1424 Maximum Candies You Can Get From Boxes

Maximum Candies You Can Get from Boxes

You have n boxes labeled from 0 to n - 1. You are given four arrays: status, candies, keys, and containedBoxes where:

status[i] is 1 if the ith box is open and 0 if the ith box is closed,
candies[i] is the number of candies in the ith box,
keys[i] is a list of the labels of the boxes you can open after opening the ith box.
containedBoxes[i] is a list of the boxes you found inside the ith box.

You are given an integer array initialBoxes that contains the labels of the boxes you initially have. 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 do not 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.

 

Constraints:

n == status.length == candies.length == keys.length == containedBoxes.length
1 <= n <= 1000
status[i] is either 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= n
0 <= keys[i][j] < n
All values of keys[i] are **unique**.
0 <= containedBoxes[i].length <= n
0 <= containedBoxes[i][j] < n
All values of containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= n
0 <= initialBoxes[i] < n

from typing import List

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)
        has_box = [False] * n
        has_key = [False] * n
        visited = [False] * n

        # Mark which boxes we have from the beginning
        for box in initialBoxes:
            has_box[box] = True

        total_candies = 0

        def try_open_boxes():
            nonlocal total_candies
            changed = True
            while changed:
                changed = False
                for i in range(n):
                    # Only visit boxes we have, haven't opened, and are now openable
                    if has_box[i] and not visited[i] and (status[i] == 1 or has_key[i]):
                        visited[i] = True
                        total_candies += candies[i]
                        changed = True

                        # Collect keys
                        for k in keys[i]:
                            if not has_key[k]:
                                has_key[k] = True
                                # we may now be able to open a box we skipped earlier

                        # Collect boxes
                        for b in containedBoxes[i]:
                            if not has_box[b]:
                                has_box[b] = True

        try_open_boxes()
        return total_candies