<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 m=grid.size();
        int n=grid[0].size();
        if(n==0){
            return -1;
        }
        int tot=0, cnt=0;
        queue<pair<int, int>> rottenOrange;
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j]!=0){
                    tot++;
                }
                if(grid[i][j]==2){
                    rottenOrange.push(make_pair(i,j));
                }
            }
        }
        int dx[4]={1,-1,0,0};
        int dy[4]={0,0,1,-1};
        int ans=0;
        while(!rottenOrange.empty()){
            int k=rottenOrange.size();
            cnt+=k;
            while(k--){
                int x=rottenOrange.front().first;
                int y=rottenOrange.front().second;
                rottenOrange.pop();
                for(int i=0; i<4; i++){
                    int nx=x+dx[i];
                    int ny=y+dy[i];
                    if(nx<0 || ny<0 || nx>=m || ny>=n || grid[nx][ny]!=1){
                        continue;
                    }
                    grid[nx][ny]=2;
                    rottenOrange.push(make_pair(nx, ny));
                }
            }
            if(!rottenOrange.empty()){
                ans++;
            }
        }
        return tot==cnt? ans: -1;
    }
};
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