[Solved] 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.

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Question

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:

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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:

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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.

Python Solution

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m = len(obstacleGrid[0])
        n = len(obstacleGrid)

        for i in range(n):
            for j in range(m):
                if obstacleGrid[i][j]==0:
                    obstacleGrid[i][j]=1
                else:
                    obstacleGrid[i][j]=0

        for j in range(m):
            if obstacleGrid[0][j]==0 and j!=m-1:
                for k in range(j,m):
                    obstacleGrid[0][k]=0
        
        for i in range(n):
            if obstacleGrid[i][0]==0 and i!=n-1:
                for k in range(i,n):
                    obstacleGrid[k][0]=0

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

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