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Sum of elements from an array having even parity

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Given an array arr[], the task is to calculate the sum of the elements from the given array which has even parity i.e. the number of set bits is even using bitwise operator.

Examples:  

Input: arr[] = {2, 4, 3, 5, 9} 
Output: 17 
Only 3(0011), 5(0101) and 9(1001) have even parity 
So 3 + 5 + 9 = 17

Input: arr[] = {1, 5, 4, 16, 10} 
Output: 15 

Approach: Initialize a variable sum = 0 and traverse the array from 0 to n – 1 while counting the number of set bits in arr[i] using Brian Kernighan’s Algorithm. If the count is even then update sum = sum + arr[i]. Print the sum in the end.

Below is the implementation of the above approach: 

C++




// C++ program to find the sum of the elements
// from an array which have even parity
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if x has even parity
bool checkEvenParity(int x)
{
    // We basically count set bits
    int parity = 0;
    while (x != 0) {
        x = x & (x - 1);
        parity++;
    }
 
    if (parity % 2 == 0)
        return true;
    else
        return false;
}
 
// Function to return the sum of the elements
// from an array which have even parity
long sumlist(int a[], int n)
{
    long sum = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If a[i] has even parity
        if (checkEvenParity(a[i]))
            sum += a[i];
    }
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 4, 3, 5, 9 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << sumlist(arr, n);
    return 0;
}


Java




// Java program to find the sum of the elements
// from an array which have even parity
 
import java.io.*;
 
class GFG {
 
// Function that returns true if x has even parity
static boolean checkEvenParity(int x)
{
    // We basically count set bits
    int parity = 0;
    while (x != 0) {
        x = x & (x - 1);
        parity++;
    }
 
    if (parity % 2 == 0)
        return true;
    else
        return false;
}
 
// Function to return the sum of the elements
// from an array which have even parity
static long sumlist(int a[], int n)
{
    long sum = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If a[i] has even parity
        if (checkEvenParity(a[i]))
            sum += a[i];
    }
    return sum;
}
 
// Driver code
 
    public static void main (String[] args) {
            int arr[] = { 2, 4, 3, 5, 9 };
 
    int n =arr.length;
 
    System.out.println(sumlist(arr, n));
    }
}
// This code is contributed by  inder_verma..


Python3




# Python3 program to find the sum of the elements
# from an array which have even parity
 
# Function that returns true if x
# has even parity
def checkEvenParity(x):
     
    # We basically count set bits
    parity = 0
    while (x != 0):
        x = x & (x - 1)
        parity += 1
 
    if (parity % 2 == 0):
        return True
    else:
        return False
 
# Function to return the sum of the elements
# from an array which have even parity
def sumlist(a, n):
    sum = 0
    for i in range(n):
         
        # If a[i] has even parity
        if (checkEvenParity(a[i])):
            sum += a[i]
    return sum
 
# Driver code
if __name__ == '__main__':
    arr = [ 2, 4, 3, 5, 9 ]
    n = len(arr)
    print(sumlist(arr, n))
 
# This code is contributed by 29AjayKumar.


C#




// C# program to find the sum of the elements
// from an array which have even parity
 
using System ;
 
class GFG {
 
// Function that returns true if x has even parity
static bool checkEvenParity(int x)
{
    // We basically count set bits
    int parity = 0;
    while (x != 0) {
        x = x & (x - 1);
        parity++;
    }
 
    if (parity % 2 == 0)
        return true;
    else
        return false;
}
 
// Function to return the sum of the elements
// from an array which have even parity
static long sumlist(int []a, int n)
{
    long sum = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If a[i] has even parity
        if (checkEvenParity(a[i]))
            sum += a[i];
    }
    return sum;
}
 
// Driver code
 
    public static void Main () {
            int []arr = { 2, 4, 3, 5, 9 };
 
    int n =arr.Length;
 
    Console.WriteLine(sumlist(arr, n));
    }
    // This code is contributed by Ryuga
}


PHP




<?php
// PHP program to find the sum of the
// elements from an array which have
// even parity
 
// Function that returns true
// if x has even parity
function checkEvenParity($x)
{
    // We basically count set bits
    $parity = 0;
    while ($x != 0)
    {
        $x = ($x & ($x - 1));
        $parity++;
    }
 
    if ($parity % 2 == 0)
        return true;
    else
        return false;
}
 
// Function to return the sum of the elements
// from an array which have even parity
function sumlist($a, $n)
{
    $sum = 0;
 
    for ($i = 0; $i < $n; $i++)
    {
 
        // If a[i] has even parity
        if (checkEvenParity($a[$i]))
            $sum += $a[$i];
    }
    return $sum;
}
 
// Driver code
$arr = array( 2, 4, 3, 5, 9 );
$n = sizeof($arr);
echo sumlist($arr, $n);
     
// This code is contributed by ajit.
?>


Javascript




<script>
 
// Javascript program to find the sum of the elements
// from an array which have even parity
 
// Function that returns true if x has even parity
function checkEvenParity(x)
{
     
    // We basically count set bits
    let parity = 0;
    while (x != 0)
    {
        x = x & (x - 1);
        parity++;
    }
   
    if (parity % 2 == 0)
        return true;
    else
        return false;
}
 
// Function to return the sum of the elements
// from an array which have even parity
function sumlist(a, n)
{
    let sum = 0;
 
    for(let i = 0; i < n; i++)
    {
         
        // If a[i] has even parity
        if (checkEvenParity(a[i]))
            sum += a[i];
    }
    return sum;
}
 
// Driver code
let arr = [ 2, 4, 3, 5, 9 ];
let n = arr.length;
 
document.write(sumlist(arr, n));
 
// This code is contributed by unknown2108
 
</script>


Output

17

Time complexity: O(nlogn)
Auxiliary space: O(1)



Last Updated : 10 Nov, 2022
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