63. Unique Paths II

Medium

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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 1 : Using Dictionary

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        self.grid = obstacleGrid
        self.d = {}
        return self.dfs(0,0)
    
    def dfs(self, row, col):
        if (row,col) in self.d:
            return self.d[(row,col)]
        if row >= len(self.grid) or col >= len(self.grid[0]) or self.grid[row][col] == 1:
            return 0
        if row == len(self.grid)-1 and col == len(self.grid[0])-1:
            return 1
        val = self.dfs(row+1, col) + self.dfs(row, col+1)
        self.d[(row,col)] = val
        return val

Solution 2 : With In-Built @lru_cache

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        self.grid = obstacleGrid
        return self.dfs(0,0)
    
    @lru_cache
    def dfs(self, row, col):
        if row >= len(self.grid) or col >= len(self.grid[0]) or self.grid[row][col] == 1:
            return 0
        if row == len(self.grid)-1 and col == len(self.grid[0])-1:
            return 1
        dfs1 = self.dfs(row+1, col)
        dfs2 = self.dfs(row, col+1)
        return dfs1 + dfs2