Unique Paths II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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.

C++ Unique Paths II LeetCode Solution

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];
    }
};

Java Unique Paths II LeetCode Solution

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];
}
 

Python 3 Unique Paths II LeetCode Solution

def uniquePathsWithObstacles1(self, obstacleGrid):
    if not obstacleGrid:
        return 
    r, c = len(obstacleGrid), len(obstacleGrid[0])
    dp = [[0 for _ in xrange(c)] for _ in xrange(r)]
    dp[0][0] = 1 - obstacleGrid[0][0]
    for i in xrange(1, r):
        dp[i][0] = dp[i-1][0] * (1 - obstacleGrid[i][0])
    for i in xrange(1, c):
        dp[0][i] = dp[0][i-1] * (1 - obstacleGrid[0][i])
    for i in xrange(1, r):
        for j in xrange(1, c):
            dp[i][j] = (dp[i][j-1] + dp[i-1][j]) * (1 - obstacleGrid[i][j])
    return dp[-1][-1]

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Sliding Window-73
Linked List-69
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