<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>

from collections import deque
class Solution:
    def orangesRotting(self, grid: List[List[int]]) -> int:
        n,m=len(grid),len(grid[0])
        q=deque()
        freshcnt=0
        vis=[[0 for i in range(m)] for i in range(n)]
        for i in  range(n):
            for j in range(m):
                if grid[i][j]==2:
                    q.append(((i,j),0))
                    vis[i][j]=2
                if grid[i][j]==1:
                    freshcnt+=1
        maxtime=0
        drow=[-1,0,1,0]
        dcol=[0,1,0,-1]
        cnt=0
        while q:
            coord,time=q.popleft()
            r,c=coord
            maxtime=max(time,maxtime)
            for i in range(4):
                nrow=r+drow[i]
                ncol=c+dcol[i]
                if nrow>=0 and nrow<n and ncol>=0 and ncol<m and vis[nrow][ncol]==0 and grid[nrow][ncol]==1:
                    q.append(((nrow,ncol),time+1))
                    vis[nrow][ncol]=2
                    cnt+=1
        if cnt!=freshcnt:
            return -1
        return maxtime
            
            
        
        
        
16300