Count array elements whose highest power of 2 less than or equal to that number is present in the given array

Given an array arr[] consisting of N positive integers, the task is to find the number of array elements whose highest power of 2 less than or equal to that number is present in the array.

Examples:

Input: arr[] = {3, 4, 6, 9}
Output: 2
Explanation:
There are 2 array elements (4 and 6), whose highest power of 2 is less than it (i.e. 4) are present in the array.

Input: arr[] = {3, 9, 10, 8, 1}
Output: 3

Naive Approach: The given problem can be solved by count those elements whose highest power of 2 exists in the given array and that can be found by traversing the array again. After checking for all the elements, print the total count obtained.

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by using an unordered_map to keep the count of visited elements and update the count of resultant elements accordingly. Follow the steps below to solve the given problem:

• Initialize a variable, say count that stores the count of elements whose highest power of 2 less than or equals to that is present in the array.
• Initialize a map M and store the frequency of array elements.
• Traverse the given array and for each element, if the frequency of the highest power of 2 of arr[i] not exceeding the element exists in the map then increment the value of count by 1.
• After completing the above steps, print the value of count as the resultant count of elements.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to count array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
int countElement(int arr[], int N)
{
    // Stores the resultant count
    // of array elements
    int count = 0;

    // Stores frequency of
    // visited array elements
    unordered_map<int, int> m;

    // Traverse the array
    for (int i = 0; i < N; i++) {
        m[arr[i]]++;
    }

    for (int i = 0; i < N; i++) {

        // Calculate log base 2
        // of the element arr[i]
        int lg = log2(arr[i]);

        // Highest power of 2 whose
        // value is at most arr[i]
        int p = pow(2, lg);

        // Increment the count by 1
        if (m[p]) {
            count++;
        }
    }

    // Return the resultant count
    return count;
}

// Driver Code
int main()
{
    int arr[] = { 3, 4, 6, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << countElement(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.HashMap;
import java.io.*;

class GFG{

static int log2(int N)
{

    // Calculate log2 N indirectly
    // using log() method
    int result = (int)(Math.log(N) /
                       Math.log(2));

    return result;
}

// Function to count array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
static int countElement(int arr[], int N)
{

    // Stores the resultant count
    // of array elements
    int count = 0;

    // Stores frequency of
    // visited array elements
   HashMap<Integer, Integer> m = new HashMap<>();

    // Traverse the array
    for(int i = 0; i < N; i++)
    {
        if (m.containsKey(arr[i]))
        {
            m.put(arr[i], m.get(arr[i]) + 1);
        }
        else
        {
            m.put(arr[i], 1);
        }
    }

    for(int i = 0; i < N; i++)
    {

        // Calculate log base 2
        // of the element arr[i]
        int lg = log2(arr[i]);

        // Highest power of 2 whose
        // value is at most arr[i]
        int p = (int)Math.pow(2, lg);

        // Increment the count by 1
        if (m.containsKey(p))
        {
            count++;
        }
    }

    // Return the resultant count
    return count;
}

// Driver Code
public static void main (String[] args)
{
    int arr[] = { 3, 4, 6, 9 };
    int N = arr.length;

    System.out.println(countElement(arr, N));
}
}

// This code is contributed by Potta Lokesh

Python3

# Python program for the above approach
from math import log2

# Function to count array elements
# whose highest power of 2 is less
# than or equal to that number is
# present in the given array
def countElement(arr, N):
    # Stores the resultant count
    # of array elements
    count = 0

    # Stores frequency of
    # visited array elements
    m = {}

    # Traverse the array
    for i in range(N):
        m[arr[i]] = m.get(arr[i], 0) + 1


    for i in range(N):
        # Calculate log base 2
        # of the element arr[i]
        lg = int(log2(arr[i]))

        # Highest power of 2 whose
        # value is at most arr[i]
        p = pow(2, lg)

        # Increment the count by 1
        if (p in m):
            count += 1

    # Return the resultant count
    return count

# Driver Code
if __name__ == '__main__':
    arr= [3, 4, 6, 9]
    N = len(arr)
    print (countElement(arr, N))

# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach

using System;
using System.Collections.Generic;

class GFG{

// Function to count array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
static int countElement(int []arr, int N)
{
    // Stores the resultant count
    // of array elements
    int count = 1;

    // Stores frequency of
    // visited array elements
    Dictionary<int,int> m = new Dictionary<int,int>();

    // Traverse the array
    for (int i = 0; i < N; i++) {
        if(m.ContainsKey(arr[i]))
         m[arr[i]]++;
        else
         m.Add(arr[i],1);
    }

    for(int i = 0; i < N; i++) {

        // Calculate log base 2
        // of the element arr[i]
        int lg = (int)Math.Log(arr[i]);

        // Highest power of 2 whose
        // value is at most arr[i]
        int p = (int)Math.Pow(2, lg);

        // Increment the count by 1
        if (m.ContainsKey(p)) {
            count++;
        }
    }

    // Return the resultant count
    return count;
}

// Driver Code
public static void Main()
{
    int []arr = { 3, 4, 6, 9 };
    int N = arr.Length;
    Console.Write(countElement(arr, N));

}
}

// This code is contributed by bgangwar59.

Javascript

<script>

// Javascript program for the above approach


// Function to count array elements
// whose highest power of 2 is less
// than or equal to that number is
// present in the given array
function countElement(arr, N) {
    // Stores the resultant count
    // of array elements
    let count = 0;

    // Stores frequency of
    // visited array elements
    let m = new Map();

    // Traverse the array
    for (let i = 0; i < N; i++) {
        if (m.has(arr[i])) {
            m.set(arr[i], m.get(arr[i]) + 1)
        } else {
            m.set(arr[i], 1)
        }
    }

    for (let i = 0; i < N; i++) {

        // Calculate log base 2
        // of the element arr[i]
        let lg = Math.floor(Math.log2(arr[i]));

        // Highest power of 2 whose
        // value is at most arr[i]
        let p = Math.pow(2, lg);

        // Increment the count by 1
        if (m.get(p)) {
            count++;
        }
    }

    // Return the resultant count
    return count;
}

// Driver Code

let arr = [3, 4, 6, 9];
let N = arr.length
document.write(countElement(arr, N));

// This code is contributed by _saurabh_jaiswal.
</script>

Output:

2

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