# LC 63. Unique Paths II

### [Problem link](https://leetcode.com/problems/unique-paths-ii/)

###### tags: `leedcode` `python` `c++` `medium` `DP`

You are given an <code>m x n</code> integer array <code>grid</code>. There is a robot initially located at the **top-left corner** (i.e., <code>grid[0][0]</code>). The robot tries to move to the **bottom-right corner** (i.e., <code>grid[m - 1][n - 1]</code>). The robot can only move either down or right at any point in time.

An obstacle and space are marked as <code>1</code> or <code>0</code> respectively in <code>grid</code>. 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 <code>2 * 10⁹</code>.

**Example 1:**
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/robot1.jpg" style="width: 242px; height: 242px;" />

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:**
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/robot2.jpg" style="width: 162px; height: 162px;" />

Input: obstacleGrid = [[0,1],[0,0]]
Output: 1

**Constraints:**

- <code>m == obstacleGrid.length</code>
- <code>n == obstacleGrid[i].length</code>
- <code>1 <= m, n <= 100</code>
- <code>obstacleGrid[i][j]</code> is <code>0</code> or <code>1</code>.

## Solution 1 - DP
#### Python

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])
        dp = [[0 for _ in range(n)] for _ in range(m)]

        if obstacleGrid[0][0] == 1:
            return 0

        for i in range(m):
            if obstacleGrid[i][0] == 1:
                break
            dp[i][0] = 1

        for i in range(n):
            if obstacleGrid[0][i] == 1:
                break
            dp[0][i] = 1

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

        return dp[-1][-1]

#### C++

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        vector<vector<int>> dp(m, vector<int>(n, 0));

        for (int i = 0; i < m; i++) {
            if (obstacleGrid[i][0] == 1) {
                break;
            }
            dp[i][0] = 1;
        }
        for (int i = 0; i < n; i++) {
            if (obstacleGrid[0][i] == 1) {
                break;
            }
            dp[0][i] = 1;
        }

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    continue;
                }
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
};

## Solution 2 - DP
#### Python

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])
        dp = [0] * n

        for i in range(n):
            if obstacleGrid[0][i] == 1:
                break
            dp[i] = 1

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

        return dp[-1]

#### C++

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        vector<int> dp(n, 0);

        for (int i = 0; i < n; i++) {
            if (obstacleGrid[0][i] == 1) {
                break;
            }
            dp[i] = 1;
        }

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

>### Complexity
>m = obstacleGrid.length
>n = obstacleGrid[i].length
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(mn) | O(mn) |
>| Solution 2 | O(mn) | O(n) |

## Note
solutions 1: 2 dimensional dp
solutions 2: 1 dimensional dp