LeetCode-in-Java

994. Rotting Oranges

Medium

You are given an m x n grid where each cell can have one of three values:

• 0 representing an empty cell,
• 1 representing a fresh orange, or
• 2 representing a rotten orange.

Every minute, any fresh orange that is 4-directionally adjacent to a rotten orange becomes rotten.

Return the minimum number of minutes that must elapse until no cell has a fresh orange. If this is impossible, return -1.

Example 1:

Input: grid = [[2,1,1],[1,1,0],[0,1,1]]

Output: 4

Example 2:

Input: grid = [[2,1,1],[0,1,1],[1,0,1]]

Output: -1

Explanation: The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.

Example 3:

Input: grid = [[0,2]]

Output: 0

Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 10
• grid[i][j] is 0, 1, or 2.

Solution

import java.util.LinkedList;
import java.util.Queue;

public class Solution {
    public int orangesRotting(int[][] grid) {
        Queue<int[]> queue = new LinkedList<>();
        int row = grid.length;
        int col = grid[0].length;
        int countActive = 0;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (grid[i][j] == 2) {
                    queue.add(new int[] {i, j});
                }
                if (grid[i][j] != 0) {
                    countActive++;
                }
            }
        }
        if (countActive == 0) {
            return 0;
        }
        int countCurrent = 0;
        int count = 0;
        int[] dx = {0, 0, 1, -1};
        int[] dy = {1, -1, 0, 0};
        while (!queue.isEmpty()) {
            int size = queue.size();
            count += size;
            for (int i = 0; i < size; i++) {
                int[] arr = queue.poll();
                for (int j = 0; j < 4; j++) {
                    int x = arr[0] + dx[j];
                    int y = arr[1] + dy[j];
                    if (x < 0 || y < 0 || x >= row || y >= col || grid[x][y] != 1) {
                        continue;
                    }
                    grid[x][y] = 2;
                    queue.add(new int[] {x, y});
                }
            }
            if (!queue.isEmpty()) {
                countCurrent++;
            }
        }
        return countActive == count ? countCurrent : -1;
    }
}

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