<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 row=grid.size();
      int col=grid[0].size();
      queue<pair<int,int>>q;
      int freshcount=0;
      vector<vector<int>> dim={{-1,0},{1,0},{0,-1},{0,1}};
      for(int i=0;i<row;i++){
        for(int j=0;j<col;j++){
            if(grid[i][j]==2){
                q.push({i,j});
            }
            else if(grid[i][j]==1)freshcount++;
        }
      }

        if(freshcount==0) return 0;
        int time=0;
        while(!q.empty()){
            int size=q.size();
            for(int i=0;i<size;i++){
                pair<int,int> front=q.front();
                int x=front.first;
                int y=front.second;
                q.pop();
                for(auto dir:dim){
                    int newX=x+dir[0];
                    int newY=y+dir[1];
                    if(newX>=0 && newX<row && newY>=0 && newY<col && grid[newX][newY]==1){
                        grid[newX][newY]=2;
                        q.push({newX,newY});
                        freshcount--;
                    }
                }
            }
            time++;
        }
        if(freshcount>0) return -1;
        return time-1;
    }
};
17700