**source:**https://leetcode.com/problems/unique-paths-ii/**C/C++**

**Solution to LeetCode**problem

**63**.

**Unique Paths II**.

## 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[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 test cases are generated so that the answer will be less than or equal to `2 * 10`

.^{9}

## Examples

**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:

- Right -> Right -> Down -> Down
- 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

Same solution as for **problem 62** with two differences:

- Check if the goal position has an obstacle.
- Need to check if the board/grid has a
`1`

at the current position.

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class Solution {
private:
vector<vector<int>> dp;
int paths(vector<vector<int>>& grid, int r, int c) {
if (r+1 == grid.size() && c+1 == grid[0].size())
return 1;
if (r == grid.size() || c == grid[0].size())
return 0;
if (grid[r][c] == 1)
return 0;
if (dp[r][c] != 0)
return dp[r][c];
dp[r][c] = 0;
dp[r][c] += paths(grid, r + 1, c);
dp[r][c] += paths(grid, r, c + 1);
return dp[r][c];
}
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
if (obstacleGrid[obstacleGrid.size() - 1][obstacleGrid[0].size() - 1] == 1)
return 0;
dp = vector(obstacleGrid.size(), vector<int>(obstacleGrid[0].size(), 0));
return paths(obstacleGrid, 0, 0);
}
};