1
0
mirror of https://gitee.com/coder-xiaomo/leetcode-problemset synced 2025-01-11 02:58:13 +08:00
Code Issues Projects Releases Wiki Activity GitHub Gitee
leetcode-problemset/leetcode/problem/shortest-path-in-a-grid-with-obstacles-elimination.html

35 lines
1.9 KiB
HTML
Raw Normal View History

2022-03-27 18:27:43 +08:00
<p>You are given an <code>m x n</code> integer matrix <code>grid</code> where each cell is either <code>0</code> (empty) or <code>1</code> (obstacle). You can move up, down, left, or right from and to an empty cell in <strong>one step</strong>.</p>
<p>Return <em>the minimum number of <strong>steps</strong> to walk from the upper left corner </em><code>(0, 0)</code><em> to the lower right corner </em><code>(m - 1, n - 1)</code><em> given that you can eliminate <strong>at most</strong> </em><code>k</code><em> obstacles</em>. If it is not possible to find such walk return <code>-1</code>.</p>
<p>&nbsp;</p>
2023-12-09 18:42:21 +08:00
<p><strong class="example">Example 1:</strong></p>
2022-03-27 18:27:43 +08:00
<img alt="" src="https://assets.leetcode.com/uploads/2021/09/30/short1-grid.jpg" style="width: 244px; height: 405px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1
<strong>Output:</strong> 6
<strong>Explanation:</strong>
The shortest path without eliminating any obstacle is 10.
The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -&gt; (0,1) -&gt; (0,2) -&gt; (1,2) -&gt; (2,2) -&gt; <strong>(3,2)</strong> -&gt; (4,2).
</pre>
2023-12-09 18:42:21 +08:00
<p><strong class="example">Example 2:</strong></p>
2022-03-27 18:27:43 +08:00
<img alt="" src="https://assets.leetcode.com/uploads/2021/09/30/short2-grid.jpg" style="width: 244px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1],[1,1,1],[1,0,0]], k = 1
<strong>Output:</strong> -1
<strong>Explanation:</strong> We need to eliminate at least two obstacles to find such a walk.
</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;= 40</code></li>
<li><code>1 &lt;= k &lt;= m * n</code></li>
<li><code>grid[i][j]</code> is either <code>0</code> <strong>or</strong> <code>1</code>.</li>
<li><code>grid[0][0] == grid[m - 1][n - 1] == 0</code></li>
</ul>