63 Unique Paths II

This problem is a sequel to 62 Unique Paths - so make sure to check that problem out first.

This problem is the same as it’s predecessor, just that there are some obstacles in a grid. We just have to modify the part where if there’s a obstacle at grid[i][j] we stop our search.

Code

Python3

  def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
 
      @functools.cache
      def compute(i, j):
          if i < 0 or j < 0 or obstacleGrid[i][j] == 1:
              '''
              if the obstacle check is happening - that means both the first two checks for i and j were false. Meaning, by the time obstacle is checked - both i and j are valid and in-bound
              '''
              return 0
 
          if i == 0 and j == 0:
              return 1  # 1 indicates we reached (0,0)
 
          return compute(i-1, j) + compute(i, j-1)
 
      m, n = len(obstacleGrid), len(obstacleGrid[0])
 
      return compute(m-1, n-1)

Big O Analysis

• Runtime

  The runtime complexity here is $O(N)$ same as the previous problem. It’s actually $O(2^N)$ but caching makes it linear time.

• Memory

  The memory usage is $O(N)$ since we use the cache function that uses some dictionary internally for caching/memoising.

— A

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