<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:
    void dfs(vector<vector<int>>& grid,int r,int c,int min){
        if(r<0 || c<0 || r>grid.size()-1 || c>grid[r].size()-1 || grid[r][c]==0 || (grid[r][c]<min && grid[r][c]>1)){
            return;
        }
        grid[r][c]=min;
        dfs(grid,r-1,c,min+1);
        dfs(grid,r,c+1,min+1);
        dfs(grid,r,c-1,min+1);
        dfs(grid,r+1,c,min+1);
        return;
    }
    int orangesRotting(vector<vector<int>>& grid) {
        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[i].size();j++){
                if(grid[i][j]==2){
                    dfs(grid,i,j,2);
                }
            }
        }
        int m=2;
        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[i].size();j++){
                if(grid[i][j]==1)
                return -1;
                m=max(m,grid[i][j]);
            }
        }
        return m-2;
    }
};
15700