Unique Paths II

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:

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

 

Constraints:

  • m == obstacleGrid.length
  • n == obstacleGrid[i].length
  • 1 <= m, n <= 100
  • obstacleGrid[i][j] is 0 or 1.
SOLUTION:
class Solution:
    memo = {}
    
    def uniquePathsRec(self, m, n, i, j, obstacleGrid):
        if i == m - 1 and j == n - 1:
            if obstacleGrid[i][j] != 1:
                return 1
            return 0
        if obstacleGrid[i][j] == 1:
            return 0
        if (i, j) not in self.memo:
            self.memo[(i, j)] = 0
            if i + 1 < m:
                self.memo[(i, j)] += self.uniquePathsRec(m, n, i + 1, j, obstacleGrid)
            if j + 1 < n:
                self.memo[(i, j)] += self.uniquePathsRec(m, n, i, j + 1, obstacleGrid)
        return self.memo[(i , j)]  
            
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        self.memo = {}
        return self.uniquePathsRec(len(obstacleGrid), len(obstacleGrid[0]), 0, 0, obstacleGrid)

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