[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.

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:

robot1
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:

robot2
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]
Abhishek Sharma
Abhishek Sharma

Started my Data Science journey in my 2nd year of college and since then continuously into it because of the magical powers of ML and continuously doing projects in almost every domain of AI like ML, DL, CV, NLP.

Articles: 520

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