Find element with the maximum set bits in an array
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: 7
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++ 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
}
// C++ implementation of the approach 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 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));
}
}
# 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# 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 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
?>
<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)
