Unique Paths II
<p>You are given an <code>m x n</code> integer array <code>grid</code>. There is a robot initially located at the <b>top-left corner</b> (i.e., <code>grid[0][0]</code>). The robot tries to move to the <strong>bottom-right corner</strong> (i.e., <code>grid[m - 1][n - 1]</code>). The robot can only move either down or right at any point in time.</p>

<p>An obstacle and space are marked as <code>1</code> or <code>0</code> respectively in <code>grid</code>. A path that the robot takes cannot include <strong>any</strong> square that is an obstacle.</p>

<p>Return <em>the number of possible unique paths that the robot can take to reach the bottom-right corner</em>.</p>

<p>The testcases are generated so that the answer will be less than or equal to <code>2 * 10<sup>9</sup></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/0000-0099/0063.Unique%20Paths%20II/images/robot1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -&gt; Right -&gt; Down -&gt; Down
2. Down -&gt; Down -&gt; Right -&gt; Right
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0063.Unique%20Paths%20II/images/robot2.jpg" style="width: 162px; height: 162px;" />
<pre>
<strong>Input:</strong> obstacleGrid = [[0,1],[0,0]]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == obstacleGrid.length</code></li>
	<li><code>n == obstacleGrid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>obstacleGrid[i][j]</code> is <code>0</code> or <code>1</code>.</li>
</ul>
Array; Dynamic Programming; Matrix
class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i >= m or j >= n or obstacleGrid[i][j]:
                return 0
            if i == m - 1 and j == n - 1:
                return 1
            return dfs(i + 1, j) + dfs(i, j + 1)

        m, n = len(obstacleGrid), len(obstacleGrid[0])
        return dfs(0, 0)
