Physical Address

304 North Cardinal St.
Dorchester Center, MA 02124

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]
Unique Paths II LeetCode Solution Review:

In our experience, we suggest you solve this Unique Paths II LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Unique Paths II LeetCode Solution

Find on LeetCode

Conclusion:

I hope this Unique Paths II LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions

Leave a Reply

Your email address will not be published. Required fields are marked *