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# Unique Paths II LeetCode Solution

## Problem – Unique Paths II LeetCode Solution

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`.

### Unique Paths II LeetCode Solution in Java

``````public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int width = obstacleGrid[0].length;
int[] dp = new int[width];
dp[0] = 1;
for (int[] row : obstacleGrid) {
for (int j = 0; j < width; j++) {
if (row[j] == 1)
dp[j] = 0;
else if (j > 0)
dp[j] += dp[j - 1];
}
}
return dp[width - 1];
}
``````

### Unique Paths II LeetCode Solution in C++

``````class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int m = obstacleGrid.size() , n = obstacleGrid[0].size();
vector<vector<int>> dp(m+1,vector<int>(n+1,0));
dp[0][1] = 1;
for(int i = 1 ; i <= m ; ++i)
for(int j = 1 ; j <= n ; ++j)
if(!obstacleGrid[i-1][j-1])
dp[i][j] = dp[i-1][j]+dp[i][j-1];
return dp[m][n];
}
};
``````

### Unique Paths II LeetCode Solution in JavaScript

``````var uniquePathsWithObstacles = function(OG) {
if (OG[0][0]) return 0
let m = OG.length, n = OG[0].length
let dp = Array.from({length: m}, el => new Uint32Array(n))
dp[0][0] = 1
for (let i = 0; i < m; i++)
for (let j = 0; j < n; j++)
if (OG[i][j] || (!i && !j)) continue
else dp[i][j] = (i ? dp[i-1][j] : 0) + (j ? dp[i][j-1] : 0)
return dp[m-1][n-1]
};
``````

### Unique Paths II LeetCode Solution in Python

``````class Solution:
def uniquePathsWithObstacles(self, OG: List[List[int]]) -> int:
if OG[0][0]: return 0
m, n = len(OG), len(OG[0])
dp = [[0] * n for _ in range(m)]
dp[0][0] = 1
for i in range(m):
for j in range(n):
if OG[i][j] or (i == 0 and j == 0): continue
dp[i][j] = (dp[i-1][j] if i else 0) + (dp[i][j-1] if j else 0)
return dp[m-1][n-1]
``````
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