Grokking the Coding Interview: Patterns for Coding Questions
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Problem Challenge 3: Count of Structurally Unique Binary Search Trees (hard)

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A

Anonymous

· 2 years ago


class Solution:
  def countTrees(self, n):
    
    def solve(start, end):
      if start > end:
        return 1
        
      count = 0 
      
      for i in range(start, end + 1):
        lefts = solve(start, i - 1)
        rights = solve(i + 1, end)

        count += lefts * rights
      
      return count

    return solve(1, n)
A

Anonymous

· 3 years ago

Could you please explain further why it is O(N^2) time complexity when we use memoization? Thank you!

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A

Anonymous

· 4 years ago

time limit exceeded on leetcode

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Anonymous

· 4 years ago

Why is the space complexity for non memoized version 2^N? It appears to me that the space required is used by the recursion call stack which is at most the depth of n - 1?

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