63. Unique Paths II

Edited by Charles

Medium · Python · Dynamic Programming

Problem

Read the original problem on LeetCode.

The official page contains the problem statement, examples, and constraints.

Explanation, examples & other approaches

Core idea

Recall 62. Unique Paths for the core idea. Then we need to deal with paths that connect the obstacles.

```python

Boundary Example

obstacleGrid[0][1] = [0, 0, 1, 0, 0]

Initialization

grid[0][0] = 1

Boundary Values

grid[0][1] = grid[0][0] = 1

Original study notes follow, including intermediate snippets and earlier attempts.

Note:

1. Recall 62. Unique Paths for the core idea.
2. 1. consider exceptions: start grid[0][0] and grid[-1][-1] end are obstacles. No feasible path.
   2. grid is initialized as a 2-dimensional array of zeros.
   3. for regular cases, initialize the start grid[0][0] as 1.

   4. iterate boundaries, and the grid components on the boundaries are identical to their previous components if the current position does not have obstacles; once an obstacle pops up, the value passing will stop and all the incoming boundary components will be zeroes. Take below as an example.

      ### Boundary Example
      obstacleGrid[0][1] = [0, 0, 1, 0, 0]
      ### Initialization
      grid[0][0] = 1
      ### Boundary Values
      grid[0][1] = grid[0][0] = 1
      
      ### Since obstacleGrid[0][2] == 1
      ### No value passing from the previous component, i.e. grid[0][i] == grid[0][i - 1]
      grid[0][2] # keeps the original value 0
      
      ### For i >= 3, restore grid[0][i] == grid[0][i - 1] since there is no more obstacles.
      ### However, only 0 can be passed since there is an obstacle at [0][2].
      

   Then we need to deal with paths that connect the obstacles.

   1. for any position off the boundaries and has an obstacle, simply skip it since no path should be connect to this position.

Study record

Field	Value
Attempts	1
Mastery	❚
The Latest Attempt	February 2, 2026

Python solution

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        if obstacleGrid[0][0] == 1 or obstacleGrid[-1][-1] == 1:
            return 0

        m, n = len(obstacleGrid), len(obstacleGrid[0])
        grid = [[0] * n for _ in range(m)]
        grid[0][0] = 1

        for i in range(1, m):
            if obstacleGrid[i][0] == 0:
                grid[i][0] = grid[i - 1][0]

        for j in range(1, n):
            if obstacleGrid[0][j] == 0:
                grid[0][j] = grid[0][j - 1]

        for i in range(1, m):
            for j in range(1, n):
                if obstacleGrid[i][j] == 1:
                    continue

                grid[i][j] = grid[i - 1][j] + grid[i][j - 1]

        return grid[-1][-1]

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