Sunday, October 6, 2024
Google search engine
HomeData Modelling & AISum of elements from an array having even parity

Sum of elements from an array having even parity

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)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

RELATED ARTICLES

Most Popular

Recent Comments