<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:
        num_min=-1
        fresh,rot=1,2
        n,m=len(grid),len(grid[0])
        num_fresh=0
        q=deque()
        for i in range(n):
            for j in range(m):
                if grid[i][j]==rot:
                    q.append((i,j))
                elif grid[i][j]==fresh:
                    num_fresh+=1
        if num_fresh==0: return 0
        while q:
            num_min+=1
            qlen=len(q)
            for _ in range(qlen):
                ic,jc=q.popleft()
                for dr,dc in [[0,-1],[0,1],[1,0],[-1,0]]:
                    r,c=ic+dr,jc+dc
                    if 0<=r<n and 0<=c<m and grid[r][c]==fresh:
                        grid[r][c]=rot
                        q.append((r,c))
                        num_fresh-=1
        if num_fresh>0: return -1
        return num_min
16200