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Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
from collections import deque
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
l = len(status)
found = [False] * l
run_q = deque([])
wait_q = deque([])
tmp_q = deque([])
for box in initialBoxes:
if status[box] == 1:
run_q.append(box)
else:
wait_q.append(box)
res = 0
while run_q:
box = run_q.popleft()
res += candies[box]
for key in keys[box]:
status[key] = 1
for found_box in containedBoxes[box]:
wait_q.append(found_box)
while wait_q:
sleeping_box = wait_q.popleft()
if status[sleeping_box] == 1:
run_q.append(sleeping_box)
else:
tmp_q.append(sleeping_box)
while tmp_q:
wait_q.append(tmp_q.popleft())
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
boxes = set(initialBoxes)
bfs = [x for x in initialBoxes if status[x]]
for i in bfs:
for j in containedBoxes[i]:
boxes.add(j)
if status[j]:
bfs.append(j)
for k in keys[i]:
if status[k] == 0 and k in boxes:
bfs.append(k)
status[k] = 1
return sum([candies[i] for i in boxes if status[i]])
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
import queue
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
ans = 0
q = queue.Queue()
notopen = set()
op = set()
for x in initialBoxes:
q.put(x)
while not q.empty():
# print(q.queue)
cur = q.get()
if status[cur] == 1:
ans += candies[cur]
for k in keys[cur]:
status[k] = 1
if k in notopen and k not in op:
q.put(k)
op.add(k)
for c in containedBoxes[cur]:
q.put(c)
else:
notopen.add(cur)
return ans
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
key_set = set()
box_set = set()
used = set()
queue = collections.deque(initialBoxes)
ans = 0
while queue:
box = queue.popleft()
if box in used:
continue
if status[box] == 1 or box in key_set:
used.add(box)
if box in key_set:
key_set.remove(box)
if box in box_set:
box_set.remove(box)
ans += candies[box]
for futureBox in containedBoxes[box]:
box_set.add(futureBox)
for futureKey in keys[box]:
key_set.add(futureKey)
for futureBox in box_set:
if futureBox in key_set or status[futureBox] == 1:
queue.append(futureBox)
# for newBox in toBeOpenBox:
# queue.append(newBox)
# if newBox in key_set:
# key_set.remove(newBox)
# if newBox in box_set:
# box_set.remove(newBox)
else:
box_set.add(box)
return ans
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
n = len(status)
self.count = 0
queue = collections.deque([i for i in initialBoxes if status[i]==1])
boxSet = set(initialBoxes)
keySet = set(i for i in range(n) if status[i] == 1)
opened = set()
while (queue):
l = len(queue)
for i in range(l):
cur = queue.popleft()
self.count += candies[cur]
opened.add(cur)
for key in keys[cur]:
keySet.add(key)
if key not in opened and key in boxSet and key not in queue:
queue.append(key)
for box in containedBoxes[cur]:
boxSet.add(box)
if box not in opened and box in keySet and box not in queue:
queue.append(box)
return self.count
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
def dfs(i):
if i in visited:
return
if status[i] == 0:
missedKey.add(i)
return
visited.add(i)
self.ans += candies[i]
for k in keys[i]:
status[k] = 1
if k in missedKey:
missedKey.discard(k)
dfs(k)
for j in containedBoxes[i]:
dfs(j)
visited = set()
missedKey = set()
self.ans = 0
for i in initialBoxes:
dfs(i)
return self.ans
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
def dfs(i):
if i in visited: return
elif status[i] == 0:
missedKeys.add(i)
return
visited.add(i)
self.ans += candies[i]
for k in keys[i]:
status[k] = 1
if k in missedKeys:
missedKeys.discard(k)
dfs(k)
for j in containedBoxes[i]:
dfs(j)
self.ans = 0
visited= set()
missedKeys = set()
for i in initialBoxes:
dfs(i)
return self.ans
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
#The queue only contains boxes which is accessible
q=collections.deque()
keys_set=set() #Set for all available keys
boxes_set=set() #set for all un-opened boxes
#Initialize the queue to process box
for box in initialBoxes:
if(status[box]==1 or box in keys_set): #Check if this box is open or there is key available
q.append(box)
for key in keys[box]:
keys_set.add(key)
if(key in boxes_set): #Check if the new key can open a box in boxes_set
q.append(key)
boxes_set.remove(key)
else:
boxes_set.add(box)
res=0
visited=set()
#use DFS to populate the queue
while q:
boxi=q.popleft()
#Avoid duplicated couting of candies
if(boxi in visited):
continue
else:
visited.add(boxi)
res +=candies[boxi]
#Check all neighbors of current boxi
for nei in containedBoxes[boxi]:
if(status[nei]==1 or nei in keys_set): #Check if this box is open or there is a key for this box
q.append(nei)
for key in keys[nei]:
keys_set.add(key)
if(key in boxes_set): #check if the new key can open a box in boxes_set
q.append(key)
boxes_set.remove(key)
else:
boxes_set.add(nei)
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
from collections import deque
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
q = deque(initialBoxes)
ownedkeys = set()
allkz = set()
for k in keys:
for l in k:
allkz.add(l)
# for i, b in enumerate(status):
# if not b and b not in allkz:
# unreachable.add(b)
opened = set(i for i, b in enumerate(status) if not b and i not in allkz) # treat unreachable as already opened
res = 0
while q:
curBox = q.popleft()
if curBox not in opened:
if status[curBox] or curBox in ownedkeys:
opened.add(curBox)
for k in keys[curBox]:
ownedkeys.add(k)
res += candies[curBox]
q += containedBoxes[curBox]
else:
q.append(curBox)
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
locked_boxes = set()
unused_keys = set()
candy_count = 0
cur_box = []
for box in initialBoxes:
if status[box] or box in unused_keys:
candy_count += candies[box]
for i in keys[box]:
if not status[i]:
if i not in locked_boxes:
unused_keys.add(i)
else:
locked_boxes.remove(i)
cur_box.append(i)
for i in containedBoxes[box]:
if not status[i] and i not in unused_keys:
locked_boxes.add(i)
else:
cur_box.append(i)
if i in unused_keys:
unused_keys.remove(i)
if box in unused_keys:
unused_keys.remove(box)
else:
locked_boxes.add(box)
while cur_box:
box = cur_box.pop()
candy_count += candies[box]
for i in keys[box]:
if not status[i]:
if i not in locked_boxes:
unused_keys.add(i)
else:
cur_box.append(i)
for i in containedBoxes[box]:
if not status[i] and i not in unused_keys:
locked_boxes.add(i)
else:
cur_box.append(i)
if i in unused_keys:
unused_keys.remove(i)
if box in unused_keys:
unused_keys.remove(box)
return candy_count
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
from collections import deque
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
# Step 1: Traverse through intialBoxes using it sort of as our stack
# With each box, add candies to result, open every box we have a key for, add every contained box onto stack
# use array to keep track of seen boxes
# what to do about keys? could have box that isn't opened but will be opened later on
# possible soln: skip that box, come back to it later on, if we reach a point where every box on stack can't be opened we end
n = len(status)
seen, queue, result = [False] * n, deque(initialBoxes), 0
num_locked = 0
for box in initialBoxes:
seen[box] = True
while queue:
box = queue.popleft()
# Check if we can process box yet.
if not status[box]:
queue.append(box)
num_locked += 1
if num_locked == len(queue):
break
continue
num_locked = 0
result += candies[box]
# Open every box we have a key for.
for k in keys[box]:
status[k] = 1
# Add all contained boxes to queue.
for cb in containedBoxes[box]:
if not seen[cb]:
queue.append(cb)
seen[cb] = True
return result
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
if not initialBoxes: return 0
if 1 not in status: return 0
boxes = []
result = 0
while True:
flag = False
for b in initialBoxes:
if status[b] == 1:
flag = True
result += candies[b]
boxes += containedBoxes[b]
for j in keys[b]: status[j] = 1
initialBoxes.remove(b)
initialBoxes += boxes
boxes = []
if not flag: break
return result
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
# O(n^2) time, O(n) space
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
OPEN, CLOSED = 1, 0
q = deque(initialBoxes)
keys_obtained = set(i for i in range(len(status)) if status[i] == OPEN)
count = 0
while set(q) & keys_obtained:
q2 = deque()
while q:
box = q.popleft()
if box in keys_obtained:
count += candies[box]
keys_obtained |= set(keys[box])
q.extend(containedBoxes[box])
else:
q2.append(box)
q = q2
return count
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
reached = set()
visited = set()
q = initialBoxes
yummy = 0
while q:
new_q = []
for box in q:
reached.add(box)
if status[box] and box not in visited:
visited.add(box)
new_q.extend(containedBoxes[box])
new_q.extend(unlockedBox for unlockedBox in keys[box] if unlockedBox in reached)
yummy += candies[box]
for unlockedBox in keys[box]:
status[unlockedBox] = 1
q = new_q
return yummy
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
locked_boxes = set()
unused_keys = set()
candy_count = 0
cur_box = []
for box in initialBoxes:
if status[box] or box in unused_keys:
candy_count += candies[box]
for i in keys[box]:
if not status[i]:
if i not in locked_boxes:
unused_keys.add(i)
else:
locked_boxes.remove(i)
cur_box.append(i)
for i in containedBoxes[box]:
if not status[i] and i not in unused_keys:
locked_boxes.add(i)
else:
cur_box.append(i)
if i in unused_keys:
unused_keys.remove(i)
if box in unused_keys:
unused_keys.remove(box)
else:
locked_boxes.add(box)
while cur_box:
box = cur_box.pop()
candy_count += candies[box]
for i in keys[box]:
if not status[i]:
if i not in locked_boxes:
unused_keys.add(i)
else:
locked_boxes.remove(i)
cur_box.append(i)
for i in containedBoxes[box]:
if not status[i] and i not in unused_keys:
locked_boxes.add(i)
else:
cur_box.append(i)
if i in unused_keys:
unused_keys.remove(i)
if box in unused_keys:
unused_keys.remove(box)
return candy_count
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
q = collections.deque()
for box in initialBoxes:
q.append(box)
keys_found = set()
ans = 0
found = True
while q and found:
size = len(q)
found = False
for _ in range(size):
cur = q.popleft()
if status[cur] == 0 and cur not in keys_found:
q.append(cur)
else:
found = True
ans += candies[cur]
for box in containedBoxes[cur]:
q.append(box)
for key in keys[cur]:
keys_found.add(key)
return ans
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
OPEN, CLOSED = 1, 0
q = deque(initialBoxes)
keys_obtained = set(i for i in range(len(status)) if status[i] == OPEN)
count = 0
while set(q) & keys_obtained:
q2 = deque()
while q:
box = q.popleft()
if box in keys_obtained:
count += candies[box]
keys_obtained |= set(keys[box])
q.extend(containedBoxes[box])
else:
q2.append(box)
q = q2
return count
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
queue = []
visited = set()
closedBoxes = set()
foundedKeys = set()
for box in initialBoxes:
if status[box] == 1:
queue.append(box)
else:
closedBoxes.add(box)
res = 0
while queue:
cur = queue.pop(0)
res += candies[cur]
for box in containedBoxes[cur]:
if status[box] == 1:
queue.append(box)
elif (status[box] == 0 and box in foundedKeys):
queue.append(box)
foundedKeys.remove(box)
else:
closedBoxes.add(box)
for key in keys[cur]:
if key in closedBoxes:
queue.append(key)
closedBoxes.remove(key)
else:
foundedKeys.add(key)
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
boxes = set(initialBoxes)
openedBox = set()
bfs = [i for i in boxes if status[i]]
#BFS every open box
while bfs:
size = len(bfs)
for _ in range(size):
box = bfs.pop(0)
openedBox.add(box)
# Get all boxs in given box, push to queue if it's open
for eachContainedBox in containedBoxes[box]:
if status[eachContainedBox]:
bfs.append(eachContainedBox)
boxes.add(eachContainedBox)
# check all the keys in given box, push box to queue if there is key to the box
for eachKey in keys[box]:
if status[eachKey] == 0 and eachKey in boxes:
bfs.append(eachKey)
# it's necessary to update status, because above you need use status[eachKey] == 0 to filter out the box that you have alreay taken candies
status[eachKey] = 1
return sum(candies[box] for box in openedBox)
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
# def maxCandies(status,candies,keys,containedBoxes,initialBoxes):
# def dfs(i,status,candies,keys,containedBoxes):
q=[]
res=0
vis=set()
collected=set()
for i in initialBoxes:
q.append(i)
while q:
cur=q.pop(0)
vis.add(cur)
if status[cur]!=1 or cur in collected:
continue
res+=candies[cur]
collected.add(cur)
for i in keys[cur]:
status[i]=1
if i in vis:
q.append(i)
for j in containedBoxes[cur]:
q.append(j)
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
res, bcache, kcache = 0, set(initialBoxes), set()
for _ in range(1000):
tmp = set()
for b in bcache:
if status[b] == 1 or b in kcache:
tmp |= set(containedBoxes[b])
kcache |= set(keys[b])
res += candies[b]
else:
tmp.add(b)
bcache = tmp
# print(f'{bcache} {kcache}')
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
done = set()
to_open = set()
current_keys = set()
opened = list()
for box in initialBoxes:
if status[box] == 1:
opened.append(box)
else:
to_open.add(box)
total = 0
while opened:
box = opened.pop()
if box in done:
continue
total += candies[box]
done.add(box)
for openable in keys[box]:
if openable in to_open:
opened.append(openable)
to_open.remove(openable)
else:
current_keys.add(openable)
for contained in containedBoxes[box]:
if status[contained] == 1:
opened.append(contained)
elif contained in current_keys:
opened.append(contained)
current_keys.remove(contained)
else:
to_open.add(contained)
return total
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status, candies, keys, containedBoxes, initialBoxes):
ret=0
collected_keys=set()
opened=set()
queue=initialBoxes
while queue:
cont=False
for i in range(len(queue)):
box_index=queue.pop(0)
if status[box_index] or box_index in collected_keys:
if box_index not in opened:
cont=True
opened.add(box_index)
ret+=candies[box_index]
queue+=containedBoxes[box_index]
collected_keys.update(keys[box_index])
else:
queue.append(box_index)
if not cont: break
return ret
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
from collections import deque
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
q = deque(initialBoxes)
ownedkeys = set()
allkz = set()
for k in keys:
for l in k:
allkz.add(l)
# for i, b in enumerate(status):
# if not b and b not in allkz:
# unreachable.add(b)
opened = set()
for i, b in enumerate(status):
if not b and i not in allkz:
opened.add(i)
res = 0
while q:
curBox = q.popleft()
if curBox not in opened:
if status[curBox] or curBox in ownedkeys:
opened.add(curBox)
for k in keys[curBox]:
ownedkeys.add(k)
res += candies[curBox]
for z in containedBoxes[curBox]:
q.append(z)
else:
q.append(curBox)
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
if len(initialBoxes) == 0:
return 0
else:
boxdict = {}
candysum = 0
for initialbox in initialBoxes:
boxdict[initialbox] = [0,0]
p = len(status)**2
while p > 0:
for i in range(len(containedBoxes)):
if i in boxdict:
for containedbox in containedBoxes[i]:
boxdict[containedbox] = [0,0]
p -= 1
for j in range(len(keys)):
for boxedkeys in keys[j]:
if boxedkeys in boxdict:
boxdict[boxedkeys][1] = 1
for l in range(len(status)):
if status[l] == 1 and l in boxdict:
boxdict[l][0] = 1
for key in boxdict:
if (boxdict[key][0] == 1 and boxdict[key][1] == 1) or (boxdict[key][0] == 1 or boxdict[key][1] == 1):
candysum += candies[key]
return candysum
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
boxes = set(initialBoxes)
bfs = [i for i in initialBoxes if status[i]]
for i in bfs:
for j in containedBoxes[i]:
boxes.add(j)
if status[j]:
bfs.append(j)
for j in keys[i]:
if j in boxes and status[j]==0:
bfs.append(j)
status[j] = 1
return sum(candies[i] for i in bfs)
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
q = collections.deque()
n = len(status)
for i in initialBoxes:
q.append(i)
key = set()
res = 0
while q:
length = len(q)
hasOpen = False
for i in range(length):
box = q.popleft()
if status[box]==0 and box not in key:
q.append(box)
else:
hasOpen = True
res += candies[box]
for k in keys[box]:
key.add(k)
for x in containedBoxes[box]:
q.append(x)
if not hasOpen:
break
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
n = len(status)
self.count = 0
queue = collections.deque([i for i in initialBoxes if status[i]==1])
for i in queue:
status[i] = 2
boxSet = set(initialBoxes)
keySet = set(i for i in range(n) if status[i]==1)
while (queue):
l = len(queue)
for i in range(l):
cur = queue.popleft()
self.count += candies[cur]
for box in containedBoxes[cur]:
boxSet.add(box)
if box in keySet and status[box] != 2:
queue.append(box)
status[box] = 2
print(box)
for key in keys[cur]:
keySet.add(key)
if key in boxSet and status[key] != 2:
queue.append(key)
status[key] = 2
print(key)
return self.count
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
#Approach 1: use set
#The queue only contains boxes which is accessible
# q=collections.deque()
# keys_set=set() #Set for all available keys
# boxes_set=set() #set for all un-opened boxes
# #Initialize the queue to process box
# for box in initialBoxes:
# if(status[box]==1 or box in keys_set): #Check if this box is open or there is key available
# q.append(box)
# for key in keys[box]:
# keys_set.add(key)
# if(key in boxes_set): #Check if the new key can open a box in boxes_set
# q.append(key)
# boxes_set.remove(key)
# else:
# boxes_set.add(box)
# res=0
# visited=set()
# #use DFS to populate the queue
# while q:
# boxi=q.popleft()
# #Avoid duplicated couting of candies
# if(boxi in visited):
# continue
# else:
# visited.add(boxi)
# res +=candies[boxi]
# #Check all neighbors of current boxi
# for nei in containedBoxes[boxi]:
# if(status[nei]==1 or nei in keys_set): #Check if this box is open or there is a key for this box
# q.append(nei)
# for key in keys[nei]:
# keys_set.add(key)
# if(key in boxes_set): #check if the new key can open a box in boxes_set
# q.append(key)
# boxes_set.remove(key)
# else:
# boxes_set.add(nei)
# return res
#Approach 2: use list
haskeys=status[:]
seen=[0]*len(status)
q=collections.deque() #Only contains boxes that are accessible
for box in initialBoxes:
seen[box]=1
if(haskeys[box]):
q.append(box) #Include the one only accessible
res=0
while q:
boxi=q.popleft()
res+=candies[boxi]
#Since eachbox is contained in one box at most
#it will not result in duplicated elements in q
for nei in containedBoxes[boxi]:
seen[nei]=1
if(haskeys[nei]): q.append(nei)
#Iterate through all available keys,
#inlcude the box has been seen but doesn't have key to the queue
for key in keys[boxi]:
if(seen[key] and haskeys[key]==0): #Meaning it has not been included in queue
q.append(key)
haskeys[key]=1
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
maxCandy = 0
keyset = set()
numbox = len(status)
q = list(initialBoxes)
unopenedBOXES = []
while q:
for _ in range(len(q)):
thisbox = q.pop(0)
if status[thisbox] == 1 or thisbox in keyset:
maxCandy += candies[thisbox]
candies[thisbox] = 0
for key in keys[thisbox]:
keyset.add(key)
else:
unopenedBOXES.append(thisbox)
for x in containedBoxes[thisbox]:
q.append(x)
for unopenedBOX in unopenedBOXES:
if unopenedBOX in keyset:
maxCandy += candies[unopenedBOX]
return maxCandy
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
locked_boxes = set()
unused_keys = set()
candy_count = 0
cur_box = []
for box in initialBoxes:
if status[box]:
cur_box.append(box)
else:
locked_boxes.add(box)
while cur_box:
box = cur_box.pop()
candy_count += candies[box]
for i in keys[box]:
if not status[i]:
if i not in locked_boxes:
unused_keys.add(i)
else:
locked_boxes.remove(i)
cur_box.append(i)
for i in containedBoxes[box]:
if not status[i] and i not in unused_keys:
locked_boxes.add(i)
else:
cur_box.append(i)
if i in unused_keys:
unused_keys.remove(i)
if box in unused_keys:
unused_keys.remove(box)
return candy_count
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
# count the candies after open all possible boxes
boxes = set(initialBoxes)
# status as initial set of keys on hand
bfs = [i for i in boxes if status[i]]
for i in bfs:
# collect boxes
for j in containedBoxes[i]:
boxes.add(j)
if status[j]:
bfs.append(j)
# collect keys
for j in keys[i]:
# open a previously obtained box
if status[j] == 0 and j in boxes:
bfs.append(j)
# use the status to represent keys obtained
status[j] = 1
return sum(candies[i] for i in bfs)
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
locked_boxes = set()
unused_keys = set()
candy_count = 0
cur_box = []
for box in initialBoxes:
if status[box]:
cur_box.append(box)
else:
locked_boxes.add(box)
while cur_box:
box = cur_box.pop()
status[box] = 1
candy_count += candies[box]
for i in keys[box]:
if not status[i]:
if i not in locked_boxes:
unused_keys.add(i)
else:
locked_boxes.remove(i)
cur_box.append(i)
for i in containedBoxes[box]:
if not status[i] and i not in unused_keys:
locked_boxes.add(i)
else:
cur_box.append(i)
if i in unused_keys:
unused_keys.remove(i)
if box in unused_keys:
unused_keys.remove(box)
return candy_count
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
from collections import deque
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
cur_queue = deque(initialBoxes)
next_queue = deque([])
key_set = set([])
# opened_set = set([])
total = 0
while True:
temp_queue = cur_queue.copy()
while temp_queue:
c = temp_queue.popleft()
if status[c] == 1 or c in key_set:
# opened_set.add(c)
key_set = key_set.union(keys[c])
total += candies[c]
for s in containedBoxes[c]:
next_queue.append(s)
else:
next_queue.append(c)
# print(total, next_queue, cur_queue)
if cur_queue == next_queue:
break
cur_queue = next_queue
next_queue = deque([])
return total
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
res, bcache, kcache = 0, set(initialBoxes), set()
for _ in range(999):
tmp = set()
for b in bcache:
if status[b] == 1 or b in kcache:
tmp |= set(containedBoxes[b])
kcache |= set(keys[b])
res += candies[b]
else:
tmp.add(b)
bcache = tmp
# print(f'{bcache} {kcache}')
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
maximum = 0
cur = []
closeddic = dict()
size = 0
for i in initialBoxes:
if(status[i] == 1):
cur.append(i)
size+=1
else:
closeddic[i] = 1
consumed = dict()
keysdic = dict()
while(size>0):
nex = []
nsize = 0
for i in cur:
maximum+=candies[i]
consumed[i] = True
for j in containedBoxes[i]:
if((status[j] == 1 or j in keysdic) and j not in consumed):
nex.append(j)
nsize+=1
elif(status[j] == 0):
closeddic[j] = 1
for j in keys[i]:
keysdic[j] = 1
if(j in closeddic and j not in consumed):
del closeddic[j]
nex.append(j)
nsize+=1
size = nsize
cur = nex
return maximum
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
keyHold = set([idx for idx, b in enumerate(status) if b == 1])
boxHold = set(initialBoxes)
opened = set()
while True:
flag = False
newKey = set()
newBox = set()
for b in boxHold:
if b in keyHold:
opened.add(b)
flag = True
for k in keys[b]:
newKey.add(k)
for nb in containedBoxes[b]:
if nb not in opened:
newBox.add(nb)
if not flag:
break
for k in newKey:
keyHold.add(k)
for b in newBox:
boxHold.add(b)
for b in opened:
boxHold.discard(b)
return sum([c for idx, c in enumerate(candies) if idx in opened])
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
open =initialBoxes
x_keys = set()
locked = set()
result = 0
seen = set()
while (len(open)!=0):
i= open.pop(0)
if i in seen:
continue
for key in keys[i]:
x_keys.add(key)
for box in containedBoxes[i]:
if box in x_keys or status[box]==1:
open.append(box)
else:
locked.add(box)
for box in locked:
if box in x_keys:
open.append(box)
result+=candies[i]
seen.add(i)
return result
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
# topo sort
parent = collections.defaultdict(int)
for i,row in enumerate(containedBoxes):
for j in row:
parent[j] = i
n = len(status)
bfs = collections.deque([(i,False) for i in initialBoxes])
res = 0
seen = set()
while bfs:
#print(bfs)
box,st = bfs.popleft()
if box in seen:
continue
if status[box]==0:
if st:
return res
bfs.append((box,True))
continue
seen.add(box)
res+=candies[box]
for key in keys[box]:
status[key] = 1
for cbox in containedBoxes[box]:
bfs.append((cbox,False))
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
q=initialBoxes
has_key=[0]*len(status)
has_open=[0]*len(status)
res=0
while q:
qq=[]
for s in q:
if status[s] or has_key[s]:
res+=candies[s]
has_open[s]=1
for b in containedBoxes[s]:
if has_open[b]: continue
qq.append(b)
for k in keys[s]:
has_key[k]=1
else:
qq.append(s)
if tuple(qq)==tuple(q): break
q,qq=qq,[]
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
from collections import deque
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
q = deque(initialBoxes)
visited = set()
res = 0
while q:
itr, opened = len(q), False # To detect cycle
opened = False
while (itr):
itr -= 1
v = q.popleft()
if status[v]: # Open box, (key is available or is open)
opened = True
res += candies[v]
visited.add(v)
for x in keys[v]:
status[x] = 1
for x in containedBoxes[v]:
if x not in visited:
q.append(x)
elif v not in visited: # Open when key is available
q.append(v)
if not opened:
return res # Exit cycle detected
return res
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
if not initialBoxes: return 0
if 1 not in status: return 0
boxes = []
result = 0
while True:
flag = False
for b in initialBoxes:
if status[b] == 1:
flag = True
result += candies[b]
boxes.extend(containedBoxes[b])
for j in keys[b]: status[j] = 1
initialBoxes.remove(b)
initialBoxes += boxes
boxes = []
if not flag: break
return result
|
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length
|
class Solution:
def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int:
# for initial boxes:
# - accumulate candies
# - for each key, pre-unlock them
# - add containedBoxes to queue
#
# for all contained boxes that's open:
# - accumulate candies
# - for each key, pre-unlock them
# ...
#
# keep track of a list of opened boxes to explore, and a list of closed boxes to explore
# - whenever open a box, pre-unlock their keys
# - if box in list of closed Q, move to open
# - whenever find an enclosed box, put to either open or closed
# - continue until we have no more open boxes
open_boxes = set()
closed_boxes = set()
# TODO: may need an explored flag for each box
for b in initialBoxes:
if status[b]:
open_boxes.add(b)
else:
closed_boxes.add(b)
total_candies = 0
while len(open_boxes) > 0:
b = open_boxes.pop()
total_candies += candies[b]
for key in keys[b]:
status[key] = 1
if key in closed_boxes:
closed_boxes.remove(key)
open_boxes.add(key)
for bb in containedBoxes[b]:
if status[bb]:
open_boxes.add(bb)
else:
closed_boxes.add(bb)
return total_candies
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
dp = {0: 0}
def racecar(self, target: int) -> int:
if target in self.dp:
return self.dp[target]
n = target.bit_length()
if 2**n - 1 == target:
self.dp[target] = n
else:
self.dp[target] = self.racecar(2**n - 1 - target) + n + 1
for m in range(n - 1):
self.dp[target] = min(self.dp[target], self.racecar(target - 2**(n - 1) + 2**m) + n + m + 1)
return self.dp[target]
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
dist = [float('inf')] * (target + 1)
dist[0] = 0 # the minimum number of steps to achieve i with starting speed being 1
#dist[1] = 1
#dist[2] = 4
#dist[3] = 2
for t in range(1, target+1):
k = t.bit_length()
up_limit = 2**k-1
if t == up_limit: # right on the target by only move forward with no reverse
dist[t] = k
# between two critical points (k-1 steps and k steps)
# the last move is forward move
# in this case there should be even number of turns, which results in odd number of consecutive acceleration period. To recurse into the original problem (starting speed is 1 and point to the target) we need to remove the first two turns (or last two turns)
for j in range(1, k):
for q in range(j):
dist[t] = min(dist[t], dist[t-2**j + 2**q]+j+1+q+1) # the last move is forward
dist[t] = min(dist[t], dist[2**k-1-t]+k+1) # the last move is backward
return dist[target]
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = deque([(0, 1, 0)])
encountered = set()
while q:
pos, spd, length = q.popleft()
if (pos, spd) in encountered: continue
else: encountered.add((pos, spd))
if pos == target:
return length
if (pos + spd, spd * 2) not in encountered:
if abs(pos + spd) <= 2 * target + 1:
q.append((pos + spd, spd * 2, length + 1))
revSpd = -1 if spd > 0 else 1
if (pos, revSpd) not in encountered:
q.append((pos, revSpd, length + 1))
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
from collections import deque
class Solution:
def racecar(self, target: int) -> int:
steps = 0
queue = deque([(0, 1)])
cache = {(0,1)}
while queue:
for _ in range(len(queue)):
p, s = queue.popleft()
if p == target:
return steps
p1,s1 = p+s, s*2
p2 = p
s2 = -1 if s > 0 else 1
if -20000 <= p1 <= 20000 and (p1,s1) not in cache:
cache.add((p1,s1))
queue.append((p1,s1))
if -20000 <= p2 <= 20000 and (p2,s2) not in cache:
cache.add((p2,s2))
queue.append((p2,s2))
steps += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
nodes = [(0, 1)]
seen, length = set([(0, 1), (0, -1)]), 0
while nodes:
new_nodes = []
for pos, speed in nodes:
if pos == target:
return length
# A
for new_pos, new_speed in [
(pos + speed, speed * 2),
(pos, -1 if speed > 0 else 1),
]:
if abs(new_speed) > 2 * target or abs(new_pos) > 2 * target:
continue
if (new_pos, new_speed) not in seen:
seen.add((new_pos, new_speed))
new_nodes.append((new_pos, new_speed))
length += 1
nodes = new_nodes
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = deque([(0, 1, 0)])
encountered = set()
while q:
pos, spd, length = q.popleft()
if (pos, spd) in encountered: continue
else: encountered.add((pos, spd))
if pos == target:
return length
if (pos + spd, spd * 2) not in encountered:
if abs(pos + spd) <= 2 * target + 1 and abs(spd) <= 2 * target:
q.append((pos + spd, spd * 2, length + 1))
revSpd = -1 if spd > 0 else 1
if (pos, revSpd) not in encountered:
q.append((pos, revSpd, length + 1))
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution1:
def __init__(self):
self.dp = {0: 0}
def racecar(self, target: int) -> int:
if target in self.dp:
return self.dp[target]
n = target.bit_length()
if 2 ** n - 1 == target:
self.dp[target] = n
else:
self.dp[target] = self.racecar(2 ** n - 1 - target) + n + 1
for m in range(n - 1):
self.dp[target] = min(self.dp[target],
self.racecar(target - 2 ** (n - 1) + 2 ** m) + n + m + 1)
return self.dp[target]
class Solution2:
def racecar(self, target: int) -> int:
# #analysis + DP
memo = [-1] * (target * 2)
def DP(s):
if memo[s] > 0: return memo[s]
duishu = math.log2(s+1)
if int(duishu) == duishu:
memo[s] = duishu
return duishu
left = int(duishu)
right = int(duishu) + 1
memo[s] = right + 1 + DP(2**right - s - 1)
for i in range(left):
huizou = 2**i - 1
memo[s] = min(memo[s], left + i + 2 + DP(s - 2**left + 1 + huizou))
return memo[s]
return int(DP(target))
class Solution:
def racecar(self, target: int) -> int:
#classic BFS
visited = {(0,1)}
queue = collections.deque()
queue.append((0,1))
steps = 0
while queue:
size = len(queue)
for _ in range(size):
s, v = queue.popleft()
if s == target: return steps
# 'A'
cur_s = s + v
cur_v = v * 2
if (cur_s, cur_v) not in visited and s >= 0:
queue.append((cur_s, cur_v))
visited.add((cur_s, cur_v))
#'R'
cur_s = s
cur_v = -1 if v > 0 else 1
if (cur_s, cur_v) not in visited and s >= 0:
queue.append((cur_s, cur_v))
visited.add((cur_s, cur_v))
steps += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target):
q, cnt, used = [(0, 1)], 0, {(0, 1)}
while q:
new = []
for pos, speed in q:
if pos == target:
return cnt
elif pos > 20000 or -20000 > pos:
continue
if (pos + speed, speed * 2) not in used:
new.append((pos + speed, speed * 2))
used.add((pos + speed, speed * 2))
if speed > 0 and (pos, -1) not in used:
new.append((pos, -1))
used.add((pos, -1))
elif speed < 0 and (pos, 1) not in used:
new.append((pos, 1))
used.add((pos, 1))
q = new
cnt += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
visited = {(0,1)}
queue = collections.deque()
queue.append((0,1))
steps = 0
while queue:
size = len(queue)
for _ in range(size):
s, v = queue.popleft()
if s == target: return steps
# 'A'
cur_s = s + v
cur_v = v * 2
if (cur_s, cur_v) not in visited and s >= 0:
queue.append((cur_s, cur_v))
visited.add((cur_s, cur_v))
#'R'
cur_s = s
cur_v = -1 if v > 0 else 1
if (cur_s, cur_v) not in visited and s >= 0:
queue.append((cur_s, cur_v))
visited.add((cur_s, cur_v))
steps += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = [(0,1)]
vis = set([(0,1)])
ans=0
while q:
new_q = []
for p,s in q:
if p==target:
return ans
p1,s1 = p+s,2*s
p2,s2 = p, -1 if s>0 else 1
# if (p1,s1) not in vis:
# vis.add((p1,s1))
new_q.append((p1,s1))
if (p2,s2) not in vis:
vis.add((p2,s2))
new_q.append((p2,s2))
q = new_q
ans+=1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target):
K = target.bit_length() + 1
barrier = 2**K
pq = [(0, target)]
dist = [float('inf')] * (2 * barrier + 1)
dist[target] = 0
while pq:
steps, targ = heapq.heappop(pq)
if dist[targ] > steps: continue
for k in range(K+1):
walk = (2**k) - 1
steps2, targ2 = steps + k + 1, walk - targ
if walk == targ: steps2 -= 1 #No \"R\" command if already exact
if abs(targ2) <= barrier and steps2 < dist[targ2]:
heapq.heappush(pq, (steps2, targ2))
dist[targ2] = steps2
return dist[0]
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = [(0,1)]
vis = {(0,1)}
ans=0
while q:
new_q = []
for p,s in q:
if p==target:
return ans
p1,s1 = p+s,2*s
p2,s2 = p, -1 if s>0 else 1
# if (p1,s1) not in vis:
# vis.add((p1,s1))
new_q.append((p1,s1))
if (p2,s2) not in vis:
vis.add((p2,s2))
new_q.append((p2,s2))
q = new_q
ans+=1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
queue = deque([(0, 1, 0)])
seen = {(0, 1)}
while queue:
pos, speed, cost = queue.popleft()
if pos == target:
return cost
# A
new_pos, new_speed = pos + speed, speed * 2
if new_pos > 0 and (new_pos, new_speed) not in seen:
seen.add((new_pos, new_speed))
queue.append((new_pos, new_speed, cost + 1)) # A
# R
reverse_speed = -1 if speed > 0 else 1
if (pos, reverse_speed) not in seen:
seen.add((pos, reverse_speed))
queue.append((pos, reverse_speed, cost + 1))
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar1(self, target: int) -> int:
# bfs 能够快速想起来的方法！剪枝的策略不理解！
q = collections.deque()
q.append([0, 1])
visited = set()
visited.add('0_1')
visited.add('0_-1')
steps = 0
while q:
qsize = len(q)
for i in range(qsize):
pos, speed = q.popleft()
pos1 = pos + speed
speed1 = speed * 2 # 加速！注意第一个可能性的构造方法：首先是新的pos1，然后是新的速度！
if pos1 == target:
return steps + 1
p1 = [pos1, speed1]
#key1 = '{}_{}'.format(pos1, speed1)
if abs(speed1) < 2*target and abs(pos1) < 2*target and not key1 in visited: #相当于说，如果>= 2*target的时候，就剪枝了！
q.append(p1) #如果 < ??? # 这里没有把p1加入visited???
#visited.add(key1) # this makes things slower... yet still can pass!
speed2 = -1 if speed > 0 else 1 # 掉头！注意是维持pos位置不变，速度如果原来是》0，则变成-1，如果是《0则变成1！！
p2 = [pos, speed2]
key2 = '{}_{}'.format(pos, speed2)
if key2 not in visited:
q.append(p2)
visited.add(key2)
steps += 1
return -1
def racecar(self, target):
k = target.bit_length() + 1
barrier = 1<<k
pq = [(0, target)]
dist = [float('inf')] * (2 * barrier + 1)
dist[target] = 0
while pq:
steps, targ = heapq.heappop(pq)
if dist[targ] > steps: continue
for k2 in range(k+1):
walk = (1<<k2) - 1
steps2, targ2 = steps + k2 + 1, walk - targ
if walk == targ:
steps2 -= 1
if abs(targ2) <= barrier and steps2 < dist[targ2]:
heapq.heappush(pq, (steps2, targ2))
dist[targ2] = steps2
return dist[0]
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
queue = collections.deque([(1, 0)])
visited = set(queue)
count = 0
while queue:
for _ in range(len(queue)):
speed, pos = queue.popleft()
if pos < -sqrt(target):
continue
if target == pos:
return count
if (speed*2, pos+speed) not in visited:
visited.add((speed*2, pos+speed))
queue.append((speed*2, pos+speed))
if speed > 0:
if (-1, pos) not in visited:
visited.add((-1, pos))
queue.append((-1, pos))
else:
if (1, pos) not in visited:
visited.add((1, pos))
queue.append((1, pos))
count += 1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
layer = collections.deque([(0,1)])
visited = set()
ans = 0
while len(layer) > 0:
nxt_layer = collections.deque()
while len(layer) > 0:
p, s = layer.popleft()
if p > target*2 and s > target*2:
continue
if p == target:
return ans
nxt_a = (p+s, 2*s)
if nxt_a not in visited:
nxt_layer.append(nxt_a)
visited.add(nxt_a)
nxt_r = (p, -1 if s > 0 else 1)
if nxt_r not in visited:
nxt_layer.append(nxt_r)
visited.add(nxt_r)
layer= nxt_layer
ans+=1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
if target == 0:
return 0
pq = [(0, 0, -1)] # steps, pos, speed
visited = set([(0, -1)])
while pq:
steps, pos, speed = heapq.heappop(pq)
next_steps, next_pos, next_speed = steps + 1, pos + speed, speed * 2
if -next_pos == target:
return next_steps
if -2 * target < -next_pos < 2 * target and -2 * target < -next_speed < 2 * target and not (next_pos, next_speed) in visited:
# if -next_pos < 2 * target and -next_speed < 2 * target and not (next_pos, next_speed) in visited:
heapq.heappush(pq, (next_steps, next_pos, next_speed))
visited.add((next_pos, next_speed))
if speed > 0 and not (pos, -1) in visited:
heapq.heappush(pq, (next_steps, pos, -1))
visited.add((pos, -1))
elif speed < 0 and not (pos, 1) in visited:
heapq.heappush(pq, (next_steps, pos, 1))
visited.add((pos, 1))
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
queue = collections.deque([(1, 0)])
visited = set(queue)
count = 0
while queue:
for _ in range(len(queue)):
speed, pos = queue.popleft()
if pos < -(target/2):
continue
if target == pos:
return count
if (speed*2, pos+speed) not in visited:
visited.add((speed*2, pos+speed))
queue.append((speed*2, pos+speed))
if speed > 0:
if (-1, pos) not in visited:
visited.add((-1, pos))
queue.append((-1, pos))
else:
if (1, pos) not in visited:
visited.add((1, pos))
queue.append((1, pos))
count += 1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
if target == 0:
return 1
self.ans = 0
self.bfsHelper(target)
return self.ans
def bfsHelper(self, target):
queue = []
queue.append((0, 1))
visited = set()
key = 'pos_' + str(0) + 'speed_' + str(1)
visited.add(key)
while len(queue) > 0:
size = len(queue)
for i in range(size):
curPos = queue[0][0]
curSpeed = queue[0][1]
queue.pop(0)
if curPos == target:
return self.ans
newPos = curPos + curSpeed
newSpeed = curSpeed * 2
key = 'pos_' + str(newPos) + 'speed_' + str(newSpeed)
if key not in visited and newPos > 0 and newPos < 2 * target:
visited.add(key)
queue.append((newPos, newSpeed))
newPos = curPos
newSpeed = -1 if curSpeed > 0 else 1
key = 'pos_' + str(newPos) + 'speed_' + str(newSpeed)
if key not in visited and newPos > 0 and newPos < 2 * target:
visited.add(key)
queue.append((newPos, newSpeed))
self.ans += 1
self.ans = -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
import bisect
class Solution:
def racecar(self, target: int) -> int:
speed = 1
dis = [0]
while dis[-1] < target:
dis.append(dis[-1] + speed)
speed *= 2
num_moves = self.recur(target, dis, {})
return num_moves
def recur(self, target: int, dis: list, memo: dict) -> int:
if target in memo:
return memo[target]
if target in dis:
return dis.index(target)
index = bisect.bisect_left(dis, target) - 1
moves_right = self.recur(dis[index + 1] - target, dis, memo) + 2 + index
moves_left = self.recur(target - dis[index], dis, memo) + 2 + index
for i in range(1, index):
new_moves_left = index + 1 + i + 1 + self.recur(target - dis[index] + dis[i], dis, memo)
moves_left = min(moves_left, new_moves_left)
memo[target] = min(moves_right, moves_left)
return min(moves_right, moves_left)
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = deque([(0, 1, 0)])
encountered = set()
while q:
pos, spd, length = q.popleft()
if (pos, spd) in encountered: continue
else: encountered.add((pos, spd))
if pos == target:
return length
if (pos + spd, spd * 2) not in encountered:
if pos + spd <= 2 * target + 1:
q.append((pos + spd, spd * 2, length + 1))
revSpd = -1 if spd > 0 else 1
if (pos, revSpd) not in encountered:
q.append((pos, revSpd, length + 1))
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
record = collections.defaultdict(int)
cur = (0, 1) #(position, speed)
queue = collections.deque()
queue.append(cur)
record[cur] = 0
while True:
for _ in range(len(queue)):
pos, speed = queue.popleft()
step = record[(pos, speed)]
if pos == target:
return step
if pos + speed == target:
return 1 + step
s1 = (pos + speed, speed * 2)
s2 = (pos, -1) if speed>0 else (pos, 1)
if s1 not in record:
queue.append(s1)
record[s1] = step + 1
if s2 not in record:
queue.append(s2)
record[s2] = step + 1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = [(0, 1)]
steps = 0
visited = set(q)
while q:
nxt = []
for pos, speed in q:
if pos + speed == target:
return steps + 1
nxt_pos = pos + speed
nxt_speed = speed * 2
if (nxt_pos, nxt_speed) not in visited:
visited.add((nxt_pos, nxt_speed))
nxt.append((nxt_pos, nxt_speed))
nxt_pos = pos
nxt_speed = -1 if speed > 0 else 1
if (nxt_pos, nxt_speed) not in visited:
visited.add((nxt_pos, nxt_speed))
nxt.append((nxt_pos, nxt_speed))
q = nxt
steps += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
if target == 0:
return 0
# BFS solution
level = 0
q = deque([(0, 1)]) # tuple of (pos, speed)
visited = {(0, 1),}
while q:
length = len(q)
for _ in range(length):
p, s = q.popleft()
# use A
new_p, new_s = p + s, 2*s
if new_p == target:
return level + 1
if (new_p, new_s) not in visited:
visited.add((new_p, new_s))
q.append((new_p, new_s))
# use R
new_p, new_s = p, -s//abs(s)
if (new_p, new_s) not in visited:
visited.add((new_p, new_s))
q.append((new_p, new_s))
level += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
import collections
class Solution:
def racecar(self, target: int) -> int:
# bfs
q = collections.deque([(0, 0, 1)]) # move, pos, vel
while q:
move, pos, vel = q.popleft()
if pos == target:
return move
# A
q.append((move+1, pos+vel, vel*2))
# R, considered in 4 cases, get rid of excessive search
if (pos > target and vel > 0) or (pos < target and vel < 0) or (pos + vel > target and vel > 0) or (pos + vel < target and vel < 0): #reverse
q.append((move+1, pos, -1 if vel > 0 else 1))
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
from collections import deque
class Solution:
def racecar(self, target: int) -> int:
if target == 0:
return 0
visited = set()
queue = deque()
queue.append((0, 1))
visited.add((0, 1))
step = 1
while queue:
for _ in range(len(queue)):
cur_pos, cur_speed = queue.popleft()
next_pos, next_speed = cur_pos + cur_speed, cur_speed * 2
if next_pos == target:
return step
if (next_pos, next_speed) not in visited:
queue.append((next_pos, next_speed))
visited.add((next_pos, next_speed))
if cur_speed > 0:
next_speed = -1
else:
next_speed = 1
if (cur_pos, next_speed) not in visited:
queue.append((cur_pos, next_speed))
visited.add((cur_pos, next_speed))
step += 1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
front = [(0,1)]
visited = {(0,1):0}
while front!=[]:
nf = [];
for (pos,speed) in front:
aa = [(pos+speed, speed*2),(pos, -1)] if speed>0 else [(pos+speed, speed*2),(pos, 1)]
for v in aa:
if not v in visited:
visited[v] = visited[(pos,speed)]+1
nf.append(v)
if v[0] == target:
return visited[v]
front = nf
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = [(0, 1)]
steps = 0
visited = set(q)
while q:
nxt = []
for pos, speed in q:
if pos + speed == target:
return steps + 1
if (pos+speed, speed*2) not in visited:
visited.add((pos+speed, speed*2))
nxt.append((pos+speed, speed*2))
nxt_speed = -1 if speed > 0 else 1
if (pos, nxt_speed) not in visited:
visited.add((pos, nxt_speed))
nxt.append((pos, nxt_speed))
q = nxt
steps += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
queue = deque([(0, 1)])
dists = {(0, 1): 0}
while queue:
x, v = queue.popleft()
d = dists[(x, v)]
if x+v == target:
return d+1
y, w = x+v, v << 1
if (y, w) not in dists:
dists[(y, w)] = d+1
queue.append((y, w))
opv = -1 if v > 0 else 1
if (x, opv) not in dists:
dists[(x, opv)] = d+1
queue.append((x, opv))
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def __init__(self):
self.dp = {0: 0}
def racecar(self, t: int) -> int:
if t in self.dp:
return self.dp[t]
n = t.bit_length()
if 2**n - 1 == t:
self.dp[t] = n
else:
# go pass via n A's, then turn back
self.dp[t] = self.racecar(2**n - 1 - t) + n + 1
# stop after (n - 1) A's, turn back, go m A's, then turn back
for m in range(n - 1):
self.dp[t] = min(self.dp[t], self.racecar(t - 2**(n - 1) + 2**m) + n + m + 1)
return self.dp[t]
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target):
target *= -1
q, used = [(0, 0, -1)], {(0, -1)}
while q:
cnt, pos, speed = heapq.heappop(q)
if pos == target:
return cnt
elif pos > 20000 or -20000 > pos:
continue
if (pos + speed, speed * 2) not in used:
heapq.heappush(q, (cnt + 1, pos + speed, speed * 2))
used.add((pos + speed, speed * 2))
if speed < 0 and (pos, 1) not in used:
heapq.heappush(q, (cnt + 1, pos, 1))
used.add((pos, 1))
elif speed > 0 and (pos, -1) not in used:
heapq.heappush(q, (cnt + 1, pos, -1))
used.add((pos, -1))
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
from collections import deque
class Solution:
def racecar(self, target: int) -> int:
steps = 0
queue = deque([(0, 1)])
cache = {(0,1)}
while queue:
for _ in range(len(queue)):
p, s = queue.popleft()
if p == target:
return steps
p1,s1 = p+s, s*2
p2 = p
s2 = -1 if s > 0 else 1
if (p1,s1) not in cache:
cache.add((p1,s1))
queue.append((p1,s1))
if (p2,s2) not in cache:
cache.add((p2,s2))
queue.append((p2,s2))
steps += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
dq = deque([(0, 1)])
res = 0
seen = {(0,1)}
while dq:
length = len(dq)
for _ in range(length):
p, s = dq.popleft()
rs = -1 if s>0 else 1
for np, ns in [[p+s, 2*s], [p, rs]]:
if np==target:
return res+1
if (np, ns) not in seen:
seen.add((np, ns))
dq.append((np, ns))
res += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
visited = {(0,1)}
queue = collections.deque()
queue.append((0,1))
steps = 0
while queue:
size = len(queue)
for _ in range(size):
s, v = queue.popleft()
if s == target: return steps
# 'A'
cur_s = s + v
cur_v = v * 2
if (cur_s, cur_v) not in visited:
queue.append((cur_s, cur_v))
visited.add((cur_s, cur_v))
#'R'
cur_s = s
cur_v = -1 if v > 0 else 1
if (cur_s, cur_v) not in visited:
queue.append((cur_s, cur_v))
visited.add((cur_s, cur_v))
steps += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = collections.deque()
q.append((0, 1))
visited = set()
visited.add((0, 1))
step = 0
while q:
size = len(q)
for _ in range(size):
pos, speed = q.popleft()
if pos == target:
return step
# 'A'
if (pos + speed, speed * 2) not in visited:
visited.add((pos + speed, speed * 2))
q.append((pos + speed, speed * 2))
# 'R'
if (pos > target and speed > 0) or (pos < target and speed < 0) or (pos + speed > target and speed > 0) or (pos + speed < target and speed < 0):
speed = -1 if speed > 0 else 1
if (pos, speed) not in visited:
visited.add((pos, speed))
q.append((pos, speed))
step += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = deque([(0, 1)])
visited = set([(0, 1)])
step = 0
while q:
l = len(q)
step += 1
for _ in range(l):
pos, speed = q.popleft()
if pos == target:
return step - 1
new_pos, new_speed = pos + speed, speed * 2
if (new_pos, new_speed) not in visited:
q.append((new_pos, new_speed))
visited.add((new_pos, new_speed))
new_pos, new_speed = pos, -1 if speed > 0 else 1
if (new_pos, new_speed) not in visited:
q.append((new_pos, new_speed))
visited.add((new_pos, new_speed))
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = [(0,1)]
vis = {(0,1)}
ans=0
while q:
new_q = []
for p,s in q:
if p==target:
return ans
p1,s1 = p+s,2*s
p2,s2 = p, -1 if s>0 else 1
if (p1,s1) not in vis:
vis.add((p1,s1))
new_q.append((p1,s1))
if (p2,s2) not in vis:
vis.add((p2,s2))
new_q.append((p2,s2))
q = new_q
ans+=1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = [(0,1)]
vis = set([(0,1)])
ans=0
while q:
new_q = []
for p,s in q:
if p==target:
return ans
p1,s1 = p+s,2*s
p2,s2 = p, -1 if s>0 else 1
if (p1,s1) not in vis:
vis.add((p1,s1))
new_q.append((p1,s1))
if (p2,s2) not in vis:
vis.add((p2,s2))
new_q.append((p2,s2))
q = new_q
ans+=1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
from collections import deque
class Solution:
def racecar(self, target: int) -> int:
q = deque([(0, 1, 0)])
visited = set([(0, 1)])
while q:
pos, speed, steps = q.popleft()
if pos == target:
return steps
nxt_pos = pos + speed
nxt_speed = speed * 2
if (nxt_pos, nxt_speed) not in visited:
visited.add((nxt_pos, nxt_speed))
q.append((nxt_pos, nxt_speed, steps+1))
nxt_pos = pos
nxt_speed = -1 if speed > 0 else 1
if (nxt_pos, nxt_speed) not in visited:
visited.add((nxt_pos, nxt_speed))
q.append((nxt_pos, nxt_speed, steps+1))
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
# BFS + trim
if target==0: return 0
q = {(0,1)} # (pos, speed)
visited= {(0,1)}
steps = -1
while q:
steps += 1
next_q = set()
for pos, speed in q:
if pos==target: return steps
# 1. If A
state_A = (pos+speed, speed*2)
if state_A not in visited:
visited.add(state_A)
next_q.add(state_A)
# 2. If R
state_R = (pos, -1 if speed>0 else 1)
if state_R not in visited:
visited.add(state_R)
next_q.add(state_R)
q = next_q
return steps
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
stack = collections.deque([(0, 1, 0)])
seen = set([(0, 1)])
while stack:
pos, speed, steps = stack.popleft()
if pos == target:
return steps
new = (pos + speed, speed * 2)
if new not in seen:
seen.add(new)
stack.append((new[0], new[1], steps + 1))
new = (pos, -1 if speed > 0 else 1)
if new not in seen:
seen.add(new)
stack.append((new[0], new[1], steps + 1))
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
import numpy as np
class Solution:
def racecar(self, target: int) -> int:
def dp(t):
if ans[t] > 0:
return ans[t]
n = int(np.ceil(np.log2(t + 1)))
if 1 << n == t + 1:
ans[t] = n
return ans[t]
ans[t] = n + 1 + dp((1 << n) - 1 - t)
for m in range(n):
cur = (1 << (n - 1)) - (1 << m)
ans[t] = min(ans[t], n + m + 1 + dp(t - cur))
return ans[t]
ans = [0 for _ in range(target + 1)]
return dp(target)
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, T: int, depth = 0) -> int:
q = collections.deque([[ 0, 1 ]])
seen = set('0,1')
while True:
k = len(q)
while k:
[ pos, vel ] = q.popleft()
if pos == T:
return depth # \\U0001f3af target T found
cand = []
if abs(T - (pos + vel) < T):
cand.append([ pos + vel, 2 * vel ])
cand.append([ pos, 1 if vel < 0 else -1 ])
for pos, vel in cand:
if f'{pos},{vel}' not in seen:
q.append([ pos, vel ])
seen.add(f'{pos},{vel}')
k -= 1
depth += 1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
from collections import deque
class Solution:
def racecar(self, target: int) -> int:
dic = {}
qu = deque([(0,1)])
dic[(0, 1)] = 0
while qu:
front = qu.popleft()
pos, spd = front
if pos == target:
return dic[front]
speedup = (pos + spd, spd*2)
if speedup not in dic:
qu.append(speedup)
dic[speedup] = dic[front] + 1
fwd2bwd = (pos, -1)
if spd > 0 and fwd2bwd not in dic:
qu.append(fwd2bwd)
dic[fwd2bwd] = dic[front] + 1
bwd2fwd = (pos, 1)
if spd < 0 and (pos, 1) not in dic:
qu.append((pos, 1))
dic[bwd2fwd] = dic[front] + 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = collections.deque()
q.append((0, 1))
visited = set()
visited.add((0, 1))
step = 0
while q:
size = len(q)
for _ in range(size):
pos, speed = q.popleft()
if pos == target:
return step
# 'A'
if (pos + speed, speed * 2) not in visited:
visited.add((pos + speed, speed * 2))
q.append((pos + speed, speed * 2))
# 'R'
if speed > 0:
speed = -1
else:
speed = 1
if (pos, speed) not in visited:
visited.add((pos, speed))
q.append((pos, speed))
step += 1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, T: int) -> int:
q = collections.deque([[ 0, 1 ]])
seen = set('0,1')
depth = 0
while True:
k = len(q)
while k:
[ pos, vel ] = q.popleft()
if pos == T:
return depth # \\U0001f3af target T found
cand = []
if abs(T - (pos + vel) < T):
cand.append([ pos + vel, 2 * vel ])
cand.append([ pos, 1 if vel < 0 else -1 ])
for pos, vel in cand:
if f'{pos},{vel}' not in seen:
q.append([ pos, vel ])
seen.add(f'{pos},{vel}')
k -= 1
depth += 1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
queue = deque([(0, 1, 0)])
seen = {(0, 1)}
while queue:
pos, speed, cost = queue.popleft()
if pos == target:
return cost
# A
new_pos, new_speed = pos + speed, speed * 2
if (new_pos, new_speed) not in seen:
seen.add((new_pos, new_speed))
queue.append((new_pos, new_speed, cost + 1)) # A
# R
reverse_speed = -1 if speed > 0 else 1
if (pos, reverse_speed) not in seen:
seen.add((pos, reverse_speed))
queue.append((pos, reverse_speed, cost + 1))
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, T: int, depth = 0) -> int:
q = collections.deque([[ 0, 1 ]])
seen = set('0,1')
while True:
k = len(q)
while k:
[ pos, vel ] = q.popleft()
if pos == T:
return depth # \\U0001f3af target T found
cand = []
if abs(T - (pos + vel) < T):
cand.append([ pos + vel, 2 * vel ])
cand.append([ pos, 1 if vel < 0 else -1 ])
for pos, vel in cand:
if f'{pos},{vel}' not in seen:
q.append([ pos, vel ])
seen.add(f'{pos},{vel}')
k -= 1
depth += 1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
queue = [(0, 1)]
step = 0
visited = {(0, 1)}
while queue:
tmp = queue
queue = []
for pos, speed in tmp:
if pos == target:
return step
if not (pos + speed, speed * 2) in visited:
queue.append((pos + speed, speed * 2))
visited.add((pos + speed, speed * 2))
speed = -1 if speed > 0 else 1
if not (pos, speed) in visited:
queue.append((pos, speed))
visited.add((pos, speed))
step += 1
return 0
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
pq = [(0, 0, -1)]
visited = set([(0, -1)])
while pq:
steps, pos, speed = heapq.heappop(pq)
next_steps = steps + 1
next_pos = pos + speed
next_speed = speed * 2
if -next_pos == target:
return next_steps
if -next_pos < 2 * target and -next_speed < 2 * target and not (next_pos, next_speed) in visited:
heapq.heappush(pq, (next_steps, next_pos, next_speed))
visited.add((next_pos, next_speed))
if speed < 0 and not (pos, 1) in visited:
heapq.heappush(pq, (next_steps, pos, 1))
visited.add((pos, 1))
elif speed > 0 and not (pos, -1) in visited:
heapq.heappush(pq, (next_steps, pos, -1))
visited.add((pos, -1))
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
nodes = [(0, 1)]
seen, length = set([(0, 1)]), 0
while nodes:
new_nodes = []
for pos, speed in nodes:
if pos == target:
return length
# A
for new_pos, new_speed in [
(pos + speed, speed * 2),
(pos, -1 if speed > 0 else 1),
]:
if (new_pos, new_speed) not in seen:
seen.add((new_pos, new_speed))
new_nodes.append((new_pos, new_speed))
length += 1
nodes = new_nodes
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
# dp[i] solution for target = i
dp = [ 0, 1, 4 ] + [ float('inf') ] * (target - 2)
for i in range(3, target + 1):
k = math.log2(i + 1)
if int(k) == k:
dp[i] = int(k)
continue
k = int(k)
# 1. go to 2 ** m - 1 and R^2 costing m + 2 (m <= k)
for m in range(0, k):
dp[i] = min(dp[i], dp[i - 2 ** k + 2 ** m] + k + m + 2)
# 2. go to 2 ** (k + 1) - 1 and R costing k + 1 + 1
if i > 2 ** (k + 1) - 1 - i:
dp[i] = min(dp[i], dp[2 ** (k + 1) - 1 - i] + k + 1 + 1)
return dp[target]
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
q = collections.deque([])
q.append((0, 1, 0))
visited = set()
visited.add((0, 1))
while q:
pos, speed, steps = q.popleft()
if pos == target:
return steps
if (pos + speed, speed * 2) not in visited:
q.append((pos + speed, speed * 2, steps + 1))
visited.add((pos + speed, speed * 2))
if speed > 0:
if (pos, -1) not in visited:
q.append((pos, -1, steps + 1))
visited.add((pos, -1))
else:
if (pos, 1) not in visited:
q.append((pos, 1, steps + 1))
visited.add((pos, 1))
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
from collections import deque
class Solution:
def racecar(self, target: int) -> int:
q = deque([(0, 1)])
visited = set([(0, 1)])
res = 0
while q:
size = len(q)
for _ in range(size):
pos, speed = q.popleft()
if pos == target:
return res
if (pos + speed, speed * 2) not in visited:
q.append((pos + speed, speed * 2))
visited.add((pos + speed, speed * 2))
speed = 1 if speed < 0 else -1
if (pos, speed) not in visited:
q.append((pos, speed))
visited.add((pos ,speed))
res += 1
return res
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
queue = [[(0, 1)]] #(position, speed)
visited = {(0,1)}
seq = 0
while queue:
curLevel = queue.pop()
newLevel = []
for pos, speed in curLevel:
if pos == target:
return seq
for newPos, newSpeed in [(pos+speed, speed*2), (pos, -1 if speed > 0 else 1)]:
if (newPos, newSpeed) not in visited:
visited.add((newPos, newSpeed))
newLevel.append((newPos, newSpeed))
if newLevel:
queue.append(newLevel)
seq += 1
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
initial_state = (0, 1, 0)
bfs = [initial_state]
valid_actions = ['A', 'R']
visited = set([(0, 1)])
for position, speed, count in bfs:
if position == target:
return count
for action in valid_actions:
if action == 'A':
_position = position + speed
_speed = speed * 2
elif action == 'R':
_position = position
if speed > 0:
_speed = -1
else:
_speed = 1
if (_position, _speed) not in visited:
visited.add((_position, _speed))
bfs.append((_position, _speed, count+1))
return -1
|
Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)
Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).
When you get an instruction "A", your car does the following: position += speed, speed *= 2.
When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)
For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.
Now for some target position, say the length of the shortest sequence of instructions to get there.
Example 1:
Input:
target = 3
Output: 2
Explanation:
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.
Example 2:
Input:
target = 6
Output: 5
Explanation:
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.
Note:
1 <= target <= 10000.
|
class Solution:
def racecar(self, target: int) -> int:
# bfs
bq=collections.deque([(0,1,0)])
visited=set((0,1))
while bq:
p,s,step= bq.popleft()
if p==target:
return step
if (p+s,2*s) not in visited:
bq.append((p+s,2*s,step+1))
visited.add((p+s,2*s))
ns= 1 if s<0 else -1
if (p,ns) not in visited:
bq.append((p,ns,step+1))
visited.add((p,ns))
|
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