Solution Review: Problem Challenge 3

We'll cover the following

- - Count of Structurally Unique Binary Search Trees (hard)
- - Solution
  - Code
- - - Time complexity
    - Space complexity
  - Memoized version

Count of Structurally Unique Binary Search Trees (hard) #

Given a number ‘n’, write a function to return the count of structurally unique Binary Search Trees (BST) that can store values 1 to ‘n’.

Example 1:

Input: 2
Output: 2
Explanation: As we saw in the previous problem, there are 2 unique BSTs storing numbers from 1-2.

Example 2:

Input: 3
Output: 5
Explanation: There will be 5 unique BSTs that can store numbers from 1 to 5.

Solution #

This problem is similar to Structurally Unique Binary Search Trees. Following a similar approach, we can iterate from 1 to ‘n’ and consider each number as the root of a tree and make two recursive calls to count the number of left and right sub-trees.

Code #

Here is what our algorithm will look like:

Java

Python3

C++

import java.util.*;
 
class TreeNode {
  int val;
  TreeNode left;
  TreeNode right;
 
  TreeNode(int x) {
    val = x;
  }
};
 
class CountUniqueTrees {
  public int countTrees(int n) {
    if (n <= 1)
      return 1;
    int count = 0;
    for (int i = 1; i <= n; i++) {
      // making 'i' root of the tree
      int countOfLeftSubtrees = countTrees(i - 1);
      int countOfRightSubtrees = countTrees(n - i);
      count += (countOfLeftSubtrees * countOfRightSubtrees);
    }
    return count;
  }
 
  public static void main(String[] args) {
    CountUniqueTrees ct = new CountUniqueTrees();
    int count = ct.countTrees(3);
    System.out.print("Total trees: " + count);
  }
}

Output

4.638s

Total trees: 5

Time complexity #

The time complexity of this algorithm will be exponential and will be similar to Balanced Parentheses. Estimated time complexity will be $O(n*2^n)$ but the actual time complexity ( $O(4^n/\sqrt{n})$ ) is bounded by the Catalan number and is beyond the scope of a coding interview. See more details here.

Space complexity #

The space complexity of this algorithm will be exponential too, estimated $O(2^n)$ but the actual will be ( $O(4^n/\sqrt{n})$ .

Memoized version #

Our algorithm has overlapping subproblems as our recursive call will be evaluating the same sub-expression multiple times. To resolve this, we can use memoization and store the intermediate results in a HashMap. In each function call, we can check our map to see if we have already evaluated this sub-expression before. Here is the memoized version of our algorithm, please see highlighted changes:

Java

Python3

C++

import java.util.*;
 
class TreeNode {
  int val;
  TreeNode left;
  TreeNode right;
 
  TreeNode(int x) {
    val = x;
  }
};
 
class CountUniqueTrees {
  Map<Integer, Integer> map = new HashMap<>();
 
  public int countTrees(int n) {
    if (map.containsKey(n))
      return map.get(n);
 
    if (n <= 1)
      return 1;
    int count = 0;
    for (int i = 1; i <= n; i++) {
      // making 'i' root of the tree
      int countOfLeftSubtrees = countTrees(i - 1);
      int countOfRightSubtrees = countTrees(n - i);
      count += (countOfLeftSubtrees * countOfRightSubtrees);
    }
    map.put(n, count);
    return count;
  }
 
  public static void main(String[] args) {
    CountUniqueTrees ct = new CountUniqueTrees();
    int count = ct.countTrees(3);
    System.out.print("Total trees: " + count);
  }
}
 

Output

1.869s

Total trees: 5

The time complexity of the memoized algorithm will be $O(n^2)$ , since we are iterating from ‘1’ to ‘n’ and ensuring that each sub-problem is evaluated only once. The space complexity will be $O(n)$ for the memoization map.

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Problem Challenge 3

Introduction