## DEV Community

Abhishek Chaudhary

Posted on

# Unique Paths II

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:

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`.

SOLUTION:

``````class Solution:
memo = {}

def uniquePathsRec(self, m, n, i, j, obstacleGrid):
if i == m - 1 and j == n - 1:
if obstacleGrid[i][j] != 1:
return 1
return 0
if obstacleGrid[i][j] == 1:
return 0
if (i, j) not in self.memo:
self.memo[(i, j)] = 0
if i + 1 < m:
self.memo[(i, j)] += self.uniquePathsRec(m, n, i + 1, j, obstacleGrid)
if j + 1 < n:
self.memo[(i, j)] += self.uniquePathsRec(m, n, i, j + 1, obstacleGrid)
return self.memo[(i , j)]

def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
self.memo = {}
return self.uniquePathsRec(len(obstacleGrid), len(obstacleGrid[0]), 0, 0, obstacleGrid)
``````