<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<pair<pair<int,int>,int>> q;
        vector<vector<int>> vis(n,vector<int>(m,0));

        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 drow[] = {-1,0,+1,0};
        int dcol[] = {0,+1,0,-1};
        int tm = 0;

        while(!q.empty())
        {
            int r = q.front().first.first;
            int c = q.front().first.second;
            int t = q.front().second;
            q.pop();
            tm = max(tm,t);

            for(int i=0;i<4;i++)
            {
                int nrow = r + drow[i];
                int ncol = c + dcol[i];

                if(nrow >= 0 && nrow < n && ncol >= 0 && ncol < m && vis[nrow][ncol] != 2 && grid[nrow][ncol] == 1)
                {
                    q.push({{nrow,ncol},t+1});
                    vis[nrow][ncol] = 2;
                }
            }
        }

        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 tm;
    }
};
16400