<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 tot=0,days=0,cnt=0;
        int m=grid.size();
        int n=grid[0].size();
        queue<pair<int,int>>pq;
        int dx[]={0,0,1,-1};
        int dy[]={1,-1,0,0};
        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)
                {
                    pq.push({i,j});
                }
            }
        }
        while(!pq.empty())
        {
            int k=pq.size();
            cnt+=k;
            while(k--)
            {
                int x=pq.front().first;
                int y=pq.front().second;
                pq.pop();
                for(int i=0;i<4;i++)
                {
                    int nx=dx[i]+x;
                    int ny=dy[i]+y;
                    if(nx<0||ny<0||nx>=m||ny>=n||grid[nx][ny]!=1)
                    {
                        continue;
                    }
                    grid[nx][ny]=2;
                    pq.push({nx,ny});
                }
            }
            if(!pq.empty())
            days++;
        }
        return tot==cnt?days:-1;
    }
};
16000