Number of elements smaller than root using preorder traversal of a BST

Given a preorder traversal of a BST. The task is to find the number of elements less than root. 

Examples:

Input: preorder[] = {3, 2, 1, 0, 5, 4, 6}
Output: 3

Input: preorder[] = {5, 4, 3, 2, 1}
Output: 4

For a binary search tree, a preorder traversal is of the form: 

root, { elements in left subtree of root }, { elements in right subtree of root }

Simple approach:

1. Traverse the given preorder.
2. Check if the current element is greater than root.
3. If yes then return the indexOfCurrentElement – 1 as the no. elements smaller than root will be all the elements that occurs before the currentelement except root.

Implementation:

C++

// C++ implementation of above approach
#include <iostream>
using namespace std;

// Function to find the first index of the element
// that is greater than the root
int findLargestIndex(int arr[], int n)
{
    int i, root = arr[0];

    // Traverse the given preorder
    for(i = 0; i < n-1; i++)
    {
        // Check if the number is greater than root
        // If yes then return that index-1
        if(arr[i] > root)
           return i-1;
    }

}

// Driver Code
int main()
{
    int preorder[] = {3, 2, 1, 0, 5, 4, 6};
    int n = sizeof(preorder) / sizeof(preorder[0]);

     cout << findLargestIndex(preorder, n);

    return 0;
}

Java

// Java implementation of
// above approach

class GFG
{
// Function to find the first
// index of the element that
// is greater than the root
static int findLargestIndex(int arr[],
                            int n)
{
    int i, root = arr[0];

    // Traverse the given preorder
    for(i = 0; i < n - 1; i++)
    {
        // Check if the number is
        // greater than root
        // If yes then return
        // that index-1
        if(arr[i] > root)
        return i-1;
    }
    return 0;
}

// Driver Code
public static void main(String ags[])
{
    int preorder[] = {3, 2, 1, 0, 5, 4, 6};
    int n = preorder.length;

    System.out.println(findLargestIndex(preorder, n));
}
}

// This code is contributed
// by Subhadeep Gupta

Python3

# Python3 implementation of above approach

# Function to find the first index of
# the element that is greater than the root
def findLargestIndex(arr, n):

    i, root = arr[0], arr[0];

    # Traverse the given preorder
    for i in range(0, n - 1):

        # Check if the number is greater than
        # root. If yes then return that index-1
        if(arr[i] > root):
            return i - 1;

# Driver Code
preorder= [3, 2, 1, 0, 5, 4, 6];
n = len(preorder)

print(findLargestIndex(preorder, n));

# This code is contributed
# by Akanksha Rai

C#

// C# implementation of above approach
using System;

class GFG
{

// Function to find the first
// index of the element that
// is greater than the root
static int findLargestIndex(int []arr,
                            int n)
{
    int i, root = arr[0];

    // Traverse the given preorder
    for(i = 0; i < n - 1; i++)
    {
        // Check if the number is
        // greater than root. If yes
        // then return that index-1
        if(arr[i] > root)
        return i - 1;
    }
    return 0;
}

// Driver Code
static public void Main()
{
    int []preorder = {3, 2, 1, 0, 5, 4, 6};
    int n = preorder.Length;

    Console.WriteLine(findLargestIndex(preorder, n));
}
}

// This code is contributed
// by Subhadeep Gupta

PHP

<?php
// PHP implementation of above approach

// Function to find the first index of
// the element that is greater than the root
function findLargestIndex( $arr, $n)
{
    $i; $root = $arr[0];

    // Traverse the given preorder
    for($i = 0; $i < $n - 1; $i++)
    {
        // Check if the number is greater than
        // root. If yes, then return that index-1
        if($arr[$i] > $root)
        return $i - 1;
    }

}

// Driver Code
$preorder = array(3, 2, 1, 0, 5, 4, 6);
$n = count($preorder);
echo findLargestIndex($preorder, $n);

// This code is contributed
// by 29AjayKumar
?>

Javascript

<script>

// JavaScript implementation of above approach

// Function to find the first index of the element
// that is greater than the root
function findLargestIndex(arr, n)
{
    var i, root = arr[0];

    // Traverse the given preorder
    for(i = 0; i < n-1; i++)
    {
        // Check if the number is greater than root
        // If yes then return that index-1
        if(arr[i] > root)
           return i-1;
    }

}

// Driver Code
var preorder = [3, 2, 1, 0, 5, 4, 6];
var n = preorder.length;
document.write( findLargestIndex(preorder, n));

</script>

Output

3

Time complexity: O(n)

Efficient approach (Using Binary Search): Here the idea is to make use of an extended form of binary search. 

The steps are as follows:

1. Go to mid. Check if the element at mid is greater than root. If yes then we recurse on the left half of array.
2. Else if the element at mid is lesser than root and element at mid+1 is greater than root we return mid as our answer.
3. Else we recurse on the right half of array to repeat the above steps.

Below is the implementation of the above idea. 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the smaller elements
int findLargestIndex(int arr[], int n)
{
    int root = arr[0], lb = 0, ub = n-1;
    while(lb < ub)
    {

        int mid = (lb + ub)/2;

        // Check if the element at mid
        // is greater than root.
        if(arr[mid] > root)
            ub = mid - 1;
        else
        {
          // if the element at mid is lesser
          //  than root and element at mid+1
          // is greater
           if(arr[mid + 1] > root)
              return mid;
            else lb = mid + 1;
        }
     }
     return lb;
}

// Driver Code
int main()
{
    int preorder[] = {3, 2, 1, 0, 5, 4, 6};
    int n = sizeof(preorder) / sizeof(preorder[0]);

     cout << findLargestIndex(preorder, n);

    return 0;
}

Java

// Java implementation
// of above approach
import java.util.*;

class GFG
{

// Function to count the
// smaller elements
static int findLargestIndex(int arr[],
                            int n)
{
    int root = arr[0],
        lb = 0, ub = n - 1;
    while(lb < ub)
    {

        int mid = (lb + ub) / 2;

        // Check if the element at
        // mid is greater than root.
        if(arr[mid] > root)
            ub = mid - 1;
        else
        {

            // if the element at mid is
            // lesser than root and
            // element at mid+1 is greater
            if(arr[mid + 1] > root)
                return mid;
            else lb = mid + 1;
        }
    }
    return lb;
}

// Driver Code
public static void main(String args[])
{
    int preorder[] = {3, 2, 1, 0, 5, 4, 6};
    int n = preorder.length;

    System.out.println(
           findLargestIndex(preorder, n));
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of above approach

# Function to count the smaller elements
def findLargestIndex(arr, n):
    root = arr[0];
    lb = 0;
    ub = n - 1;
    while(lb < ub):

        mid = (lb + ub) // 2;

        # Check if the element at mid
        # is greater than root.
        if(arr[mid] > root):
            ub = mid - 1;
        else:

            # if the element at mid is lesser
            # than root and element at mid+1
            # is greater
            if(arr[mid + 1] > root):
                return mid;
            else:
                lb = mid + 1;
    return lb;

# Driver Code
preorder = [3, 2, 1, 0, 5, 4, 6];
n = len(preorder);

print(findLargestIndex(preorder, n));

# This code is contributed by mits

C#

// C# implementation
// of above approach
using System;

class GFG
{

// Function to count the
// smaller elements
static int findLargestIndex(int []arr,
                            int n)
{
    int root = arr[0],
        lb = 0, ub = n - 1;
    while(lb < ub)
    {

        int mid = (lb + ub) / 2;

        // Check if the element at
        // mid is greater than root.
        if(arr[mid] > root)
            ub = mid - 1;
        else
        {

            // if the element at mid is
            // lesser than root and
            // element at mid+1 is greater
            if(arr[mid + 1] > root)
                return mid;
            else lb = mid + 1;
        }
    }
    return lb;
}

// Driver Code
public static void Main(String []args)
{
    int []preorder = {3, 2, 1, 0, 5, 4, 6};
    int n = preorder.Length;

    Console.WriteLine(
        findLargestIndex(preorder, n));
}
}

// This code contributed by Rajput-Ji

PHP

<?php
// PHP implementation of above approach

// Function to count the smaller elements
function findLargestIndex($arr, $n)
{
    $root = $arr[0];
    $lb = 0;
    $ub = $n - 1;
    while($lb < $ub)
    {

        $mid = (int)(($lb + $ub) / 2);

        // Check if the element at mid
        // is greater than root.
        if($arr[$mid] > $root)
            $ub = $mid - 1;
        else
        {
            // if the element at mid is lesser
            // than root and element at mid+1
            // is greater
            if($arr[$mid + 1] > $root)
                return $mid;
            else
                $lb = $mid + 1;
        }
    }
    return $lb;
}

// Driver Code
$preorder = array(3, 2, 1, 0, 5, 4, 6);
$n = count($preorder);

echo findLargestIndex($preorder, $n);

// This code is contributed by mits
?>

Javascript

<script>

// JavaScript implementation of above approach

// Function to count the smaller elements
function findLargestIndex(arr, n)
{
    var root = arr[0], lb = 0, ub = n-1;
    while(lb < ub)
    {

        var mid = parseInt((lb + ub)/2);

        // Check if the element at mid
        // is greater than root.
        if(arr[mid] > root)
            ub = mid - 1;
        else
        {
          // if the element at mid is lesser
          //  than root and element at mid+1
          // is greater
           if(arr[mid + 1] > root)
              return mid;
            else lb = mid + 1;
        }
     }
     return lb;
}

// Driver Code
var preorder = [3, 2, 1, 0, 5, 4, 6];
var n = preorder.length;
document.write( findLargestIndex(preorder, n));


</script>

Output

3

Time Complexity: O(logn)