> For the complete documentation index, see [llms.txt](https://coding-9.gitbook.io/untitled/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://coding-9.gitbook.io/untitled/dynamic-programming/dp-on-grids/63.-unique-paths-ii.md).

# 63. Unique Paths II

## Problem Statement

<br>

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`.

&#x20;

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/11/04/robot1.jpg)

<pre><code><strong>Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
</strong><strong>Output: 2
</strong><strong>Explanation: There is one obstacle in the middle of the 3x3 grid above.
</strong>There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right
</code></pre>

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/11/04/robot2.jpg)

<pre><code><strong>Input: obstacleGrid = [[0,1],[0,0]]
</strong><strong>Output: 1
</strong></code></pre>

&#x20;

**Constraints:**

* `m == obstacleGrid.length`
* `n == obstacleGrid[i].length`
* `1 <= m, n <= 100`
* `obstacleGrid[i][j]` is `0` or `1`.

## Intuition

```
If there is an obstacle, 

Dont write any condition

else similar to PAths1
```

### Links

<https://leetcode.com/problems/unique-paths-ii/description/>

### Video Links

<https://www.youtube.com/watch?v=TmhpgXScLyY&ab_channel=takeUforward>

### Approach 1:

```
Memoization
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    int path(vector<vector<int>>& arr, int i, int j, vector<vector<int>> &dp){
        if(i==0 and j==0)
            return arr[0][0] == 0 ? 1 : 0;

        if(i<0 or j<0)
            return 0;

        if(dp[i][j] != -1)
            return dp[i][j];

        int l = 0, r = 0;

        if(arr[i][j] == 0){
            if( i>0 and arr[i-1][j] == 0)
                l = path(arr,i-1,j,dp);

            if(j>0 and arr[i][j-1] == 0)
                r = path(arr,i,j-1,dp);
        }
        return dp[i][j] = l+r;

    }

    int uniquePathsWithObstacles(vector<vector<int>>& arr) {
        int m = arr.size();
        int n = arr[0].size();

        vector<vector<int>> dp(m,vector<int>(n,-1));

        return path(arr,m-1,n-1,dp);
    }
};
```

{% endcode %}

### Approach 2:

```
Tabulation
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& arr) {
        int m = arr.size();
        int n = arr[0].size();

        vector<vector<int>> dp(m,vector<int>(n,-1));
        dp[0][0] = arr[0][0] == 0 ? 1 : 0;

        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                if(i==0 and j==0)   continue;

                int l = 0, r = 0;

                if(arr[i][j] == 0){
                    if(i>0 and arr[i-1][j] == 0)
                        l = dp[i-1][j];

                    if(j>0 and arr[i][j-1] == 0)
                        r = dp[i][j-1];
                }
                dp[i][j] = l+r;
            }
        }
        

        return dp[m-1][n-1];
    }
};
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###