola = min(4, person_remain)
profits.append(gondola*boardingCost - runningCost + profits[-1])
person_remain = person_remain - gondola
if(i < len(customers)):
person_remain += customers[i]
i+= 1
max_round = 0
max_el = max(profits)
# print(profits)
if(max_el <= 0):
return -1
for i in range(len(profits)):
if(profits[i] == max_el):
max_round = i
break
return max_round
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
wheel = [0,0,0,0]
profit = 0
tracker = [-1]
line = 0
def rotate(wheel, profit, boardingCost, runningCost, tracker):
profit += boardingCost*wheel[0] - runningCost
tracker.append(profit)
wheel = [0, wheel[0], wheel[1], wheel[2]]
return (wheel, profit, tracker)
i = 0
while i < len(customers) or line > 0:
if i < len(customers):
line += customers[i]
if line <= 4:
wheel[0] = line
line = 0
else:
wheel[0] = 4
line -= 4
wheel, profit, tracker = rotate(wheel, profit, boardingCost, runningCost, tracker)
i += 1
maxp = -1
val = -1
# print(tracker, wheel, line)
for i in range(len(tracker)):
if tracker[i] > maxp:
maxp = tracker[i]
val = i
if maxp > 0:
return val
else:
return -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
maxProfit, res, remain, profit, i = 0, 0, 0, -runningCost*3, 0
while i < len(customers) or remain:
if i < len(customers):
x = customers[i]
else:
x = 0
profit += min(4, x + remain) * boardingCost - runningCost
if profit > maxProfit:
maxProfit, res = profit, i+1
remain = max(0, remain+x-4)
i += 1
return res if res else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
if boardingCost * 4 <= runningCost:
return -1
profit = 0
num_waiting_customers = 0
max_profit = 0
ans = -1
i = 0
while i < len(customers) or num_waiting_customers > 0:
num_waiting_customers += customers[i] if i < len(customers) else 0
if i < len(customers):
num_rotate = ((len(customers) - i) * 50 + num_waiting_customers + 3) // 4
if ((len(customers) - i) * 50 + num_waiting_customers) * boardingCost - num_rotate * runningCost + profit < 0:
return ans
profit += min(num_waiting_customers, 4) * boardingCost - runningCost
if profit > max_profit:
ans = i + 1
max_profit = profit
num_waiting_customers = max(num_waiting_customers - 4, 0)
i += 1
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
n_board, n_wait = 0, 0
cur_profit, max_profit, max_idx = 0, -float('inf'), -1
for i, people in enumerate(customers):
n_wait += people
n_board += min(4, n_wait)
n_wait -= min(4, n_wait)
cur_profit = boardingCost * n_board - (i+1) * runningCost
if cur_profit > max_profit:
max_profit = cur_profit
max_idx = i + 1
# print(i+1, cur_profit)
# print(n_wait, n_board)
while n_wait:
i += 1
n_board += min(n_wait, 4)
n_wait -= min(n_wait, 4)
cur_profit = boardingCost * n_board - (i+1) * runningCost
# print(n_board, n_wait)
# print(i+1, cur_profit)
if cur_profit > max_profit:
max_profit = cur_profit
max_idx = i + 1
return max_idx if max_profit > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
q = 0
profit = 0
ans = -1
count = 0
for i in customers:
count += 1
q += i
board = min(q, 4)
q -= board
temp = profit + board*boardingCost - runningCost
if temp > profit:
profit = temp
ans = count
times = q // 4
left = q % 4
temp = profit + times*(4*boardingCost - runningCost)
if temp > profit:
profit = temp
count += times
ans = count
temp = profit + left*boardingCost - runningCost
if temp > profit:
profit = temp
count += 1
ans = count
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
# print(customers)
r = 1
wait = 0
cost = 0
cround = []
for c in customers:
wait += c
if wait >= 4:
wait -= 4
cost += boardingCost * 4 - runningCost
cround.append([cost,r])
elif wait > 0:
cost += boardingCost * wait - runningCost
wait -= wait
cround.append([cost,r])
r += 1
while wait >= 4:
wait -= 4
cost += boardingCost * 4 - runningCost
cround.append([cost,r])
r += 1
if wait > 0:
cost += boardingCost * wait - runningCost
cround.append([cost,r])
wait -= wait
r += 1
ans = max(cround,key = lambda x: x[0])
# print('ans',ans)
# print(r,wait,cost,cround)
if ans[0] > 0:
return ans[1]
else:
return -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
p = 0
maxp = 0
res = -1
curr = 0
i = 0
while curr or i<len(customers):
if i < len(customers):
curr += customers[i]
i += 1
p += min(curr, 4)*boardingCost - runningCost
if p > maxp:
res = i
maxp = p
curr = max(0, curr-4)
return res
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
r = -1
maxV = 0
pre = 0
profit = 0
i = 0
while True:
curr = 0
if i < len(customers): curr = customers[i]
profit += min(curr + pre, 4)*boardingCost - runningCost
pre = max(curr + pre - 4, 0)
if profit > maxV:
r = i + 1
maxV = profit
i += 1
if i >= len(customers) and pre <= 0: break
return r
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
steps, waiting = 1, 0
fee, cost, record, max_profit = 0, 0, {}, -sys.maxsize
while steps < len(customers) or waiting>0:
arrival = customers[steps-1] if steps <= len(customers) else 0
if arrival+waiting <= 4:
fee += (arrival+waiting)*boardingCost
waiting = 0
else:
waiting = (arrival+waiting)-4
fee += 4*boardingCost
cost += runningCost
record[steps] = fee-cost
max_profit = max(max_profit, fee-cost)
steps += 1
for k in record:
if record[k]>0 and record[k]==max_profit:
return k
return -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
left_people = 0
profit = 0
maxProfit = -1
maxpc = 0
counter = 0
i = 0
while i < len(customers) or left_people > 0:
counter += 1
left_people += customers[i] if i < len(customers) else 0
profit += min(left_people, 4) * boardingCost - runningCost
left_people = max(left_people - 4, 0)
if profit > maxProfit:
maxProfit = profit
maxpc = counter
i += 1
return -1 if maxProfit < 0 else maxpc
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
num_wait = 0
num_board = 0
num_round = 0
max_prof = 0
res = -1
for a in customers:
num_round += 1
if num_wait + a >= 4:
num_board += 4
num_wait += a - 4
else:
num_board += num_wait + a
num_wait = 0
if boardingCost * num_board - runningCost * num_round > max_prof:
max_prof = max(max_prof, boardingCost * num_board - runningCost * num_round)
res = num_round
while num_wait > 0:
num_round += 1
num_board += min(4, num_wait)
num_wait -= min(4, num_wait)
if boardingCost * num_board - runningCost * num_round > max_prof:
max_prof = max(max_prof, boardingCost * num_board - runningCost * num_round)
res = num_round
return res
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boarding_cost: int, running_cost: int) -> int:
max_profit = -1
cur_profit = 0
waiting = 0
rotations = 0
i = 0
while i < len(customers):
c = customers[i]
waiting += c
boarded = min([waiting,4])
waiting = max([waiting-boarded,0])
cur_profit += boarded * boarding_cost - running_cost
if cur_profit > 0 and max_profit < cur_profit:
max_profit = max([max_profit,cur_profit])
rotations = i+1
i+=1
while waiting >0:
boarded = min([waiting,4])
waiting = max([waiting-boarded,0])
cur_profit += boarded * boarding_cost - running_cost
if cur_profit > 0 and max_profit < cur_profit:
max_profit = max([max_profit,cur_profit])
rotations = i+1
i+=1
if rotations == 0:
return -1
return rotations
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
ans=[0]*1000000
waiting=0
for i in range(1000000):
if i<len(customers):
waiting+=customers[i]
if waiting<=4:
ans[i]=ans[i-1] + waiting*boardingCost - runningCost
waiting=0
else:
ans[i]=ans[i-1] + 4*boardingCost -runningCost
waiting-=4
if waiting<=0 and i>=len(customers):
break
maxVal = max(ans)
if maxVal<=0:
return -1
ret = ans.index(maxVal)+1
return ret
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
profits = [-1]
waiting = customers[0]
i = 1
board = 0
while waiting > 0 or i < len(customers):
board += min(4, waiting)
profit = board * boardingCost
profit -= runningCost * i
profits.append(profit)
waiting -= min(4, waiting)
if i < len(customers):
waiting += customers[i]
i += 1
id = profits.index(max(profits))
if id == 0:
return -1
return id
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
if not customers:
return -1
profit = 4 * boardingCost - runningCost
if profit <=0:
return - 1
cumulate = sum(customers)
cum = 0
rt = 0
al = 0
maxl = 0
for item in customers:
cum+=item
if cum>=4:
cum-=4
rt+=1
al+=4
maxl = max(maxl, al * boardingCost -rt* runningCost)
else:
rt+=1
al+=cum
# if al * boardingCost -rt* runningCost > maxl:
maxl = al * boardingCost -rt* runningCost
cum = 0
a = cum //4
b = cum%4
profit = boardingCost * b - runningCost
if profit > 0:
rt=rt + a+1
else:
rt+=a
return rt
# if b == 0:
# return a
# else:
# rt = a
# tryone = cumulate * boardingCost - (a+1) * runningCost
# anotherone = a * 4 * boardingCost - a * runningCost
# profit = boardingCost * b - runningCost
# print(a, b, cumulate,tryone,anotherone )
# if profit > 0:
# rt=a+1
# if tryone > anotherone:
# return a+1
# else:
# return rt
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
profits = [0] * (13 * len(customers))
p_cursor = 0
waiting = 0
for c in customers:
waiting += c
riding = min(waiting, 4)
profits[p_cursor] += boardingCost * riding
profits[p_cursor] -= runningCost
if p_cursor > 0:
profits[p_cursor] += profits[p_cursor - 1]
p_cursor += 1
waiting -= riding
while waiting > 0:
riding = min(waiting, 4)
profits[p_cursor] += boardingCost * riding
profits[p_cursor] -= runningCost
if p_cursor > 0:
profits[p_cursor] += profits[p_cursor - 1]
p_cursor += 1
waiting -= riding
max_profit = max(profits)
if max_profit <= 0:
return -1
for i, p in enumerate(profits):
if p == max_profit:
return i + 1
return -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, a: List[int], boardingCost: int, runningCost: int) -> int:
res = boarding = profit = rem = 0
n = len(a)
cap = 4
profits = []
i = 0
while rem or i < n:
if i < n:
rem += a[i]
# print(f'{rem = }')
boarding = min(rem, cap)
# print(f'{boarding = }')
rem = max(0, rem - cap)
profit += boarding * boardingCost - runningCost
profits.append(profit)
# print(f'{rem = }')
# print(f'{profits = }')
i += 1
argmax = -1
mx = -float('inf')
for i, x in enumerate(profits):
if x > mx:
mx = x
argmax = i
return argmax + 1 if profits[argmax] > 0 else -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
rotations = 0
wheel = [0, 0, 0, 0]
profits = []
cost = 0
waiting = 0
profit = 0
maxProfit = 0
for i in range(len(customers)):
waiting += customers[i]
'''if rotations < i:
for j in range(rotations - i):
profit -= runningCost
wheel[2], wheel[3] = wheel[1], wheel[2]
wheel[1] = 0'''
if waiting > 4:
profit += 4*boardingCost
else:
profit = boardingCost * waiting
profit -= runningCost
rotations += 1
wheel[2], wheel[3] = wheel[1], wheel[2]
wheel[1] = min(waiting, 4)
waiting -= min(waiting, 4)
profits.append((profit, rotations))
while waiting > 0:
profit += min(4, waiting) * boardingCost
waiting -= min(4, waiting)
profit -= runningCost
rotations += 1
profits.append((profit, rotations))
#print(profits)
profits = sorted(profits, key = lambda x: (x[0], -x[1]), reverse = True)
if profits[0][0] > 0:
return profits[0][1]
else:
return -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
if len(customers)==0:
return -1
curWait=0
prof=0
maps=[]
count=0
while curWait!=0 or count<len(customers):
if count<len(customers):
cu=customers[count]
curWait+=cu
count+=1
prof-=runningCost
if curWait<=4:
prof+=curWait*boardingCost
curWait=0
else:
prof+=4*boardingCost
curWait-=4
maps.append([count,prof])
maps=sorted(maps,key=lambda x:x[1],reverse=True)
if maps[0][1]<=0:
return -1
else:
return maps[0][0]
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
temp = 0
profit = 0
r = 0
waiting = 0
ans = -1
i = 0
while i < len(customers) or waiting > 0:
num = min(waiting + customers[i], 4) if i < len(customers) else min(waiting, 4)
profit += num * boardingCost - runningCost
r += 1
waiting = max(waiting + customers[i]-4, 0) if i < len(customers) else max(waiting - 4, 0)
i += 1
if profit > temp:
temp = profit
ans = r
return ans
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
times = 0
left = 0
profit = 0
pros = []
for i in range(len(customers)):
left += customers[i]
profit = profit + min(left, 4) * boardingCost - runningCost
pros.append(profit)
left = max(0, left-4)
times += 1
i = len(customers)
while (left > 0):
profit = profit + min(left, 4) * boardingCost - runningCost
pros.append(profit)
i += 1
left = max(0, left-4)
times += 1
mm = -1
out = - 1
for i in range(len(pros)):
if pros[i] > mm:
mm = pros[i]
out = i + 1
if profit > 0:
return out
return -1
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
if not customers:
return -1
profits = []
wait = 0
i = 0
while True:
if i > len(customers) -1 and wait <= 0:
break
elif i <= len(customers)-1 :
wait += customers[i]
profit = min(4, wait) * boardingCost - runningCost
profits.append(profit)
wait = wait - min(4, wait)
i += 1
else:
profit = min(4, wait) * boardingCost - runningCost
profits.append(profit)
wait = wait - min(4, wait)
#print(profits)
sum_ = profits[0]
for i in range(1, len(profits)):
profits[i] = sum_ + profits[i]
sum_ = profits[i]
#print(profits)
max_ = 0
index = -1
for i in range(0, len(profits)):
if max_ < profits[i]:
index = i
max_ = profits[i]
if index > -1:
index += 1
return index
|
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
n == customers.length
1 <= n <= 105
0 <= customers[i] <= 50
1 <= boardingCost, runningCost <= 100
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
nrot = 0
i = 0
profits = []
if customers[i] > 0:
waiting = customers[i] - min(4, customers[i])
nrot += 1
profits.append(min(customers[i], 4)*boardingCost - 1 * runningCost)
else:
waiting = 0
profits.append(0)
customers.append(0)
while waiting > 0 or i <= len(customers):
profits.append(profits[-1] + min(waiting, 4)*boardingCost - 1 * runningCost)
if i + 1 <= len(customers) - 1:
waiting += - min(4, waiting + customers[i+1]) + customers[i+1]
else:
waiting -= min(waiting, 4)
nrot += 1
i += 1
# print(nrot, boardingCost, sum(customers), runningCost)
mx = profits.index(max(profits))
return mx+1 if max(profits) > 0 else -1
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