<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 bfs(vector<vector<int>>& grid,queue<pair<pair<int,int>,int>>q,vector<vector<int>>&vis)
    {
         int m = grid.size();
        int n = grid[0].size();
        
        vector<int> row = {0,0,-1,1};
        vector<int> col = {-1,1,0,0};
        int maxi = 0;
        while(q.size())
        {
            int a = q.front().first.first;
            int b = q.front().first.second;
            int c = q.front().second;
            q.pop();

            for(int i=0;i<4;i++)
            {
                int r = a+row[i];
                int cl = b+col[i];
                if(r>=0&&cl>=0&&r<m&&cl<n && vis[r][cl] == 0 && grid[r][cl] == 1)
                {
                    q.push({{r,cl},c+1});
                    vis[r][cl] = 1;
                    maxi = max(c+1,maxi);
                }
            }
        }

        return maxi;

    }
    int orangesRotting(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();

        queue<pair<pair<int,int>,int>> q;
        vector<vector<int>> vis(m,vector<int>(n,0));
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(grid[i][j] == 2)
                {
                    q.push({{i,j},0});
                    vis[i][j] = 1;
                }
            }
        }
        
        int maxi = bfs(grid,q,vis);


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

        return maxi;
    }
};
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