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63 - Unique Paths II

Difficulty: Medium | Pattern: Dynamic Programming (Grid) | Company tags: Amazon, Google, Microsoft

Problem Statement​

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (grid[0][0]). The robot tries to move to the bottom-right corner (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.

Example 1:

Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
Output: 2  (two paths avoid the center obstacle)

Example 2:

Input: obstacleGrid = [[0,1],[0,0]]
Output: 1

Approach: In-place DP — O(mn), O(1)​

Key insight: dp[i][j] = number of ways to reach (i,j). If obstacle, 0. Otherwise dp[i][j] = dp[i-1][j] + dp[i][j-1]. Modify grid in place.

def uniquePathsWithObstacles(obstacleGrid: list[list[int]]) -> int:
    m, n = len(obstacleGrid), len(obstacleGrid[0])
    
    if obstacleGrid[0][0] == 1 or obstacleGrid[m-1][n-1] == 1:
        return 0
    
    obstacleGrid[0][0] = 1
    
    for i in range(1, m):
        obstacleGrid[i][0] = 0 if obstacleGrid[i][0] == 1 else obstacleGrid[i-1][0]
    
    for j in range(1, n):
        obstacleGrid[0][j] = 0 if obstacleGrid[0][j] == 1 else obstacleGrid[0][j-1]
    
    for i in range(1, m):
        for j in range(1, n):
            if obstacleGrid[i][j] == 1:
                obstacleGrid[i][j] = 0
            else:
                obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1]
    
    return obstacleGrid[m-1][n-1]

Dry Run​

[[0,0,0],
 [0,1,0],
 [0,0,0]]

After filling:

[[1,1,1],
 [1,0,1],
 [1,1,2]]

Answer: 2 ✓

Edge Cases​

  • Start (0,0) is obstacle → return 0
  • End is obstacle → return 0
  • Obstacle in first row/column blocks all paths beyond it → those cells become 0
  • 1x1 grid with no obstacle → 1

Complexity​

  • Time: O(m x n)
  • Space: O(1) (in-place)