LeetCode 286: Walls and Gates Solution

Master LeetCode problem 286 (Walls and Gates), a medium 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.

286. Walls and Gates

Medium

Array Breadth-First Search Matrix

Problem Explanation

Explanation:

Algorithmic Idea:
1. We can solve this problem using Breadth First Search (BFS).
2. We start by iterating over all gates and enqueue their positions into the queue.
3. Then, we perform BFS to traverse all cells starting from the gates and updating their distance to the nearest gate.
4. During BFS, we update the distance of each cell by considering its neighboring cells and the current cell's distance.
Step-by-Step Iterations:
- Enqueue all gate positions into the queue.
- Perform BFS to update distances for all cells.
Time Complexity: O(m*n) where m and n are the dimensions of the grid.
Space Complexity: O(m*n) for the queue and visited array.

: :

Solution Code

class Solution {
    public void wallsAndGates(int[][] rooms) {
        if (rooms == null || rooms.length == 0) return;
        
        int m = rooms.length;
        int n = rooms[0].length;
        Queue<int[]> queue = new LinkedList<>();
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (rooms[i][j] == 0) {
                    queue.offer(new int[]{i, j});
                }
            }
        }
        
        int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        
        while (!queue.isEmpty()) {
            int[] curr = queue.poll();
            int row = curr[0];
            int col = curr[1];
            
            for (int[] dir : dirs) {
                int newRow = row + dir[0];
                int newCol = col + dir[1];
                
                if (newRow < 0 || newRow >= m || newCol < 0 || newCol >= n || rooms[newRow][newCol] != Integer.MAX_VALUE) {
                    continue;
                }
                
                rooms[newRow][newCol] = rooms[row][col] + 1;
                queue.offer(new int[]{newRow, newCol});
            }
        }
    }
}

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