Skip to content

63. Unique Paths II

Difficulty Topics

Description

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

unique-paths-ii.py
class Solution:
    def uniquePathsWithObstacles(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])
        dp = [[0] * cols for _ in range(rows)]
        dp[0][0] = 1 - grid[0][0]

        for j in range(1, cols):
            dp[0][j] += dp[0][j - 1] * (1 - grid[0][j])

        for i in range(1, rows):
            dp[i][0] += dp[i - 1][0] * (1 - grid[i][0])

        for i in range(1, rows):
            for j in range(1, cols):
                dp[i][j] += (dp[i - 1][j] + dp[i][j - 1]) * (1 - grid[i][j])

        return dp[-1][-1]