Table of Contents

## Question

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 * 10`

.^{9}

**Example 1:**

Input:obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]Output:2Explanation: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:**

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`

.

## Python Solution

class Solution: def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int: m = len(obstacleGrid[0]) n = len(obstacleGrid) for i in range(n): for j in range(m): if obstacleGrid[i][j]==0: obstacleGrid[i][j]=1 else: obstacleGrid[i][j]=0 for j in range(m): if obstacleGrid[0][j]==0 and j!=m-1: for k in range(j,m): obstacleGrid[0][k]=0 for i in range(n): if obstacleGrid[i][0]==0 and i!=n-1: for k in range(i,n): obstacleGrid[k][0]=0 for i in range(1,n): for j in range(1,m): if obstacleGrid[i][j]!=0: obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1] return obstacleGrid[-1][-1]