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Generate two BSTs from the given array such that maximum height among them is minimum

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  • Difficulty Level : Easy
  • Last Updated : 04 Oct, 2021
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Given an array of n integers where n is greater than 1, the task is to make two Binary Search Tree from the given array (in any order) such that the maximum height among the two trees is minimum possible and print the maximum height.
Examples: 
 

Input: arr[] = {1, 2, 3, 4, 6} 
Output:
 

Input: arr[] = { 74, 25, 23, 1, 65, 2, 8, 99 } 
Output:
 

 

Approach: The aim is to minimize the height of the maximum height binary search tree and to do so we need to divide the array elements equally among both the trees. And since the order doesn’t matter, we can easily choose any element for the first or second binary tree. Now, to minimize the height of the two tree, the trees will have to be almost complete and as equal in heights as possible. And the maximum (minimized) height will be log(n/2) or log(n/2 + 1).
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the log()
int cal_log(int n)
{
    int ans = 0;
 
    while (n) {
        n /= 2;
        ans++;
    }
 
    return ans;
}
 
// Function to return the maximum
// height of the tree
int maximumHeight(int n, int arr[])
{
    int level = cal_log(n / 2 + n % 2);
    return (level - 1);
}
 
// Driven code
int main()
{
    int arr[] = { 1, 2, 3, 4, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maximumHeight(n, arr);
 
    return 0;
}

Java




// Java implementation of the approach
import java.io.*;
 
class GFG
{
 
// Function to calculate the log()
static int cal_log(int n)
{
    int ans = 0;
 
    while (n > 0)
    {
        n /= 2;
        ans++;
    }
 
    return ans;
}
 
// Function to return the maximum
// height of the tree
static int maximumHeight(int n, int arr[])
{
    int level = cal_log(n / 2 + n % 2);
    return (level - 1);
}
 
// Driver code
public static void main (String[] args)
{
    int arr[] = { 1, 2, 3, 4, 6 };
    int n =arr.length;
    System.out.print( maximumHeight(n, arr));
}
}
 
// This code is contributed by anuj_67..

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to calculate the log()
static int cal_log(int n)
{
    int ans = 0;
 
    while (n > 0)
    {
        n /= 2;
        ans++;
    }
 
    return ans;
}
 
// Function to return the maximum
// height of the tree
static int maximumHeight(int n, int []arr)
{
    int level = cal_log(n / 2 + n % 2);
    return (level - 1);
}
 
// Driver code
public static void Main ()
{
    int []arr = { 1, 2, 3, 4, 6 };
    int n =arr.Length;
    Console.WriteLine( maximumHeight(n, arr));
}
}
 
// This code is contributed by anuj_67..

Python3




# Python implementation of the approach
 
# Function to calculate the log()
def cal_log(n):
    ans = 0;
 
    while (n):
        n //= 2;
        ans+=1;
 
 
    return ans;
 
 
# Function to return the maximum
# height of the tree
def maximumHeight(n, arr):
    level = cal_log(n // 2 + n % 2);
    return (level - 1);
 
 
# Driver code
arr = [ 1, 2, 3, 4, 6 ];
n = len(arr);
print(maximumHeight(n, arr));
 
# This code is contributed by Princi Singh

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to calculate the log()
function cal_log(n)
{
    var ans = 0;
 
    while (n) {
        n = parseInt(n/2);
        ans++;
    }
 
    return ans;
}
 
// Function to return the maximum
// height of the tree
function maximumHeight(n, arr)
{
    var level = cal_log(parseInt(n / 2) + n % 2);
    return (level - 1);
}
 
// Driven code
var arr = [ 1, 2, 3, 4, 6 ];
var n = arr.length;
document.write( maximumHeight(n, arr));
 
</script>

Output: 

1

 

Time Complexity: O(log(N))
Auxiliary Space: O(1) 


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