994. Rotting Oranges

994. Rotting Oranges (Medium)

In a given grid, each cell can have one of three values:

• the value 0 representing an empty cell;
• the value 1 representing a fresh orange;
• the value 2 representing a rotten orange.

Every minute, any fresh orange that is adjacent (4-directionally) 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 instead.

 

Example 1:

Input: [[2,1,1],[1,1,0],[0,1,1]]
Output: 4

Example 2:

Input: [[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: [[0,2]]
Output: 0
Explanation:  Since there are already no fresh oranges at minute 0, the answer is just 0.

 

Note:

1. 1 <= grid.length <= 10
2. 1 <= grid[0].length <= 10
3. grid[i][j] is only 0, 1, or 2.

Related Topics:
Breadth-first Search

Similar Questions:

• Walls and Gates (Medium)

Solution 1. BFS

// OJ: https://leetcode.com/problems/rotting-oranges/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int orangesRotting(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), ans = 0, dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
        queue<pair<int, int>> q;
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == 2) q.emplace(i, j);
            }
        }
        if (q.empty()) ans = 1;
        while (q.size()) {
            int cnt = q.size();
            ++ans;
            while (cnt--) {
                auto [x, y] = q.front();
                q.pop();
                for (auto &dir : dirs) {
                    int a = x + dir[0], b = y + dir[1];
                    if (a < 0 || a >= M || b < 0 || b >= N || A[a][b] != 1) continue;
                    A[a][b] = 2;
                    q.emplace(a, b);
                }
            }
        }
        for (auto &row : A) {
            for (int x : row) {
                if (x == 1) return -1;
            }
        }
        return ans - 1;
    }
};