<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();
        queue<vector<int>> q;

        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(grid[i][j]==2)
                {
                    q.push({0,i,j});
                }
            }
        }

        int maxtime=0;
        while(!q.empty())
        {
            vector<int> p1=q.front();
            q.pop();
            int t=p1[0];
            int x=p1[1];
            int y=p1[2];
            if(x>0 && grid[x-1][y]==1)
            {
                grid[x-1][y]=2;
                q.push({t+1,x-1,y});
            }
            if(y>0 && grid[x][y-1]==1)
            {
                grid[x][y-1]=2;
                q.push({t+1,x,y-1});
            }
            if(x<n-1 && grid[x+1][y]==1)
            {
                grid[x+1][y]=2;
                q.push({t+1,x+1,y});
            }
            if(y<m-1 && grid[x][y+1]==1)
            {
                grid[x][y+1]=2;
                q.push({t+1,x,y+1});
            }
            maxtime=max(maxtime,t);
        }

        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(grid[i][j]==1)
                {
                    return -1;
                }
            }
        }

        return maxtime;
        
    }
};
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