<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) {
        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>> pq;
        int m = grid.size();
        int n = grid[0].size();
        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){
                    pq.push({0,{i,j}});
                    vis[i][j]=1;
                }
            }
        }
        int ans = 0;
        vector<int> roww={0,-1,0,1};
        vector<int> coll={-1,0,1,0};

        while(!pq.empty()){
            auto x = pq.top();
            pq.pop();
            int time = x.first;
            int row = x.second.first;
            int col = x.second.second;
            ans = time;

            for(int k=0;k<4;k++){
                int r = row+roww[k];
                int c = col+coll[k];
                if(r>=0 && r<m & c>=0 && c<n && !vis[r][c] && grid[r][c]==1){
                    vis[r][c]=1;
                    grid[r][c]=2;
                    pq.push({time+1,{r,c}});
                }
            }
        }


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

        

    }
};
17400