994. Rotting Oranges

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.

// BFS
int orangesRotting(vector<vector<int>>& grid) {
    if (grid.empty() || grid[0].empty()) return 0;
    int m = grid.size(), n = grid[0].size(), fresh_cnt = 0;
    queue<pair<int, int> > q;
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            if (grid[i][j] == 1) ++fresh_cnt;
            else if (grid[i][j] == 2) q.emplace(i, j);
        }
    }
    if (fresh_cnt == 0) return 0;
    int res = 0;
    vector<vector<int> > dirs({{-1, 0}, {1, 0}, {0, -1}, {0, 1}});
    while (!q.empty() && fresh_cnt) {
        int sz = q.size();
        for (int k = 0; k < sz; ++k) {
            auto pt = q.front(); q.pop();
            for (auto dir : dirs) {
                int r = pt.first + dir[0];
                int c = pt.second + dir[1];
                if (r < 0 || r >= m || c < 0 || c >= n || grid[r][c] != 1) continue;
                grid[r][c] = 2;
                --fresh_cnt;
                q.emplace(r, c);
            }
        }
        ++res;
    }
    return fresh_cnt == 0 ? res : -1;
}