LeetCode 1298: Maximum Candies You Can Get from Boxes Solution

Master LeetCode problem 1298 (Maximum Candies You Can Get from Boxes), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1298. Maximum Candies You Can Get from Boxes

Hard

Array Breadth-First Search Graph

Problem Explanation

Explanation

We can model this problem as a graph where each box is a node and the keys and contained boxes are the edges.
We perform a breadth-first search (BFS) starting from the initial boxes we have.
We maintain a set of visited boxes and a queue for boxes to process.
At each step, we check if we can open the box, collect candies, and add new boxes to the queue based on keys and contained boxes.
We continue this process until there are no more boxes to process in the queue. We keep track of the total candies collected.

Time Complexity: O(n), where n is the total number of boxes
Space Complexity: O(n)

Solution Code

class Solution {
    public int maxCandies(int[] status, int[] candies, int[][] keys, int[][] containedBoxes, int[] initialBoxes) {
        int n = status.length;
        Set<Integer> visited = new HashSet<>();
        Queue<Integer> queue = new LinkedList<>();
        int totalCandies = 0;

        for (int box : initialBoxes) {
            queue.offer(box);
            visited.add(box);
        }

        while (!queue.isEmpty()) {
            int currBox = queue.poll();
            if (status[currBox] == 1 || visited.containsAll(Arrays.asList(keys[currBox]))) {
                totalCandies += candies[currBox];
                for (int newBox : containedBoxes[currBox]) {
                    queue.offer(newBox);
                    visited.add(newBox);
                }
                for (int key : keys[currBox]) {
                    visited.add(key);
                }
            } else {
                queue.offer(currBox);
            }
        }

        return totalCandies;
    }
}

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