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
class Solution {
    public int orangesRotting(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        Deque<int[]> q = new ArrayDeque<>();
        int cnt = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    ++cnt;
                } else if (grid[i][j] == 2) {
                    q.offer(new int[] {i, j});
                }
            }
        }
        final int[] dirs = {-1, 0, 1, 0, -1};
        for (int ans = 1; !q.isEmpty() && cnt > 0; ++ans) {
            for (int k = q.size(); k > 0; --k) {
                var p = q.poll();
                for (int d = 0; d < 4; ++d) {
                    int x = p[0] + dirs[d], y = p[1] + dirs[d + 1];
                    if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1) {
                        grid[x][y] = 2;
                        q.offer(new int[] {x, y});
                        if (--cnt == 0) {
                            return ans;
                        }
                    }
                }
            }
        }
        return cnt > 0 ? -1 : 0;
    }
}
