Number of Unique BST with a given key | Dynamic Programming

Given N, Find the total number of unique BSTs that can be made using values from 1 to N.

Examples:

Input: n = 3 
Output: 5
For n = 3, preorder traversal of Unique BSTs are:
1. 1 2 3
2. 1 3 2
3. 2 1 3
4. 3 1 2
5. 3 2 1

Input: 4 
Output: 14


In the previous post a O(n) solution has been discussed. In this post we will discuss a solution based on Dynamic Programming. For all possible values of i, consider i as root, then [1….i-1] numbers will fall in the left subtree and [i+1….n] numbers will fall in the right subtree. So, add (i-1)*(n-i) to the answer. The summation of the products will be the answer to the number of unique BST.

Below is the implementation for above approach:

C++

// C++ code to find number of unique BSTs
// Dynamic Programming solution
#include <bits/stdc++.h>
using namespace std;
  
// Function to find number of unique BST
int numberOfBST(int n)
{
  
    // DP to store the number of unique BST with key i
    int dp[n + 1];
    fill_n(dp, n + 1, 0);
  
    // Base case
    dp[0] = 1;
    dp[1] = 1;
  
    // fill the dp table in top-down approach.
    for (int i = 2; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
  
            // n-i in right * i-1 in left
            dp[i] = dp[i] + (dp[i - j] * dp[j - 1]);
        }
    }
  
    return dp[n];
}
  
// Driver Code
int main()
{
    int n = 3;
    cout << "Number of structurally Unique BST with " << 
    n << " keys are : " << numberOfBST(n) << "\n";
  
    return 0;
}

Java

// Java code to find number
// of unique BSTs Dynamic 
// Programming solution
import java.io.*;
import java.util.Arrays;
  
class GFG
{
    static int numberOfBST(int n)
    {
  
    // DP to store the number 
    // of unique BST with key i
    int dp[] = new int[n + 1];
    Arrays.fill(dp, 0);
  
    // Base case
    dp[0] = 1;
    dp[1] = 1;
  
    // fill the dp table in
    // top-down approach.
    for (int i = 2; i <= n; i++) 
    {
        for (int j = 1; j <= i; j++) 
        {
  
            // n-i in right * i-1 in left
            dp[i] = dp[i] + (dp[i - j] * 
                             dp[j - 1]);
        }
    }
  
    return dp[n];
}
  
// Driver Code
public static void main (String[] args) 
{
    int n = 3;
    System.out.println("Number of structurally "
                           "Unique BST with "+ n +
                                  " keys are : "
                                  numberOfBST(n));
}
}
  
// This code is contributed
// by shiv_bhakt.

C#

// C# code to find number
// of unique BSTs Dynamic 
// Programming solution
using System;
  
class GFG
{
    static int numberOfBST(int n)
    {
  
    // DP to store the number 
    // of unique BST with key i
    int []dp = new int[n + 1];
      
    // Base case
    dp[0] = 1;
    dp[1] = 1;
  
    // fill the dp table in
    // top-down approach.
    for (int i = 2; i <= n; i++) 
    {
        for (int j = 1; j <= i; j++) 
        {
  
            // n-i in right * i-1 in left
            dp[i] = dp[i] + (dp[i - j] * 
                             dp[j - 1]);
        }
    }
    return dp[n];
}
  
// Driver Code
public static void Main () 
{
    int n = 3;
    Console.Write("Number of structurally "
                      "Unique BST with "+ n +
                             " keys are : "
                             numberOfBST(n));
}
}
  
// This code is contributed
// by shiv_bhakt.

PHP

<?php
// PHP code to find number 
// of unique BSTs Dynamic
// Programming solution
  
// Function to find number 
// of unique BST
function numberOfBST($n)
{
  
    // DP to store the number 
    // of unique BST with key i
    $dp = array($n + 1);
    for($i = 0; $i <= $n + 1; $i++)
    $dp[$i] = 0;
  
    // Base case
    $dp[0] = 1;
    $dp[1] = 1;
  
    // fill the dp table 
    // in top-down approach.
    for ($i = 2; $i <= $n; $i++) 
    {
        for ($j = 1; $j <= $i; $j++) 
        {
  
            // n-i in right * 
            // i-1 in left
            $dp[$i] += (($dp[$i - $j]) * 
                        ($dp[$j - 1]));
        }
    }
  
    return $dp[$n];
}
  
// Driver Code
$n = 3;
echo "Number of structurally "
           "Unique BST with "
          $n , " keys are : "
              numberOfBST($n) ;
  
// This code is contributed
// by shiv_bhakt.
?>

Output:

Number of structurally Unique BST with 3 keys are : 5

Time Complexity: O(n2)



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Improved By : shiv_bhakt




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