Find element with the maximum set bits in an array - GeeksforGeeks

26 Nov, 2021
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++
Java
Python3
C#
PHP
Javascript
// 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 bitsint 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 codeint 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 approachclass 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 bitsdef 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 codeif __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 approachusing 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 bitsfunction 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 bitsfunction 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>
7
 
Time Complexity: O(n)
Auxiliary Space: O(1)
Rajput-Ji
ankthon
princiraj1992
SURENDRA_GANGWAR
rishavmahato348
subhammahato348
Arrays
Bit Magic
Mathematical
Arrays
Mathematical
Bit Magic
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