Python Solution:

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:

        m = len(obstacleGrid)
        n = len(obstacleGrid[0])
        #1. The first cell
        #2. The first column
        #3. The first row
        #4. The cells except for the first 3 cases

        # If the starting cell has an obstacle, there would be no path to the destination.
        if obstacleGrid[0][0] == 1:
            return 0
        else:
            obstacleGrid[0][0] = 1

        # First column
        for i in range(1,m):
            if obstacleGrid[i][0] == 1:
                obstacleGrid[i][0] = 0
            else:
                obstacleGrid[i][0]=obstacleGrid[i-1][0]

        # First row        
        for j in range(1, n):
            if obstacleGrid[0][j] == 1:
                obstacleGrid[0][j] = 0
            else:
                obstacleGrid[0][j]=obstacleGrid[0][j-1]

        # Starting from cell(1,1) fill up the values
        # No. of unique paths of reaching cell[i][j] = cell[i - 1][j] + cell[i][j - 1]
        for i in range(1,m):
            for j in range(1,n):
                if obstacleGrid[i][j] == 0:
                    obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1]
                else:
                    obstacleGrid[i][j] = 0

        # Return value stored in rightmost bottommost cell. That is the destination.            
        return obstacleGrid[m-1][n-1]
    
    #Time complixity: O(mn)
    #Space complexity: O(1)

 

Time Complexity: O(mn)

Space Complexity: O(1)

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