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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:
# O(n) Time | O(1) Space
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
result = 0
totalPeople = 0
totalProfit = 0
maxProfit = 0
for i, num in enumerate(customers):
totalPeople += num
onBoard = min(4, totalPeople)
totalPeople -= onBoard
totalProfit += onBoard * boardingCost - runningCost
if totalProfit > maxProfit:
maxProfit = totalProfit
result = i + 1
q, r = divmod(totalPeople, 4)
if 4 * boardingCost > runningCost:
result += q
if r * boardingCost > runningCost:
result += 1
return result if result > 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: [int], boardingCost: int, runningCost: int) -> int:
waiting=0
profits=[0]
if 4*boardingCost<runningCost:
return -1
for customer in customers:
if customer+waiting>=4:
profit=4*boardingCost-runningCost
waiting=waiting+customer-4
else:
profit = (waiting+customer) * boardingCost - runningCost
waiting=0
last=profits[-1]
profits.append(profit+last)
while waiting>0:
if waiting>=4:
profit = 4 * boardingCost - runningCost
last = profits[-1]
profits.append(profit + last)
waiting-=4
else:
profit = waiting * boardingCost - runningCost
last = profits[-1]
profits.append(profit + last)
waiting=0
return profits.index(max(profits))
|
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
answer = 0
current = 0
counter = 0
result = -1
for entry in customers:
counter += 1
wait += entry
current -= runningCost
if wait >= 4:
current += (4*boardingCost)
wait -= 4
else:
current += (wait*boardingCost)
wait = 0
if current > answer:
result = counter
answer = max(answer, current)
while wait > 0:
counter += 1
current -= runningCost
if wait >= 4:
current += (4*boardingCost)
wait -= 4
else:
current += (wait*boardingCost)
wait = 0
if current > answer:
result = counter
answer = max(answer, current)
return result
|
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 or len(customers) == 0:
return -1
left = 0
profit = 0
res = 0
max_profit = 0
idx = 0
r = 0
while idx < len(customers) or left > 0:
r += 1
if idx < len(customers):
left += customers[idx]
idx += 1
if left < 4:
profit += left * boardingCost - runningCost
left = 0
else:
left -= 4
profit += 4 * boardingCost - runningCost
if profit > max_profit:
max_profit = profit
res = r
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:
res = 0
ans = -1
rotate = 1
current = 0
rotate_times = -1
for i, number in enumerate(customers):
if number > 4:
res += number - 4
current += 4
elif number == 4:
current += number
else:
if res + number > 4:
current += 4
res = res + number - 4
else:
current += res + number
res = 0
profit = boardingCost * current - rotate*runningCost
# print(\"the profit is: \" + str(profit))
rotate += 1
if ans < profit:
ans = profit
rotate_times = rotate
# print(\"the res is: \" + str(res))
while res > 0:
if res > 4:
current += 4
res -= 4
else:
current += res
res = 0
profit = boardingCost * current - rotate*runningCost
# print(\"the profit is: \" + str(profit))
rotate += 1
if ans < profit:
ans = profit
rotate_times = rotate
return rotate_times -1 if rotate_times - 1 > 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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from collections import deque
from typing import List
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
ls = []
waiting = 0 # 라인 기다리는 사람
gondola = deque() # 곤돌라 que
curr = 0 # 현재 곤돌라에 타고있는 사람
days = 0
for i, customer in enumerate(customers):
waiting += customer
if waiting >= 4:
waiting -= 4
on_board = 4
else:
on_board = waiting
waiting = 0
curr += on_board
# profit
ls.append(curr * boardingCost - (days + 1) * runningCost)
days += 1
while waiting > 0:
if waiting >= 4:
waiting -= 4
on_board = 4
else:
on_board = waiting
waiting = 0
curr += on_board
# profit
ls.append(curr * boardingCost - (days + 1) * runningCost)
days += 1
max_val = max(ls)
if max_val < 0:
return -1
else:
return ls.index(max_val) + 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
|
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
res, maxprofit = -1, 0
curr_profit = nwait = 0
for idx,num in enumerate(customers):
nwait += num
curr_profit += (4*boardingCost if nwait >= 4 else nwait * boardingCost) - runningCost
if nwait >= 4:
nwait -= 4
else:
nwait = 0
if curr_profit > maxprofit:
res = idx+1
maxprofit = curr_profit
while nwait > 0:
idx += 1
curr_profit += (4*boardingCost if nwait >= 4 else nwait * boardingCost) - runningCost
if nwait >= 4:
nwait -= 4
else:
nwait = 0
if curr_profit > maxprofit:
res = idx+1
maxprofit = curr_profit
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:
bc, rc, n = boardingCost, runningCost, len(customers)
res, sum_v, max_v, wait, i = -1, 0, 0, 0, 0
if 4 * bc <= rc:
return -1
# customers.append(0)
while i < n or wait > 0:
wait += customers[i] if i < n else 0
cur = wait if wait < 4 else 4
wait -= cur
sum_v += cur * bc - rc
# (i,wait,cur,sum_v,max_v).p()
if sum_v > max_v:
max_v = sum_v
res = i + 1
i += 1
return res
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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:
count =0
max_profit = 0
profit =0
rot =-1
i=0
while count>0 or i<len(customers):
if i<len(customers):
new_customers = customers[i]
count+=new_customers
if count>=4:
count-=4
profit+=4*boardingCost-runningCost
else:
profit+=count*boardingCost-runningCost
count=0
i+=1
if profit>max_profit:
max_profit = profit
rot = i
#print(i,profit,count)
return rot
|
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, i, max_profit, waiting, profit, max_idx = len(customers), 0, 0, 0, 0, -1
while waiting > 0 or i<n:
if i<n:
waiting += customers[i]
i+=1
count = min(4, waiting)
waiting -= count
profit += count * boardingCost - runningCost
if profit > max_profit:
max_profit, max_idx = profit, i
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:
ans,max_profit=-1,0
waiting=0
profit=0
i=0
while waiting>0 or i<len(customers):
if i<len(customers):waiting+=customers[i]
if waiting>=4:
waiting-=4
boarded=4
else:
boarded=waiting
waiting=0
max_profit+=(boardingCost*boarded)-runningCost
if max_profit>0 and max_profit>profit:
ans=i+1
profit=max_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:
ans = -1
most = pnl = waiting = 0
for i, x in enumerate(customers):
waiting += x # more people waiting in line
waiting -= (chg := min(4, waiting)) # boarding
pnl += chg * boardingCost - runningCost
if most < pnl: ans, most = i+1, pnl
q, r = divmod(waiting, 4)
if 4*boardingCost > runningCost: ans += q
if r*boardingCost > runningCost: ans += 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:
best = 0
bestindex = 0
current = 0
currentindex = 0
nc = 0
for i in range(len(customers)):
nc += customers[i]
canadd = min(4,nc)
nc -= canadd
current += canadd*boardingCost
current -= runningCost
currentindex+=1
# print(current)
if current > best:
best = current
bestindex = currentindex
while nc > 0:
canadd = min(4,nc)
nc -= canadd
current += canadd*boardingCost
current -= runningCost
currentindex+=1
# print(current)
if current > best:
best = current
bestindex = currentindex
if best == 0:
bestindex = -1
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:
curr, ans, waiting, profit = 0, 0, 0, 0
for turn, customers in enumerate(customers):
waiting += customers
boarding = min(4, waiting)
waiting -= boarding
profit += (boardingCost * boarding) - runningCost
if profit > curr:
curr, ans = profit, turn+1
else:
j = turn
while waiting > 0:
j += 1
boarding = min(4, waiting)
waiting -= boarding
profit += (boardingCost * boarding) - runningCost
if profit > curr:
curr, ans = profit, j + 1
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:
max_profit = -sys.maxsize
result = - sys.maxsize
pending = 0
running_counter = 0
current_onboarding = 0
for customer in customers:
pending += customer
running_counter += 1
if pending > 0:
real_boarding = min(4, pending)
current_onboarding += real_boarding
profit = current_onboarding * boardingCost - running_counter * runningCost
if max_profit < profit:
max_profit = profit
result = running_counter
pass
pending -= real_boarding
pass
pass
while pending:
running_counter += 1
real_boarding = min(4, pending)
current_onboarding += real_boarding
pending -= real_boarding
profit = current_onboarding * boardingCost - running_counter * runningCost
if max_profit < profit:
max_profit = profit
result = running_counter
pass
pass
if max_profit <= 0:
return -1
return result
|
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:
b = w = i = 0
n = len(customers)
prof = 0
mprof = 0
pos = -1
while i < n or w:
if i < n:
w += customers[i]
i += 1
b = min(w, 4)
w -= b
prof += b * boardingCost - runningCost
if prof > mprof:
mprof = prof
pos = i
return pos
|
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:
run = 0
maxRun = 0
profit = 0
maxProfit = 0
total = 0
i = 0
n = len(customers)
while total > 0 or i < n:
if i < n:
total += customers[i]
i += 1
group = min(4, total)
total -= group
profit = profit + group*boardingCost - runningCost
run += 1
if profit > maxProfit:
maxProfit = profit
maxRun = run
return maxRun 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, lis: List[int], b: int, r: int) -> int:
n = len(lis)
q=pro=ans=tot=c=fin=0
c=1
for i in range(n):
q+=lis[i]
t = min(4,q)
q-=t
tot+=t
pro = tot*b - r*(c)
if pro>ans:
ans=pro
fin=c
# print(pro,ans,tot)
c+=1
while q>0:
t = min(4,q)
q-=t
tot+=t
pro = tot*b - r*(c)
if pro>ans:
ans=pro
fin=c
# print(pro,ans,tot)
c+=1
if fin==0:
fin=-1
return fin
|
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
answ = -1
waiting = 0
profit = 0
i = 0
while i < len(customers) or waiting > 0:
if i < len(customers):
waiting += customers[i]
if waiting >= 4:
profit += 4*boardingCost - runningCost
waiting -= 4
elif waiting > 0:
profit += waiting*boardingCost - runningCost
waiting = 0
else:
profit -= runningCost
if max_profit < profit:
max_profit = profit
answ = i + 1
i += 1
return answ
|
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_waiting = 0
max_till_now = 0
prev_cost = 0
answer = -1
for index, customer in enumerate(customers):
on_board = min(customer + customers_waiting, 4)
customers_waiting = customer + customers_waiting - 4 if on_board == 4 else 0
cost = prev_cost + on_board * boardingCost - runningCost
prev_cost = cost
if cost > max_till_now:
max_till_now = cost
answer = index + 1
while customers_waiting:
index += 1
on_board = min(customers_waiting, 4)
customers_waiting = customers_waiting - 4 if on_board == 4 else 0
cost = prev_cost + on_board * boardingCost - runningCost
prev_cost = cost
if cost > max_till_now:
max_till_now = cost
answer = index + 1
return answer
|
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 typing import List
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
def boarders():
waiters = 0
for c in customers:
waiters += c
cur = min(4, waiters)
waiters -= cur
yield cur
while waiters > 0:
cur = min(4, waiters)
waiters -= cur
yield cur
max = 0
max_idx = -1
cur = 0
for i, b in enumerate(boarders()):
cur += (b * boardingCost) - runningCost
if cur > max:
max = cur
max_idx = i + 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:
if (boardingCost << 2) <= runningCost:
return -1
waitingCustomers = 0
curProfit = 0
maxProfit = 0
maxProfitTurns = 0
curTurn = 0
for nCust in customers:
curTurn += 1
waitingCustomers += nCust
if waitingCustomers > 0:
boardedCustomers = min(4, waitingCustomers)
waitingCustomers -= boardedCustomers
curProfit += boardedCustomers * boardingCost
curProfit -= runningCost
if curProfit > maxProfit:
maxProfit = curProfit
maxProfitTurns = curTurn
fullLoads = waitingCustomers >> 2
remLoad = waitingCustomers & 0b11
curProfit += ((fullLoads * boardingCost) << 2) - runningCost * fullLoads
curTurn += fullLoads
if curProfit > maxProfit:
maxProfit = curProfit
maxProfitTurns = curTurn
if remLoad > 0:
curProfit += remLoad * boardingCost - runningCost
curTurn += 1
if curProfit > maxProfit:
maxProfit = curProfit
maxProfitTurns = curTurn
if curProfit > maxProfit:
return curTurn
if maxProfit == 0:
return -1
else:
return maxProfitTurns
|
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, cust: List[int], board: int, run: int) -> int:
wait = 0;tot = 0;profit = 0;move = 1;ans =0;maxi = 0
for i in range(len(cust)):
tot += min(4,cust[i]+wait)
wait = max(0,cust[i]+wait-4)
profit = tot*board - move*run
if profit > maxi:
maxi = profit;ans = move
move +=1
#print(tot,wait,profit)
while wait > 0:
tot += min(4,wait)
wait -= 4
profit = tot*board - move*run
if profit > maxi:
maxi = profit;ans = move
move +=1
#print(tot,wait,profit)
if maxi > 0:
return ans
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], boarding_cost: int, running_cost: int) -> int:
customer = 0
remained = 0
max_profit = -1
run = 0
max_run = 0
index = 0
while remained or index < len(customers):
run += 1
remained += customers[index] if index < len(customers) else 0
if (remained < 4):
customer += remained
remained = 0
else:
customer += 4
remained -= 4
profit = customer * boarding_cost - run * running_cost
if (profit > max_profit):
max_profit = profit
max_run = run
index += 1
return -1 if max_profit < 0 else max_run
|
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 0
waiting = 0
op = 0
maxProfit = -1
index = 0
total_boarding = 0
for i, people in enumerate(customers):
op += 1
waiting += people
boarding = min(4, waiting)
total_boarding += boarding
currProfit = total_boarding * boardingCost - op * runningCost
# other way is to just add the profit
# currProfit += boarding * boardingCost - runningCost
waiting -= boarding
if currProfit > maxProfit:
maxProfit = currProfit
index = op
while waiting > 0:
op += 1
boarding = min(waiting, 4)
total_boarding += boarding
currProfit = total_boarding * boardingCost - op * runningCost
waiting -= boarding
if currProfit > maxProfit:
maxProfit = currProfit
index = op
if maxProfit == -1:
return -1
else:
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, c: List[int], b: int, r: int) -> int:
n = len(c)
i = 0
rest = 0
max_val, max_i = 0, -2
val = 0
while i<n or rest > 0:
if i < n:
rest += c[i]
p = min(rest, 4)
val += p * b - r
if val > max_val:
max_val = val
max_i = i
rest -= p
i += 1
return max_i + 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
|
MIN = float('-inf')
class Solution:
def minOperationsMaxProfit(self, customers, boardingCost, runningCost):
n = len(customers)
step, maxStep, maxProfit = 0, 0, MIN
i, people, queue = 0, 0, 0
while True:
if i < n:
queue += customers[i]
i += 1
p = min(4, queue)
queue -= p
people += p
step += 1
profit = people * boardingCost - step * runningCost
if profit > maxProfit:
maxProfit = profit
maxStep = step
if queue == 0 and i == n:
break
return maxStep 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:
profit=record=peeps=rots=i=0
n = len(customers)
while peeps or i < n:
if i < n:
peeps+=customers[i]
board = min(4, peeps)
profit += boardingCost*board - runningCost
peeps-=board
if profit > record:
rots = i + 1
record = profit
i+=1
return rots if rots != 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:
customers = customers[::-1]
in_line = profit = rot = best_rot = 0
max_profit = -1
while customers or in_line:
c = customers.pop() if customers else 0
in_line += c
board = min(in_line, 4)
in_line -= board
profit += board * boardingCost
rot += 1
profit -= runningCost
if profit > max_profit:
max_profit = profit
best_rot = rot
return best_rot 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:
curr_waiting = 0
curr_profit, max_profit, curr_op, max_profit_ops = 0, 0, 0, 0
for c in customers:
curr_op += 1
curr_waiting += c
num_boarded = min(4, curr_waiting)
curr_waiting -= num_boarded
curr_profit += boardingCost * num_boarded - runningCost
if curr_profit > max_profit:
max_profit = curr_profit
max_profit_ops = curr_op
while curr_waiting:
curr_op += 1
num_boarded = min(4, curr_waiting)
curr_waiting -= num_boarded
curr_profit += boardingCost * num_boarded - runningCost
if curr_profit > max_profit:
max_profit = curr_profit
max_profit_ops = curr_op
return max_profit_ops 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
|
from collections import deque
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
customers = deque(customers)
in_line = profit = rot = best_rot = 0
max_profit = -1
while customers or in_line:
c = customers.popleft() if customers else 0
in_line += c
board = min(in_line, 4)
in_line -= board
profit += board * boardingCost
rot += 1
profit -= runningCost
if profit > max_profit:
max_profit = profit
best_rot = rot
return best_rot 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:
rotations = 0
profit, max_profit = 0, 0
res = -1
def xreport(boarded:int):
nonlocal profit, rotations, max_profit, res
profit +=boarded*boardingCost - runningCost
rotations +=1
if profit>max_profit:
max_profit = profit
res = rotations
#print(f'p: {profit}, mprofit: {max_profit}, rotations:{rotations}')
return
for i in range(len(customers)-1):
if customers[i] >4:
customers[i+1]+=customers[i]-4
customers[i] =4
xreport(customers[i])
waiting = customers[-1]
while waiting>0:
if waiting>4:
waiting-=4
xreport(4)
else:
xreport(waiting)
waiting = 0
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:
profits=[]
occupy=0
cur_waiting = 0
k=0
while cur_waiting ==0:
cur_waiting=customers[k]
k+=1
m=k
counter=0
while cur_waiting >0:
if cur_waiting < 4:
occupy+=cur_waiting
cur_waiting = 0
else:
occupy+=4
cur_waiting-=4
counter+=1
profits.append(occupy*boardingCost - counter*runningCost)
if k<len(customers):
cur_waiting+=customers[k]
k+=1
z=max(profits)
if z>0:
return (profits.index(z)+m)
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:
d = {}
rotations, sum_cust = 0, 0
d[0] = 0
for i, current in enumerate(customers):
sum_cust +=current
if sum_cust>=4:
while sum_cust>=4:
rotations +=1
d[rotations] = d[rotations-1] + 4*boardingCost - runningCost
sum_cust -=4
else:
if i+1 in d:
continue
rotations+=1
d[rotations] = d[rotations-1]+ sum_cust*boardingCost - runningCost
sum_cust = 0
if sum_cust>0:
rotations+=1
d[rotations] = d[rotations-1]+ sum_cust*boardingCost - runningCost
sum_cust = 0
#print(d)
d[0] = -1
tmp, res = -1, -1
for i in d:
if tmp<d[i]:
tmp = d[i]
res = 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
|
from typing import List
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
cur = best = wheel_turns = 0
best_idx = None
waiting = 0
capacity = 4
while waiting or wheel_turns < len(customers):
# Add new waiting customers, put passengers on ride, remove passengers from waiting.
if wheel_turns < len(customers):
waiting += customers[wheel_turns]
wheel_turns += 1
passengers = min(waiting, capacity)
waiting -= passengers
# Update money.
cur += (passengers * boardingCost) - runningCost
if cur > best:
best_idx = wheel_turns
best = cur
if best_idx is None:
best_idx = -1
return best_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:
res = round = profit = max_profit = remain = i = 0
while remain or i < len(customers):
if i < len(customers):
remain += customers[i]
i += 1
on_board = min(4, remain)
remain -= on_board
profit += boardingCost * on_board - runningCost
round += 1
if profit > max_profit:
max_profit = profit
res = round
return res 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
|
import sys # O(N) , N is customers.length
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boarding_cost: int, running_cost: int) -> int:
i = 0
remains = 0
standard = sum(customers)
stack = 0
ans = -1
big = -sys.maxsize-1
while True:
if standard <= 0:
break
if i < len(customers):
remains += customers[i]
if remains >= 4:
remains -= 4
stack += 4
standard -= 4
else:
stack += remains
standard -= remains
remains = 0
tmp = stack*boarding_cost - running_cost*(i+1)
if tmp > big:
big = tmp
ans = i
i += 1
if big <= 0:
return -1
return ans+1
# member도 늘어나야 한다.
|
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 or len(customers) == 0:
return -1
cur_round = 1
max_round = -1
customer_cnt = board_cnt = cur_profit = max_profit = i = 0
while (customer_cnt > 0 or i < len(customers)):
if i < len(customers):
customer_cnt += customers[i]
i += 1
board_cnt = min(customer_cnt, 4)
customer_cnt -= board_cnt
cur_profit += (board_cnt * boardingCost) - runningCost
if cur_profit > max_profit:
max_profit = cur_profit
max_round = cur_round
cur_round += 1
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:
if runningCost > 4*boardingCost:
return -1
profit = 0
board = 0
wait = 0
ans = 0
for i in range(len(customers)):
if customers[i] > 4:
wait += customers[i]-4
board += 4
p = board*boardingCost - (i+1)*runningCost
if p > profit:
profit = p
ans = i+1
else:
add = min(wait, 4-customers[i])
wait -= add
board += add+customers[i]
p = board*boardingCost - (i+1)*runningCost
if p > profit:
profit = p
ans = i+1
i = len(customers)
while wait != 0:
add = min(wait,4)
board += add
wait -= add
p = board*boardingCost - (i+1)*runningCost
if p > profit:
profit = p
ans = i+1
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:
profit = [0]
left = 0
maxMid = -1
maxProfit = 0
for i in range(len(customers)):
toBeBoarded = left + customers[i]
if toBeBoarded > 4:
boarded = 4
left = toBeBoarded - 4
else:
boarded = customers[i]
left = 0
newProfit = boardingCost * boarded - runningCost + profit[-1]
profit.append(newProfit)
if newProfit > maxProfit:
maxProfit = newProfit
maxMid = i
# print(profit)
if left == 0:
return maxMid
elif left < 4:
lastProfit = boardingCost * left - runningCost + profit[-1]
if maxProfit >= lastProfit:
return maxMid + 1 if maxProfit > 0 else -1
else:
return len(customers) + 1 if lastProfit > 0 else -1
else:
potential = boardingCost * 4 - runningCost
if potential > 0:
rotations = left // 4
lastRun = left % 4
rotationEnd = profit[-1] + potential * rotations
lastProfit = rotationEnd + lastRun * boardingCost - runningCost
if maxProfit >= rotationEnd:
return maxMid + 1 if maxProfit > 0 else -1
else:
if rotationEnd >= lastProfit:
return len(customers) + rotations if rotationEnd > 0 else -1
else:
return len(customers) + rotations + 1 if lastProfit > 0 else -1
else:
return maxMid + 1 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:
waiting = 0
boarded = 0
maxProfit = 0
maxIndex = -1
if len(customers) == 0:
return -1
for i, customerCount in enumerate(customers):
waiting += customerCount
if waiting >= 4:
waiting -= 4
boarded += 4
else:
boarded += waiting
waiting = 0
profit = boarded * boardingCost - (i + 1) * runningCost
if profit > maxProfit:
maxIndex = i + 1
maxProfit = profit
i += 1
while waiting > 0:
if waiting >= 4:
waiting -= 4
boarded += 4
else:
boarded += waiting
waiting = 0
profit = boarded * boardingCost - (i + 1) * runningCost
if profit > maxProfit:
maxIndex = i + 1
maxProfit = profit
i += 1
if maxProfit > 0:
return maxIndex
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:
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]
board = min(4, wait)
earn += board* boardingCost - runningCost
if earn > max_earn:
res = i + 1
max_earn = earn
wait -= board
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:
remain = 0
profit = 0
maxProfit = -float('inf')
maxId = -1
for i,val in enumerate(customers):
total = remain + val
if total>4:
profit += 4*boardingCost
remain = total - 4
else:
profit += total*boardingCost
remain = 0
profit -= runningCost
if profit>maxProfit:
maxId = i
maxProfit = profit
#print(remain,profit,maxProfit)
j = maxId+1
while remain>0:
profit+=min(remain,4)*boardingCost - runningCost
if profit>maxProfit:
maxId = j
maxProfit = profit
#print(remain,profit,maxProfit,maxId)
remain = max(remain-4,0)
j = j+1
if maxProfit<=0:
return -1
else:
return maxId+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:
#greedy: put the most people on at a given turn
#why would you ever wait? I dont think you would, might as well take the instant profit
#mostly because theres no reason to leave money on the table
#when would you definitely not wait?
remaining_customers = 0
max_profit = 0
profit = 0
spin = 0
best_spin = 0
while remaining_customers > 0 or spin < len(customers):
while spin < len(customers):
remaining_customers += customers[spin]
boarded = min(4, remaining_customers)
remaining_customers -= boarded
profit = profit + boardingCost*boarded - runningCost
#print('Profit after spin {} is {}'.format(spin+1, profit))
if profit > max_profit:
max_profit = profit
best_spin = spin
spin += 1
boarded = min(4, remaining_customers)
remaining_customers -= boarded
profit = profit + boardingCost*boarded - runningCost
if profit > max_profit:
max_profit = profit
best_spin = spin
spin += 1
#print('Profit after spin {} is {}'.format(spin+1, profit))
if max_profit <= 0:
return -1
else:
return best_spin+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:
i = 0
wait = 0
profit = 0
maxProfit = 0
passengers = 0
rot = 0
finalRot = 0
while i < len(customers) or wait > 0:
if i < len(customers):
wait += customers[i]
i += 1
if wait < 5:
passengers += wait
wait = 0
else:
passengers += 4
wait -= 4
rot += 1
profit = passengers * boardingCost - rot * runningCost
if profit > maxProfit:
maxProfit = profit
finalRot = rot
if maxProfit > 0:
return finalRot
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:
waiting=peak_at=peak=delta=profit=t=0
while waiting or t<len(customers):
if t < len(customers):
waiting+=customers[t]
t+=1
delta =min(4,waiting)
profit+=delta*boardingCost-runningCost
waiting-=delta
if peak<profit:
peak=profit
peak_at=t
return peak_at if peak_at>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:
result = []
maxval = 0
Current_profit = 0
board_list = 0
total_board = 0
count = 0
i = 1
j = 1
for x in range(0,len(customers)):
#print(x)
if customers[x] != 0:
wait_list = customers[x]
i += x
break
else:
count += 1
#print(\"Waiting List & x\",wait_list,i)
while wait_list != 0:
if wait_list >= 4:
total_board += 4
wait_list -= 4
else:
total_board += wait_list
wait_list -= wait_list
#print(total_board,j)
Current_profit = total_board * boardingCost - j * runningCost
if i < len(customers):
wait_list += customers[i]
i += 1
j += 1
result.append(Current_profit)
#print(Current_profit)
#print(result)
#print(result[992])
#print(len(result))
maxval = max(result)
#print(maxval)
if maxval < 0:
return -1
else:
return result.index(maxval) + 1 + count
#return Current_profit
|
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:
# 思路copied from https://leetcode.com/problems/maximum-profit-of-operating-a-centennial-wheel/discuss/866409/Java-Simple-O(N)-greedy
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
run = 0
maxRun = 1
prof = maxProf = 0
count_ppl = 0
i = 0
while count_ppl > 0 or i < len(customers):
if i < len(customers):
count_ppl += customers[i]
i += 1
count_bd = min(4, count_ppl) # boarding people by greedy.
count_ppl -= count_bd
prof = prof + count_bd * boardingCost - runningCost
run += 1
if prof > maxProf:
maxProf = prof
maxRun = run
return maxRun if maxProf > 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
ans=0
N=len(customers)
cur1,cur2,cur3,cur4=0,0,0,0
wait=0
profit=0
out=-1
for i in range(N):
wait+=customers[i]
add=min(4,wait)
profit+=(boardingCost*add-runningCost)
wait-=add
if profit>ans:
out=i+1
ans=profit
cur1,cur2,cur3,cur4=add,cur1,cur2,cur3
while wait>0:
i+=1
add=min(4,wait)
profit+=(boardingCost*add-runningCost)
wait-=add
if profit>ans:
out=i+1
ans=profit
cur1,cur2,cur3,cur4=add,cur1,cur2,cur3
return out
|
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 = 0
mark = 0
waiting = 0
n = len(customers)
i = 0
while i < n or waiting > 0:
if i < n:
waiting = waiting + customers[i]
if waiting > 4:
waiting = waiting - 4
profit = profit + 4 * boardingCost
else:
profit = profit + waiting * boardingCost
waiting = 0
profit = profit - runningCost
if profit > max_profit:
max_profit = profit
mark = i + 1
i = i + 1
return mark 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 runningCost >= 4 * boardingCost:
return -1
profit = 0
max_profit = float('-inf')
wc = customers[0]
i = 1
rot = 1
max_rot = 0
while wc > 0 or i < len(customers):
if wc >= 4:
#print wc
wc -= 4
profit += 4 * boardingCost
elif wc < 4:
bc = wc
wc = 0
profit += bc * boardingCost
profit -= runningCost
prev = max_profit
max_profit = max(profit, max_profit)
if max_profit != prev:
max_rot = rot
if i < len(customers):
wc += customers[i]
i+=1
rot +=1
if max_profit > 0:
return max_rot
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:
rt = 0
prof = 0
maxRt = 0
maxProf = 0
wait = 0
i = 0
onborad = 0
while wait > 0 or i < len(customers):
if i < len(customers):
wait += customers[i]
i += 1
onboard = min(4, wait)
wait -= onboard
prof = prof + onboard * boardingCost - runningCost
rt += 1
if maxProf < prof:
maxProf = prof
maxRt = rt
if maxProf > 0 :
return maxRt
else:
return -1
# pro = 0
# curpro = 0
# wait = 0
# on = 0
# j = 0
# for i, n in enumerate(customers):
# if n >= 4:
# wait += n - 4
# on += 4
# else:
# on += n
# j += 1
# curpro = on* boardingCost - j *runningCost
# if curpro < 0:
# ans = -1
# elif pro < curpro:
# pro = max(pro, curpro)
# ans = j
# while wait > 0:
# if wait >= 4:
# wait -= 4
# on += 4
# else:
# on += wait
# wait = 0
# j += 1
# curpro = on* boardingCost - j*runningCost
# if curpro < 0:
# ans = -1
# elif pro < curpro:
# if j > 300:
# print(on, wait, pro, curpro, j)
# ans = j
# pro = max(pro, curpro)
# 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:
# 思路copied from https://leetcode.com/problems/maximum-profit-of-operating-a-centennial-wheel/discuss/866409/Java-Simple-O(N)-greedy
# The solution is based on simply simulating the the rotations and keep track of waiting customers
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
run = 0
maxRun = 1
prof = maxProf = 0
count_ppl = 0
i = 0
while count_ppl > 0 or i < len(customers):
if i < len(customers):
count_ppl += customers[i]
i += 1
count_bd = min(4, count_ppl) # boarding people by greedy.
count_ppl -= count_bd
prof = prof + count_bd * boardingCost - runningCost
run += 1
if prof > maxProf:
maxProf = prof
maxRun = run
return maxRun if maxProf > 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:
#counter for the customers array
i = 0
#customers waitlist
wait = 0
#profit at each iteration
profit = 0
#max profit made, we will compare it each time to profit
maxProfit = 0
#total number of passengers who paid the boarding coast
passengers = 0
#total # rotations at each iteration
rot = 0
#the # rotations for which profit = max profit
finalRot = 0
while i < len(customers) or wait > 0:
# in case we didn't finish the array customers:
if i < len(customers):
wait += customers[i]
i += 1
# in case wait has less than 5 customers, we make it equal to 0
if wait < 5:
passengers += wait
wait = 0
else:
passengers += 4
wait -= 4
#total number of rotations until now
rot += 1
#total profit until now
profit = passengers * boardingCost - rot * runningCost
#updating max profit and best number of rotations:
if profit > maxProfit:
maxProfit = profit
finalRot = rot
if maxProfit > 0:
return finalRot
#in case the profit is always <= 0
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:
waiting = served = onboard = max_round = max_profit = 0
cur_round = 1
while cur_round <= len(customers) or waiting > 0:
if cur_round <= len(customers):
waiting += customers[cur_round - 1]
if waiting > 4:
onboard = 4
waiting -= onboard
else:
onboard = waiting
waiting = 0
served += onboard
cur_profit = served * boardingCost - cur_round * runningCost
if cur_profit > max_profit:
max_profit = cur_profit
max_round = cur_round
cur_round += 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:
board = customers[0]
profit = 0
prevprofit = 0
best = -1
i=1
k = 0
while board>0 or i!=len(customers):
k+=1
sub = min(board,4)
profit += sub*boardingCost - runningCost
board-=sub
if profit>prevprofit and profit>0:
best = k
if i<len(customers):
board += customers[i]
i+=1
prevprofit = profit
return best
|
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
if boardingCost * 4 == runningCost:
return 0
i = 0
res = 0
curM = 0
cur = 0
remind = 0
served = 0
for i in range(len(customers)):
remind += customers[i]
served += min(remind, 4)
remind = max(0, remind-4)
cur = served * boardingCost - (i+ 1) *runningCost
if cur > curM:
curM = cur
res = i + 1
if remind * boardingCost - runningCost <= 0:
return res
res = len(customers)
while min(4, remind) * boardingCost - runningCost > 0:
remind -= min(4, remind)
res += 1
return res
total = sum(customers)
if (total%4) * boardingCost <= runningCost:
return total // 4
return total // 4 + 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
most = profit = waiting = 0
for i, x in enumerate(customers):
waiting += x # more people waiting in line
waiting -= (running := min(4, waiting)) # boarding
profit += running * boardingCost - runningCost
if most < profit:
ans, most = i+1, profit
q, r = divmod(waiting, 4)
if 4*boardingCost > runningCost:
ans += q
if r*boardingCost > runningCost:
ans += 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:
waiting = 0
profits = 0
bestProfits = 0
bestRotations = -1
for rotation in range(999999):
if rotation < len(customers):
waiting += customers[rotation]
elif waiting == 0:
break
bording = min(4, waiting)
waiting -= bording
profits += bording * boardingCost
profits -= runningCost
if profits > bestProfits:
bestProfits = profits
bestRotations = rotation+1
# print(rotation, waiting, bording, profits, bestProfits)
return bestRotations
|
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 = total = profit = maxp = res = i = 0
while wait or i < len(customers):
if i < len(customers):
total += customers[i]
wait = max(0, customers[i] + wait - 4)
else:
wait = max(0, wait - 4)
i += 1
profit = boardingCost * (total - wait) - runningCost * i
if profit > maxp:
maxp, res = profit, i
return res if maxp > 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:
wait = 0
i = 0
n = len(customers)
max_profit, min_rotates = 0, -1
total_boarded = 0
rotate = 0
while True:
if i == n and not wait:
break
if i < n:
wait += customers[i]
i += 1
rotate += 1
board = min(wait, 4)
wait -= board
total_boarded += board
profit = total_boarded * boardingCost - rotate * runningCost
if profit > max_profit:
max_profit = profit
min_rotates = rotate
return min_rotates
|
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
maxprofit = 0
n = len(customers)
if n == 0:
return maxprofit
waiting = customers[0]
k = 1
rounds = 1
max_rounds = -1
so_far = 0
while k < n or waiting > 0:
this_round = min(4, waiting)
waiting -= this_round
so_far += this_round
profit = so_far * boardingCost - rounds * runningCost
if profit > maxprofit:
maxprofit = profit
max_rounds = rounds
if k < n:
waiting += customers[k]
k += 1
rounds += 1
return max_rounds
|
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 typing import List
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
cur = best = idx = 0
best_idx = None
waiting = 0
capacity = 4
while waiting or idx < len(customers):
# Add new waiting customers, put passengers on ride, remove passengers from waiting.
if idx < len(customers):
waiting += customers[idx]
passengers = min(waiting, capacity)
waiting -= passengers
# Update money.
cur += (passengers * boardingCost) - runningCost
if cur > best:
best_idx = idx + 1
best = cur
idx += 1
if best_idx is None:
best_idx = -1
return best_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:
queue_customer = 0
total_profit = [0]
for c in customers:
queue_customer += c
n_board = min(4, queue_customer)
queue_customer -= n_board
total_profit.append(total_profit[-1] + n_board * boardingCost - runningCost)
while queue_customer > 0:
n_board = min(4, queue_customer)
queue_customer -= n_board
total_profit.append(total_profit[-1] + n_board * boardingCost - runningCost)
max_profit = max(total_profit)
if max_profit <= 0:
return -1
else:
for i, p in enumerate(total_profit):
if p == max_profit:
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:
max_prof = -math.inf
idx = -1 # rotation idx for max_prof
sofar = 0
profit_sofar = 0
i = 0
while True:
if i < len(customers):
sofar += customers[i]
if sofar == 0 and i >= len(customers):
break
num_onboard = min(4, sofar)
sofar -= num_onboard
profit_sofar += num_onboard * boardingCost - runningCost
# print('num_onboard: {}, profit_sofar: {}'.format(num_onboard, profit_sofar))
if profit_sofar > max_prof:
max_prof = profit_sofar
idx = i
i += 1
return idx + 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], boardingCost: int, runningCost: int) -> int:
customers.reverse()
wait = 0
board = 0
turn = 0
ans = -1
score = 0
while wait or customers:
turn += 1
if customers:
wait += customers.pop()
if wait >= 4:
board += 4
wait -= 4
else:
board += wait
wait = 0
if score < board * boardingCost - turn * runningCost:
ans = turn
score = board * boardingCost - turn * runningCost
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], bc: int, rc: int) -> int:
peak = res = idx = cust = i = 0
while cust > 0 or i < len(customers):
if i < len(customers):
cust += customers[i]
c = min(cust, 4)
res += c * bc - rc
if res > peak:
peak = res
idx = i + 1
cust -= c
i += 1
return idx if idx > 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, A, BC, RC):
ans=profit=t=0
maxprofit=0
wait=i=0
n=len(A)
while wait>0 or i<n:
if i<n:
wait+=A[i]
i+=1
t+=1
y=min(4,wait)
wait-=y
profit+=y*BC
profit-=RC
if profit>maxprofit:
maxprofit=profit
ans=t
return -1 if maxprofit<=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:
rotate = 0
total = 0
left = 0
for i in customers:
total += i
left += i
if left <= 4:
rotate += 1
left = 0
left -= 4
rotate += 1
rotate += left // 4
if left % 4 * boardingCost > runningCost:
rotate += 1
if boardingCost*total - rotate*runningCost > 0:
return rotate
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, arr: List[int], boardingCost : int, runningCost : int) -> int:
# Dividing people in the groups of <=4
grps = []
# length of customers array
n = len(arr)
# rem--> number of people waiting
rem = 0
# traversing the customers array
for i in range(n):
# total number of people available right now
avail = arr[i]+rem
# number of available people >=4 then append grp of 4 and update remaining [rem]
if avail>=4:
avail-=4
grps.append(4)
rem = avail
# number of available people <4 then make group of available people and update remaining [rem=0]
else:
rem = 0
grps.append(avail)
# make groups of 4 until remaining >=4 otherwise make <4 and break
while rem>0:
if rem>=4:
rem-=4
grps.append(4)
else:
grps.append(rem)
rem = 0
# mex--> represents maximum profit
mex = -10**10
# cost--> represents current total cost
cost = 0
# ind --> represents rotation number
ind = 0
for i in range(len(grps)):
# calculate net cost till now
cost+= boardingCost*grps[i]-runningCost
# upadte max profit and rotation number
if mex<cost:
mex = max(mex,cost)
ind = i+1
# max profit< 0
if mex<0:
return -1
# return rotation number
return 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:
# Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
# Output: 9
max_amt, curr_amt, min_turns, curr_turns = 0, 0, 0, 0
total, idx = 0, 0
while (len(customers) > idx) or total > 0:
# print(f'{curr_amt=} | {total=} | {curr_turns=} | {min_turns=}')
if len(customers) > idx:
total += customers[idx]
idx += 1
curr_turns += 1
if total >= 4: curr = 4
else: curr = total
total -= curr
curr_amt += (boardingCost * curr) - (runningCost)
if curr_amt > max_amt:
max_amt = curr_amt
min_turns = curr_turns
if not min_turns: return -1
else: return min_turns
|
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
n = len(customers)
maxtotal, maxrotations = 0, -1
total, rotations = 0, 1
i = 0
while True:
if i == n-1 and customers[n-1] == 0:
break
if customers[i]>4:
total += boardingCost*4 - runningCost
if i < n-1:
customers[i+1] += customers[i]-4
customers[i] = 0
else:
customers[i] -= 4
else:
total += customers[i]*boardingCost - runningCost
customers[i] = 0
if total>maxtotal:
maxtotal = total
maxrotations = rotations
if i<n-1: i += 1
rotations += 1
return maxrotations
|
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, board = 0, 0
ans = 1
res = 1
m = -1
def cal():
nonlocal total, ans, res, m, board
t = 4 if total >= 4 else total
total -= t
board += t
cost = board * boardingCost - ans * runningCost
if cost > m:
m = cost
res = ans
ans += 1
for c in customers:
total += c
cal()
while total > 0:
cal()
return res if m >= 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:
prof = 0; max_prof = 0; max_run = 0; run = 0; i = 0
waiting = 0
while waiting > 0 or i < len(customers):
if i < len(customers):
waiting += customers[i]
i += 1
board = min(4, waiting)
waiting -= board
prof += boardingCost * board - runningCost
run += 1
if prof > max_prof:
max_prof = prof
max_run = run
return max_run 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], boardingCost: int, runningCost: int) -> int:
n = len(customers)
i = 0
num_waiting = 0
ans = -1
max_profit = 0
taken = 0
while i < n or num_waiting:
curr = num_waiting
if i < n: curr += customers[i]
if curr > 4:
taken += 4
num_waiting = curr - 4
elif curr > 0:
taken += curr
num_waiting = 0
elif curr == 0:
num_waiting = 0
if boardingCost * taken - runningCost * (i + 1) > max_profit:
max_profit = boardingCost * taken - runningCost * (i + 1)
ans = i + 1
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:
total_customers = 0
max_profit = 0
no_rotation = 0
pending = 0
i = 0
while i<len(customers) or pending>0:
if i<len(customers):
pending += customers[i]
if pending>=4:
pending -= 4
total_customers += 4
else:
total_customers += pending
pending = 0
profit = total_customers*boardingCost - (i+1)*runningCost
if profit>max_profit:
max_profit = profit
no_rotation = (i+1)
i += 1
if no_rotation == 0:
return -1
return no_rotation
|
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:
max_pr = pr = 0; cnt = max_cnt = 0; w = 0
for x in a:
x = x + w
pr += min(x, 4) * bc - rc
cnt += 1
if pr > max_pr: max_pr, max_cnt = pr, cnt
w = max(x - 4, 0)
while w > 0:
pr += min(w, 4) * bc - rc
cnt += 1
if pr > max_pr: max_pr, max_cnt = pr, cnt
w = max(w - 4, 0)
return max_cnt if max_pr > 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
num_spins = 0
num_customers = 0
min_spins = float('inf')
max_profit = -float('inf')
be = runningCost // boardingCost
if be >= 4:
return -1
for new_customers in customers:
num_spins += 1
q += new_customers
loaded_customers = min(q, 4)
q -= loaded_customers
num_customers += loaded_customers
profit = num_customers*boardingCost - num_spins*runningCost
if profit > max_profit:
max_profit = profit
min_spins = num_spins
if q:
full, partial = divmod(q, 4)
num_spins += full
num_customers += 4*full
if partial > be:
num_spins += 1
num_customers += partial
profit = num_customers*boardingCost - num_spins*runningCost
if profit > max_profit:
min_spins = num_spins
return min_spins
|
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:
l = list()
num_of_people = 0
wait_people = 0
c = 0
for i in customers:
c += 1
if (i + wait_people) >= 4:
num_of_people += 4
wait_people = i + wait_people - 4
else:
num_of_people += i + wait_people
wait_people = 0
temp = num_of_people*boardingCost - c*runningCost
l.append(temp)
while(wait_people > 0):
c += 1
if wait_people >= 4:
num_of_people += 4
wait_people -= 4
else:
num_of_people += wait_people
wait_people = 0
temp = num_of_people*boardingCost - c*runningCost
l.append(temp)
if max(l) > 0:
return l.index(max(l))+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
|
import math
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
len_rot = len(customers)
if(4*boardingCost <= runningCost or len_rot == 0):
return -1
tot = 0
profit = 0
lis = []
for i in customers:
tot += i
profit += min(4, tot)*boardingCost - runningCost
tot -= min(4, tot)
lis.append(profit)
while(tot > 0):
profit += min(4, tot)*(boardingCost) - runningCost
tot -= min(4, tot)
lis.append(profit)
max_value = max(lis)
return lis.index(max_value)+1 if max_value > 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:
wait = arrive = done = rotation = profit = maxprofit = i = 0
while i < len(customers) or wait > 0:
if i < len(customers):
arrived = customers[i]
else:
arrived = 0
wait += arrived
if wait >= 4:
wait -= 4
done += 4
else:
done += wait
wait = 0
profit = done * boardingCost - (i+1)*runningCost
i += 1
if profit > maxprofit:
maxprofit = profit
rotation = i
if profit <= 0:
return -1
else:
return rotation
|
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:
run, maxRun, prof = 0, 1, 0
maxProf = prof
sum_, i = 0, 0
while sum_ > 0 or i < len(customers):
if i < len(customers):
sum_ += customers[i]
bd = min(sum_, 4)
sum_ -= bd
prof = prof + bd * boardingCost - runningCost
run += 1
if prof > maxProf:
maxProf = prof
maxRun = run
i += 1
return maxRun if maxProf > 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 or len(customers)==0 :
return -1
maxVal = 0
times , recordedTime = 0 ,0
served , waitting = 0 , 0
while times < len(customers) or waitting!=0:
if times<len(customers):
waitting = waitting + customers[times]
if waitting > 4:
served+=4
waitting-=4
else:
served += waitting
waitting = 0
times+=1
currentProfit = served*boardingCost - times*runningCost
if currentProfit > maxVal :
maxVal = currentProfit
recordedTime = times
return recordedTime if recordedTime>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:
gondolas = [0] * 4
curr_gondola = 0
n = len(customers)
max_profit = profit = 0
min_rotation = 0
rotations = 0
waiting = 0
i = 0
while i < n or waiting:
# selected = False:
# customers[i] += 4 - customers[i]
if i < n:
while customers[i] < 4 and waiting:
customers[i] += 1
waiting -= 1
if i < n:
customer = customers[i]
if customer > 4:
waiting += customer - 4
customer = 4
curr_gondola = (1 + curr_gondola) % 4
profit += customer * boardingCost
else:
customer = min(4, waiting)
waiting -= customer
profit += customer * boardingCost
profit -= runningCost
rotations += 1
if profit > max_profit:
min_rotation = rotations
max_profit = profit
if i < n:
i += 1
# print(i, waiting, profit, max_profit, min_rotation, customer)
# while waiting > 0:
# customer = min(4, waiting)
# waiting -= customer
# profit += customer * boardingCost
# profit -= runningCost
# rotations += 1
# if profit > max_profit:
# min_rotation = rotations
# max_profit = profit
print((max_profit, rotations, min_rotation, customers))
if max_profit <= 0:
return -1
# print(min_rotation)
return min_rotation
|
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
N=len(customers)
cur1,cur2,cur3,cur4=0,0,0,0
wait=0
profit=0
out=-1
for i in range(N):
wait+=customers[i]
add=min(4,wait)
profit+=(boardingCost*add-runningCost)
wait-=add
if profit>ans:
out=i+1
ans=profit
cur1,cur2,cur3,cur4=add,cur1,cur2,cur3
while wait>0:
i+=1
add=min(4,wait)
profit+=(boardingCost*add-runningCost)
wait-=add
if profit>ans:
out=i+1
ans=profit
cur1,cur2,cur3,cur4=add,cur1,cur2,cur3
return out
|
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:
if len(customers) == 0: return 0
if boarding_cost * 4 <= running_cost: return -1
available_customer_count = 0
customers_per_rotation = []
for i in range(len(customers)):
available_customer_count += customers[i]
customers_per_rotation.append(min(available_customer_count, 4))
available_customer_count -= customers_per_rotation[-1]
while available_customer_count > 0:
customers_per_rotation.append(min(available_customer_count, 4))
available_customer_count -= customers_per_rotation[-1]
max_profit = 0
max_turn = -1
previous_profit = 0
current_customer_count = 0
for i,customer_count in enumerate(customers_per_rotation):
current_customer_count += customer_count
profit = ((current_customer_count * boarding_cost) - (running_cost * (i+1)))
if profit > max_profit:
max_profit = profit
max_turn = i + 1
return max_turn
|
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 = 0
max_profit_rotations = -1
rotations = 0
current_profit = 0
in_line = 0
for customer in customers:
in_line += customer
current_profit += min(in_line, 4) * boarding_cost - running_cost
in_line = max(in_line - 4, 0)
rotations += 1
if current_profit > max_profit:
max_profit = current_profit
max_profit_rotations = rotations
while in_line > 0:
current_profit += min(in_line, 4) * boarding_cost - running_cost
in_line = max(in_line - 4, 0)
rotations += 1
if current_profit > max_profit:
max_profit = current_profit
max_profit_rotations = rotations
return max_profit_rotations
'''
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.
10 - 4 - 24
'''
|
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, arr: List[int], boardingCost: int, runningCost: int) -> int:
grps = []
n = len(arr)
rem = 0
for i in range(n):
avail = arr[i]+rem
if avail>=4:
avail-=4
grps.append(4)
rem = avail
else:
rem = 0
grps.append(avail)
while rem>0:
if rem>=4:
rem-=4
grps.append(4)
else:
grps.append(rem)
rem = 0
mex = -10**10
cost = 0
ind = 0
for i in range(len(grps)):
# calculate net cost till now
cost+= boardingCost*grps[i]-runningCost
# upadte max profit and rotation number
if mex<cost:
mex = max(mex,cost)
ind = i+1
# max profit< 0
if mex<0:
return -1
# return rotation number
return ind
# idx = 0
# profit = 0
# max_idx = -1
# max_profit = 0
# n_cus = 0
# for cus in customers:
# idx += 1
# profit += min(4, cus) * boardingCost - runningCost
# n_cus += max(cus-4, 0)
# if profit > max_profit:
# max_idx = idx
# max_profit = profit
# if n_cus >= 4:
# if 4*boardingCost <= runningCost:
# return max_idx
# else:
# profit += (4*boardingCost - runningCost) * (n_cus // 4)
# idx += n_cus // 4
# n_cus %= 4
# if profit > max_profit:
# max_idx = idx
# max_profit = profit
# idx += 1
# profit += n_cus * boardingCost - runningCost
# if profit > max_profit:
# max_idx = idx
# 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, boardingCost, runningCost):
# we can never make profit
if 4*boardingCost <= runningCost:
return -1
maxProfit = -1
maxRotate = 0
accuProfit = 0
rotate = 0
boardingPool = 0
# keep rotate and board the customers.
i = 0
while i < len(customers) or boardingPool > 0:
# add customers to boarding queue
if i < len(customers):
boardingPool += customers[i]
i += 1
# make profit
currBoarding = min(4, boardingPool) # maximum boarding limit is 4
boardingPool -= currBoarding
accuProfit += currBoarding * boardingCost
# we need pay runningCost
rotate += 1
accuProfit -= runningCost
if maxProfit < accuProfit:
maxProfit = accuProfit
maxRotate = rotate
return -1 if maxProfit < 0 else maxRotate
|
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 = []
times = 0
for i in range(len(customers)-1):
if customers[i] <= 4:
times +=1
curr_profit = customers[i]*boardingCost - times*runningCost
else:
times +=1
curr_profit = 4*boardingCost - runningCost
customers[i+1] = customers[i+1] + customers[i] - 4
#print(customers)
profit.append(curr_profit)
while customers[-1] > 4:
times +=1
curr_profit = 4*boardingCost - runningCost
profit.append(curr_profit)
customers[-1] -= 4
times +=1
curr_profit = customers[-1]*boardingCost - runningCost
profit.append(curr_profit)
tot_profit = []
max_profit = 0
for i in range(len(profit)):
max_profit += profit[i]
tot_profit.append(max_profit)
real_max = max(tot_profit)
if real_max < 0:
return -1
else:
return tot_profit.index(real_max)+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 = []
wait = 0
onboard = 0
i = 0
while i<len(customers):
wait += customers[i]
onboard += min(4, wait)
wait -= 4
if wait<0:
wait = 0
ans.append((boardingCost * onboard) - (runningCost*(i+1)))
i+=1
while wait:
onboard += min(4, wait)
wait -= 4
if wait<0:
wait = 0
ans.append((boardingCost * onboard) - (runningCost*(i+1)))
i+=1
val = max(ans)
if val<0:
return -1
else:
return ans.index(val)+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
|
def wheel(cust, ticket, cost):
n_waiting = 0
earn = 0
rots = 0
res = [0, 0]
i = 0
l = len(cust)
while True:
if i < l:
n_waiting += cust[i]
i += 1
seated = min (n_waiting, 4)
if i == l and seated == 0:
break
n_waiting -= seated
earn += seated * ticket - cost
rots += 1
if (earn > res[1]):
res[1] = earn
res[0] = rots
if res[0] <= 0:
return -1
return res[0]
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
return wheel(customers, boardingCost, runningCost)
|
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 cal(k,p):
if k == 0:
p = p
else:
p += k*boardingCost - runningCost
return p
ans = 0
p = 0
w = 0
i = 0
itop = -1
for n in customers:
i+=1
w += n
k = min(4, w)
p = cal(k,p)
if ans<p:
ans = p
itop = i
w-=k
while w:
i+=1
k = min(4, w)
p = cal(k,p)
if ans<p:
ans = p
itop = i
w-=k
return itop
|
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:
y = 0
n = 0
m = -math.inf
res = -1
k = 0
for x in customers:
n += x
y += min(n,4)*boardingCost-runningCost
n-=min(n,4)
k+=1
if y>m:
m = y
res = k
while n:
y += min(n,4)*boardingCost-runningCost
n-=min(n,4)
k+=1
if y>m:
m = y
res = k
return res if m>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:
nwaits = 0
best_profit = 0
best_rotations = 0
profit = 0
for i, c in enumerate(itertools.chain(customers, itertools.repeat(0))):
nwaits += c
board = min(4, nwaits)
nwaits -= board
profit += (board * boardingCost - runningCost)
if profit > best_profit:
best_profit = profit
best_rotations = i + 1
if i >= len(customers) and nwaits == 0:
break
return best_rotations if best_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:
prof = rots = wait = 0
maxp = float('-inf')
maxr = -1 # result
for c in customers:
rots += 1
wait += c
ride = min(wait, 4)
wait -= ride
prof += ride * boardingCost - runningCost
if maxp < prof:
maxp = prof
maxr = rots
while wait > 0:
rots += 1
ride = min(wait, 4)
wait -= ride
prof += ride * boardingCost - runningCost
if maxp < prof:
maxp = prof
maxr = rots
return maxr if maxp > 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:
# 4 gondalas, each gondola can sit up to 4 people
# rotates counterclockwise, costs 'runningCost'
# customers[i] is the number of new customers arriving before the ith rotation
if 4*boardingCost <= runningCost: return -1
remain = 0
for i,c in enumerate(customers):
if c == 4 or (c < 4 and remain == 0):
continue
if c < 4:
if c + remain <= 4:
c += remain
remain = 0
customers[i] = c
else:
# get new remain
remain -= (4 - c)
customers[i] = 4
else:
# collect remain
remain += (c - 4)
customers[i] = 4
sofar = 0
hi = 0
hiInd = -1
for i,c in enumerate(customers):
sofar += (c * boardingCost - runningCost)
if sofar > 0 and sofar > hi:
hi = sofar
hiInd = i + 1
rounds = remain // 4
extra = remain % 4
sofar += rounds*(4*boardingCost - runningCost)
if sofar > 0 and sofar > hi:
hi = sofar
hiInd = len(customers) + rounds
sofar += (extra*boardingCost - runningCost)
if sofar > 0 and sofar > hi:
hiInd += 1
return hiInd
|
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):
if 4*boardingCost < runningCost:
return -1
n = len(customers)
maxtotal, maxrotations = 0, -1
total, rotations = 0, 1
i = 0
while i != n-1 or customers[n-1] != 0:
if customers[i]>4:
total += boardingCost*4 - runningCost
if i < n-1:
customers[i+1] += customers[i]-4
customers[i] = 0
else:
customers[i] -= 4
else:
total += customers[i]*boardingCost - runningCost
customers[i] = 0
if total>maxtotal:
maxtotal = total
maxrotations = rotations
if i<n-1: i += 1
rotations += 1
return maxrotations
|
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, c: List[int], b: int, r: int) -> int:
cnt = 0
p = 0
m = 0
mi = -1
ind = 0
for i,cc in enumerate(c):
cnt+=cc
p+=min(4,cnt)*b-r
if p > m:
mi = i+1
m = p
cnt = max(cnt-4, 0)
ind = i
while cnt:
ind+=1
p+=min(4,cnt)*b-r
if p > m:
mi = ind+1
m = p
cnt = max(cnt-4, 0)
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:
# keep track of maximum profit
#
q_len = current_profit = 0
max_profit = ans = -1
for rotation, new_people in enumerate(customers):
q_len += new_people
current_profit += boardingCost * min(4, q_len) - runningCost
if current_profit > max_profit:
max_profit = current_profit
ans = rotation + 1
q_len = max(0, q_len - 4)
rotation = len(customers)
while q_len > 0:
current_profit += boardingCost * min(4, q_len) - runningCost
if current_profit > max_profit:
max_profit = current_profit
ans = rotation + 1
q_len = max(0, q_len - 4)
rotation += 1
return ans
|
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