# 63. Unique Paths II

#### Medium

***

You are given an `m x n` integer array `grid`. There is a robot initially located at the **top-left corner** (i.e., `grid[0][0]`). The robot tries to move to the **bottom-right corner** (i.e., `grid[m-1][n-1]`). The robot can only move either down or right at any point in time.

An obstacle and space are marked as `1` or `0` respectively in `grid`. A path that the robot takes cannot include **any** square that is an obstacle.

Return *the number of possible unique paths that the robot can take to reach the bottom-right corner*.

The testcases are generated so that the answer will be less than or equal to `2 * 109`.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/11/04/robot1.jpg)

```
Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
Output: 2
Explanation: There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/11/04/robot2.jpg)

```
Input: obstacleGrid = [[0,1],[0,0]]
Output: 1
```

 

**Constraints:**

* `m == obstacleGrid.length`
* `n == obstacleGrid[i].length`
* `1 <= m, n <= 100`
* `obstacleGrid[i][j]` is `0` or `1`.

#### Solution 1 : Using Dictionary

```python
class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        self.grid = obstacleGrid
        self.d = {}
        return self.dfs(0,0)
    
    def dfs(self, row, col):
        if (row,col) in self.d:
            return self.d[(row,col)]
        if row >= len(self.grid) or col >= len(self.grid[0]) or self.grid[row][col] == 1:
            return 0
        if row == len(self.grid)-1 and col == len(self.grid[0])-1:
            return 1
        val = self.dfs(row+1, col) + self.dfs(row, col+1)
        self.d[(row,col)] = val
        return val
```

#### Solution 2 : With In-Built @lru\_cache

```python
class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        self.grid = obstacleGrid
        return self.dfs(0,0)
    
    @lru_cache
    def dfs(self, row, col):
        if row >= len(self.grid) or col >= len(self.grid[0]) or self.grid[row][col] == 1:
            return 0
        if row == len(self.grid)-1 and col == len(self.grid[0])-1:
            return 1
        dfs1 = self.dfs(row+1, col)
        dfs2 = self.dfs(row, col+1)
        return dfs1 + dfs2
```