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leetcode/problem/rotting-oranges.html

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+<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>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>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>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>