<p>You are given an <code>m x n</code> <code>grid</code> where each cell can have one of three values:</p>

<ul>
	<li><code>0</code> representing an empty cell,</li>
	<li><code>1</code> representing a fresh orange, or</li>
	<li><code>2</code> representing a rotten orange.</li>
</ul>

<p>Every minute, any fresh orange that is <strong>4-directionally adjacent</strong> to a rotten orange becomes rotten.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/02/16/oranges.png" style="width: 650px; height: 137px;" />
<pre>
<strong>Input:</strong> grid = [[2,1,1],[1,1,0],[0,1,1]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[2,1,1],[0,1,1],[1,0,1]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,2]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> Since there are already no fresh oranges at minute 0, the answer is just 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10</code></li>
	<li><code>grid[i][j]</code> is <code>0</code>, <code>1</code>, or <code>2</code>.</li>
</ul>

class Solution {
public:
    int orangesRotting(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        int drow[] = {-1, 0, 1, 0};
        int dcol[] = {0, 1, 0, -1};
        queue<pair<pair<int, int>, int>> q;
        int vis[n][m];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==2){
                    q.push({{i, j}, 0});
                    vis[i][j]=2;
                }
                else{
                    vis[i][j]=0;
                }
            }
        }
        int t=0;
        while(!q.empty()){
            int r=q.front().first.first;
            int c=q.front().first.second;
            int time=q.front().second;
            t=max(t, time);
            q.pop();
            for (int i = 0; i < 4; i++) {
                int nr = r + drow[i];
                int nc = c + dcol[i];
                if (nr >= 0 && nr < n && nc >= 0 && nc < m && 
                    vis[nr][nc]!=2 && grid[nr][nc] == 1) {
                    vis[nr][nc] = 2;
                    q.push({{nr, nc}, time+1});
                }
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(vis[i][j]!=2&&grid[i][j]==1){
                    return -1;
                }
            }
        }
        return t;
    }
};
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