Python Solution with comments

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    class Solution:
        def uniquePathsWithObstacles(self, obstacleGrid):
            :type obstacleGrid: List[List[int]]
            :rtype: int
            :type m: int
            :type n: int
            :rtype: int
            # Here logic is same  current paths = top_paths + left_paths but the twist is now if we have block at that location then we need to set it to zero in order to block path from top and left
            m = len(obstacleGrid)
            n = len(obstacleGrid[0])
            memo = [[0]*n for elem in range(m)]
            # If source is blocked then no paths available else 1 for source
            if(obstacleGrid[0][0] == 1):
                return 0
                memo[0][0] = 1
            # If i == 0 and j == 0  continue for not setting up memo[i][j] again to zero
            # if obstacle present then set the current pos as zero else curr = left + top
            # and we are done
            for i in range(m):
                for j in range(n):
                    if(i == 0 and j == 0):
                    if(obstacleGrid[i][j] != 1):
                        memo[i][j] = memo[i-1][j]+memo[i][j-1]

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