<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 count=0;

        vector<vector<bool>> visited(grid.size(),vector<bool>(grid[0].size(),false));
        queue<pair<int,int>> q;
            int fresh=0;
        for(auto i=0; i<grid.size(); i++){
            for(auto j=0; j<grid[i].size(); j++){

                if(grid[i][j]==1){
                    fresh++;
                }
                if(grid[i][j]==2){
                    q.push({i,j});
                    visited[i][j]=true;
                }
            }
        }

        if(fresh==0) return 0;
        while(q.empty()==false){
            int n=q.size();

            for(int i=0; i<n; i++){
                auto temp=q.front();
                q.pop();
                int x=temp.first;
                int y=temp.second;
                cout<<x<<" "<<y<<endl;
                if(x+1 <grid.size() && visited[x+1][y]==false && grid[x+1][y]==1){ q.push({x+1,y});grid[x+1][y]=2; visited[x+1][y]=true;}

                if(x-1 >=0 && visited[x-1][y]==false && grid[x-1][y]==1){ q.push({x-1,y}); grid[x-1][y]=2; visited[x-1][y]=true;}

                if(y+1 < grid[x].size() && visited[x][y+1]==false && grid[x][y+1]==1){ q.push({x,y+1}); grid[x][y+1]=2; visited[x][y+1]=true; }

                if(y-1 >=0 && visited[x][y-1]==false && grid[x][y-1]==1){ q.push({x,y-1}); grid[x][y-1]=2; visited[x][y-1]=true;}
            }
            count++;
        }


        for(auto i : grid){
            for(auto j : i){
                if(j==1) return -1;
            }
        }
       

        return count-1;


        
    }
};
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