# [Solved] You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid). 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. ## 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`). 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:

```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:

```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)
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[j]==0 and j!=m-1:
for k in range(j,m):
obstacleGrid[k]=0

for i in range(n):
if obstacleGrid[i]==0 and i!=n-1:
for k in range(i,n):
obstacleGrid[k]=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]```