LeetCode - Unique Paths II

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.

 

My Solution:

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        dp = [[0] * n for _ in range(m)]
        dp[0][0] = 1 if obstacleGrid[0][0] == 0 else 0

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