Rotting Oranges
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0994.Rotting%20Oranges/images/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>
Breadth-First Search; Array; Matrix
func orangesRotting(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	q := [][2]int{}
	cnt := 0
	for i, row := range grid {
		for j, x := range row {
			if x == 1 {
				cnt++
			} else if x == 2 {
				q = append(q, [2]int{i, j})
			}
		}
	}
	dirs := [5]int{-1, 0, 1, 0, -1}
	for ans := 1; len(q) > 0 && cnt > 0; ans++ {
		for k := len(q); k > 0; k-- {
			p := q[0]
			q = q[1:]
			for d := 0; d < 4; d++ {
				x, y := p[0]+dirs[d], p[1]+dirs[d+1]
				if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 {
					grid[x][y] = 2
					q = append(q, [2]int{x, y})
					if cnt--; cnt == 0 {
						return ans
					}
				}
			}
		}
	}
	if cnt > 0 {
		return -1
	}
	return 0
}
