<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) {
        queue<pair<int,int>> q;
        vector<vector<int>> visited(grid.size(),vector<int>(grid[0].size(),0));
        int flag = 1;
        for(int i=0;i<grid.size();i++)
        {
            for(int j=0;j<grid[0].size();j++)
            {
                if(grid[i][j] == 2)
                {
                    flag = 0;
                    q.push({i,j});
                }
                if(grid[i][j] == 1)
                {
                    flag = 0;
                }
            }
        }
        if(flag)
        {
            return 0;
        }
        int count = 0;

        while(!q.empty())
        {
            int size = q.size();
            for(int i=0;i<size;i++)
            {
                pair<int,int> front = q.front();
                q.pop();
                int r = front.first;
                int c = front.second;
                if(c+1<grid[0].size() && grid[r][c+1] == 1 && !visited[r][c+1])
                {
                    grid[r][c+1] = 2;
                    visited[r][c+1]=1;
                    q.push({r,c+1});
                }
                if(c-1>=0 && grid[r][c-1] == 1 && !visited[r][c-1])
                {
                    grid[r][c-1] = 2;
                    visited[r][c-1]=1;
                    q.push({r,c-1});
                }
                if(r+1<grid.size() && grid[r+1][c] == 1 && !visited[r+1][c])
                {
                    grid[r+1][c] = 2;
                    visited[r+1][c]=1;
                    q.push({r+1,c});
                }
                if(r-1>=0 && grid[r-1][c] == 1 && !visited[r-1][c])
                {
                    grid[r-1][c] = 2;
                    visited[r-1][c]=1;
                    q.push({r-1,c});
                }
            }
            count++;
        }
        for(auto i: grid)
        {
            for(auto j: i)
            {
                if(j==1)
                {
                    return -1;
                }
            }
        }
        return count-1;
    }
};
16800