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]
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