<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 n=grid.size();
        int m=grid[0].size();
        vector<vector<int>>vis=vector<vector<int>>(n,vector<int>(m));
        queue<vector<int>>q;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==2)
           { 
            q.push({i,j,0});
            vis[i][j]=true;
            }
            }
        }
        int ans=0;
        while(!q.empty()){
            auto item=q.front();
            int i=item[0];
            int j=item[1];
            int w=item[2];
            q.pop();
            if(i-1>=0&&!vis[i-1][j]&&grid[i-1][j]==1){
            vis[i-1][j]=true;
            q.push({i-1,j,w+1});
            ans=max(ans,w+1);
            }
            if(i+1<n&&!vis[i+1][j]&&grid[i+1][j]==1){
            vis[i+1][j]=true;
            q.push({i+1,j,w+1});
            ans=max(ans,w+1);
            }
            if(j-1>=0&&!vis[i][j-1]&&grid[i][j-1]==1){
               vis[i][j-1]=true; 
               q.push({i,j-1,w+1});
               ans=max(ans,w+1);
            }
            if(j+1<m&&!vis[i][j+1]&&grid[i][j+1]==1){
                 vis[i][j+1]=true; 
               q.push({i,j+1,w+1});
               ans=max(ans,w+1);
            }
        }
        
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==1&&!vis[i][j])return -1;
        }
        }

        return ans;

    }
};
17400