# Unique Paths II Leetcode Solution

In this post, we are going to solve the Unique Paths II Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

## Problem

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 bottomright 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`.

Now, lets see the leetcode solution of Unique Paths II Leetcode Solution.

### Unique Paths II Leetcode Solution in Python

```class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
m = len(obstacleGrid)
n = len(obstacleGrid)
# dp[i][j] := unique paths from (0, 0) to (i - 1, j - 1)
dp = [ * (n + 1) for _ in range(m + 1)]
dp = 1  # Can also set dp = 1

for i in range(1, m + 1):
for j in range(1, n + 1):
if obstacleGrid[i - 1][j - 1] == 0:
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

return dp[m][n]
```

### Unique Paths II Leetcode Solutionin CPP

```class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
const int m = obstacleGrid.size();
const int n = obstacleGrid.size();
// dp[i][j] := unique paths from (0, 0) to (i - 1, j - 1)
vector<vector<long>> dp(m + 1, vector<long>(n + 1, 0));
dp = 1;  // Can also set dp = 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 Java

```class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
final int m = obstacleGrid.length;
final int n = obstacleGrid.length;
// dp[i][j] := unique paths from (0, 0) to (i - 1, j - 1)
long[][] dp = new long[m + 1][n + 1];
dp = 1; // Can also set dp = 1

for (int i = 1; i <= m; ++i)
for (int j = 1; j <= n; ++j)
if (obstacleGrid[i - 1][j - 1] == 0)
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];

return (int) dp[m][n];
}
}
```

Note: This problem Unique Paths II is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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