Leetcode 63: Unique Paths II

https://leetcode.com/problems/unique-paths-ii/

Unique Paths II

Total Accepted: 47668 Total Submissions: 169510 Difficulty: Medium

Follow up for “Unique Paths”:

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3×3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

 

My Code:

class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        if len(obstacleGrid)==0 or len(obstacleGrid[0])==0:
            return 0
        
        ways = [0] * len(obstacleGrid[0])        
        ways[0] = 1 
        for j in xrange(0, len(obstacleGrid)):
            for i in xrange(0, len(obstacleGrid[0])):
                if obstacleGrid[j][i]:
                    ways[i] = 0
                elif i:  # only handle cells after the first column
                    ways[i] = ways[i] + ways[i-1] 
        
        return ways[-1]

 

Idea:

Following my previous post, you only need 1D array to store intermediate number of ways reaching `cell[j][i]`. Other than that, the only difference is that you need to handle obstacle. It is intuitive to do that though: you set `ways[i]=0` if you see an obstacle, i.e., no way can reach a cell without obstacle.

 

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