<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:
        if not grid:
            return 0
        m,n=len(grid),len(grid[0])
        fresh=0
        queue=deque()
        for x in range(m):
            for y in range(n):
                if grid[x][y]==2:
                    queue.append((x,y))
                elif grid[x][y]==1:
                    fresh+=1

        if fresh==0:
            return 0

        directions=((1,0),(-1,0),(0,1),(0,-1))
        minute=0
        while queue:
            size=len(queue)
            for _ in range(size):
                x,y = queue.popleft()
                
                for addx,addy in directions:
                    newx,newy=x+addx,y+addy
                    if 0<=newx<m and 0<=newy<n and grid[newx][newy]==1:
                        queue.append((newx,newy))
                        grid[newx][newy]=2
                        fresh-=1
            minute+=1
        return minute-1 if fresh==0 else -1


        
16300