Solution Explanation

This problem involves finding the maximum number of candies obtainable from a set of boxes, given constraints on box opening and contents. The solution uses a breadth-first search (BFS) approach to efficiently explore all possible candy collections.

Approach:

1. Initialization:

   • We start with a queue (q) containing only the initially accessible boxes (those in initialBoxes that are already open, status[i] == 1).
   • ans stores the total candies collected. It's initialized with candies from initially open boxes.
   • has is a set tracking boxes we've encountered (regardless of whether they're open). This handles boxes found inside other boxes.
   • took is a set to keep track of boxes from which we have already collected candies. This avoids recounting candies.

2. BFS Traversal:

   • The while loop continues as long as the queue is not empty.
   • In each iteration, we dequeue a box (i).
   • Keys: We iterate through the keys (keys[i]) found in box i. If a key opens a new box (status[k] = 1) and that box has been encountered (has[k]) but not yet processed (!took[k]), we add its candies to ans, mark it as processed (took[k]), and add it to the queue for further exploration.
   • Contained Boxes: We iterate through the boxes (containedBoxes[i]) found inside box i. If a contained box has been encountered (has[j]), is open (status[j] == 1), and hasn't been processed (!took[j]), we collect its candies, mark it as processed, and enqueue it.

3. Return Value:

   • Finally, ans (the total candies collected) is returned.

Time Complexity: O(N), where N is the number of boxes. In the worst case, we might visit each box once. The operations within the loop (checking sets, adding to queue) are O(1) on average for hash sets and queues.

Space Complexity: O(N) The space is dominated by the q, has, and took data structures, each of which could potentially store up to N elements in the worst case.

Code Explanation (Python)

from collections import deque
 
class Solution:
    def maxCandies(self, status, candies, keys, containedBoxes, initialBoxes):
        q = deque([i for i in initialBoxes if status[i] == 1]) # Start with open initial boxes
        ans = sum(candies[i] for i in initialBoxes if status[i] == 1) # Initial candy count
        has = set(initialBoxes) # Set of encountered boxes
        took = {i for i in initialBoxes if status[i] == 1} # Set of processed boxes
 
        while q:
            i = q.popleft() # Get the next box
            for k in keys[i]: # Process keys
                status[k] = 1 # Open the box using the key
                if k in has and k not in took: # If encountered but not processed
                    ans += candies[k] # Collect candies
                    took.add(k) # Mark as processed
                    q.append(k) # Add to queue
            for j in containedBoxes[i]: # Process contained boxes
                has.add(j) # Mark as encountered
                if status[j] and j not in took: # If open and not processed
                    ans += candies[j] # Collect candies
                    took.add(j) # Mark as processed
                    q.append(j) # Add to queue
        return ans
 

The other languages (Java, C++, Go) follow the same logical structure, utilizing their respective queue and set implementations. The core algorithm remains consistent across all languages.