<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:
    def orangesRotting(self, grid: List[List[int]]) -> int:
        rows, columns = len(grid), len(grid[0])
        directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
        rotten_q = collections.deque() # (row, column, expiration time)
        remaining_oranges = 0

        # Get all rotten oranges
        for r in range(rows):
            for c in range(columns):
                if grid[r][c] == 1:
                    remaining_oranges += 1
                elif grid[r][c] == 2:
                    rotten_q.append((r, c, 0))

        if not remaining_oranges:
            return 0
        if not rotten_q:
            return -1
            
        while rotten_q:
            for _ in range(len(rotten_q)):
                row, column, exp_time = rotten_q.popleft()
                for row_dir, column_dir in directions:
                    r, c = row + row_dir, column + column_dir
                    if min(r, c) >= 0 and r < rows and c < columns and grid[r][c] == 1:
                        grid[r][c] = 2 # Rott this orange
                        remaining_oranges -= 1
                        if not remaining_oranges:
                            return exp_time + 1
                        rotten_q.append((r, c, exp_time + 1))

        return -1
16500