Leetcode: Unique Paths 2

Problem Description

Problem definition is taken from leetcode.

• Unique Paths 2

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).

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

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

For Unique Paths 1 problem visit Unique Paths 1

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

Solution

Solution is similar to Unique Paths 1.

In order not to confuse with obstacle values, target value is set to 2. As a result each path is counted twice and in the end result is divided by 2. Also obstacle values are not propagated.

No extra space is used. Time complexity is O(mxn).

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        grid = obstacleGrid
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])
        
        if grid[m-1][n-1]==1 or grid[0][0]==1:
            return 0
        
        grid[m-1][n-1]=2
        for i in range(m-1, -1, -1):
            for j in range(n-1, -1, -1):
                self.step(grid, i, j)
        return int(grid[0][0]/2)
    
    def step(self, grid, i,j):
        count = grid[i][j]
        if count%2==1:
            return 
        
        if j>0:
            grid[i][j-1]+=count
        if i>0:
            grid[i-1][j]+=count