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Find element with the maximum set bits in an array

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  • Difficulty Level : Easy
  • Last Updated : 26 Nov, 2021
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Given an array arr[]. The task is to find an element from arr[] which has the maximum count of set bits.
Examples: 
 

Input: arr[] = {10, 100, 1000, 10000} 
Output: 1000 
Binary(10) = 1010 (2 set bits) 
Binary(100) = 1100100 (3 set bits) 
Binary(1000) = 1111101000 (6 set bits) 
Binary(10000) = 10011100010000 (5 set bits)
Input: arr[] = {3, 2, 4, 7, 1, 10, 5, 8, 9, 6} 
Output:
 

 

Approach: Traverse the array and find the count of set bits in the current element and find the element with the maximum number of set bits.
Below is the implementation of the above approach:
 

C++




// C++ implementation of the approach
 
#include <bits/stdc++.h>
using namespace std; ;
 
// Function to return the element from the array
// which has the maximum set bits
int maxBitElement(int arr[], int n)
{
 
    // To store the required element and
    // the maximum set bits so far
    int num = 0, max = -1;
 
    for (int i = 0; i < n; i++) {
 
        // Count of set bits in
        // the current element
        int cnt = __builtin_popcount(arr[i]);
 
        // Update the max
        if (cnt > max) {
            max = cnt;
            num = arr[i];
        }
    }
    return num;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 2, 4, 7, 1, 10, 5, 8, 9, 6 };
    int n = sizeof(arr)/ sizeof(arr[0]) ;
     
    cout << maxBitElement(arr, n) << endl;
     
return 0 ;
 
// This code is contributed by AnkitRai01
}


Java




// Java implementation of the approach
class GFG {
 
    // Function to return the element from the array
    // which has the maximum set bits
    static int maxBitElement(int arr[], int n)
    {
 
        // To store the required element and
        // the maximum set bits so far
        int num = 0, max = -1;
 
        for (int i = 0; i < n; i++) {
 
            // Count of set bits in
            // the current element
            int cnt = Integer.bitCount(arr[i]);
 
            // Update the max
            if (cnt > max) {
                max = cnt;
                num = arr[i];
            }
        }
        return num;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 3, 2, 4, 7, 1, 10, 5, 8, 9, 6 };
        int n = arr.length;
        System.out.print(maxBitElement(arr, n));
    }
}


Python3




# Python 3 implementation of the approach
 
# Function to return the element from the array
# which has the maximum set bits
def maxBitElement(arr, n):
     
    # To store the required element and
    # the maximum set bits so far
    num = 0
    max = -1
 
    for i in range(n):
         
        # Count of set bits in
        # the current element
        cnt = bin(arr[i]).count('1')
 
        # Update the max
        if (cnt > max):
            max = cnt
            num = arr[i]
    return num
 
# Driver code
if __name__ == '__main__':
    arr = [3, 2, 4, 7, 1,
          10, 5, 8, 9, 6]
    n = len(arr)
     
    print(maxBitElement(arr, n))
     
# This code is contributed by
# Surendra_Gangwar


C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the element from the array
    // which has the maximum set bits
    static int maxBitElement(int []arr, int n)
    {
 
        // To store the required element and
        // the maximum set bits so far
        int num = 0, max = -1;
 
        for (int i = 0; i < n; i++)
        {
 
            // Count of set bits in
            // the current element
            int cnt = BitCount(arr[i]);
 
            // Update the max
            if (cnt > max)
            {
                max = cnt;
                num = arr[i];
            }
        }
        return num;
    }
    static int BitCount(int n)
    {
        int count = 0;
        while (n != 0)
        {
            count++;
            n &= (n - 1);
        }
        return count;
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 3, 2, 4, 7, 1, 10, 5, 8, 9, 6 };
        int n = arr.Length;
        Console.Write(maxBitElement(arr, n));
    }
}
 
// This code contributed by Rajput-Ji


PHP




<?php
// PHP implementation of the approach
 
// Function to return the element from
// the array which has the maximum set bits
function maxBitElement($arr, $n)
{
 
    // To store the required element and
    // the maximum set bits so far
    $num = 0; $max = -1;
 
    for ($i = 0; $i < $n; $i++)
    {
 
        // Count of set bits in
        // the current element
        $cnt = BitCount($arr[$i]);
 
        // Update the max
        if ($cnt > $max)
        {
            $max = $cnt;
            $num = $arr[$i];
        }
    }
    return $num;
}
 
function BitCount($n)
{
    $count = 0;
    while ($n != 0)
    {
        $count++;
        $n &= ($n - 1);
    }
    return $count;
}
     
// Driver code
$arr = array(3, 2, 4, 7, 1, 10, 5, 8, 9, 6 );
$n = count($arr);
echo(maxBitElement($arr, $n));
 
// This code contributed by PrinciRaj1992
?>


Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the element from the array
// which has the maximum set bits
function maxBitElement(arr, n)
{
 
    // To store the required element and
    // the maximum set bits so far
    let num = 0, max = -1;
 
    for (let i = 0; i < n; i++) {
 
        // Count of set bits in
        // the current element
        let cnt = BitCount(arr[i]);
 
        // Update the max
        if (cnt > max) {
            max = cnt;
            num = arr[i];
        }
    }
    return num;
}
 
function BitCount(n)
{
    let count = 0;
    while (n != 0)
    {
        count++;
        n &= (n - 1);
    }
    return count;
}
     
// Driver code
    let arr = [ 3, 2, 4, 7, 1, 10, 5, 8, 9, 6 ];
    let n = arr.length;
     
    document.write(maxBitElement(arr, n));
 
</script>


Output: 

7

 

Time Complexity: O(n)

Auxiliary Space: O(1)


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