Sum of all even occurring element in an array

Given an array of integers containing duplicate elements. The task is to find the sum of all even occurring elements in the given array. That is the sum of all such elements whose frequency is even in the array.

Examples: 

Input : arr[] = {1, 1, 2, 2, 3, 3, 3}
Output : 6
The even occurring element are 1 and 2 (both occur two times). 
Therefore sum of all 1's in the array = 1+1+2+2 = 6.

Input : arr[] = {10, 20, 30, 40, 40}
Output : 80
Element with even frequency are 40.
Sum = 40.

Approach: 

• Traverse the array and use a unordered_map in C++ to store the frequency of elements of the array such that the key of map is the array element and value is its frequency in the array.
• Then, traverse the map to find the frequency of elements and check if it is even, if it is even, then add this element to sum.

Below is the implementation of the above approach: 

C++

// C++ program to find the sum of all even
// occurring elements in an array

#include <bits/stdc++.h>
using namespace std;

// Function to find the sum of all even
// occurring elements in an array
int findSum(int arr[], int N)
{
    // Map to store frequency of elements
    // of the array
    unordered_map<int, int> mp;

    for (int i = 0; i < N; i++) {
        mp[arr[i]]++;
    }

    // variable to store sum of all
    // even occurring elements
    int sum = 0;

    // loop to iterate through map
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {

        // check if frequency is even
        if (itr->second % 2 == 0)
            sum += (itr->first) * (itr->second);
    }

    return sum;
}

// Driver Code
int main()
{
    int arr[] = { 10, 20, 20, 40, 40 };

    int N = sizeof(arr) / sizeof(arr[0]);

    cout << findSum(arr, N);

    return 0;
}

Java

// Java program to find the sum of all even
// occurring elements in an array

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;

public class GFG {
    public static int element(int[] arr, int n)
    {

        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < n; i++) {
            int count = 0;
            if (map.get(arr[i]) != null) {
                count = map.get(arr[i]);
            }
            map.put(arr[i], count + 1);
        }

        int sum = 0;
        for (Entry<Integer, Integer> entry : map.entrySet()) {
            if (entry.getValue() % 2 == 0) {
                sum += entry.getKey() * entry.getValue();
            }
        }

        return sum;
    }

    public static void main(String[] args)
    {
        int arr[] = { 1, 1, 2, 2, 3, 3, 3 };

        // sum should be = 1+1+2+2=6;
        int n = arr.length;
        System.out.println(element(arr, n));
    }
}

Python3

# Python3 program to find the sum
# of all even occurring elements
# in an array

# Function to find the sum of all even
# occurring elements in an array
def findSum(arr, N):

    # Map to store frequency of
    # elements of the array
    mp = {}

    for i in range(N):
        if arr[i] in mp:
            mp[arr[i]] += 1
        else:
            mp[arr[i]] = 1

    # Variable to store sum of all
    # even occurring elements
    Sum = 0

    # Loop to iterate through map
    for first, second in mp.items():

        # Check if frequency is even
        if (second % 2 == 0):
            Sum += (first) * (second)

    return Sum

# Driver code
arr = [ 10, 20, 20, 40, 40 ]

N = len(arr)

print(findSum(arr, N))

# This code is contributed by divyeshrabadiya07

C#

// C# program to find the sum of all even
// occurring elements in an array
using System;
using System.Collections.Generic;

class GFG {

    static int element(int[] arr, int n)
    {
        Dictionary<int, int> map = new Dictionary<int, int>();
        for (int i = 0; i < n; i++)
        {

            if(!map.ContainsKey(arr[i]))
            {
                map.Add(arr[i], 1);
            }
            else
            {
                map[arr[i]] += 1;
            }

        }

        int sum = 0;

        foreach(KeyValuePair<int, int> entry in map)
        {
            if (entry.Value % 2 == 0)
            {
                sum += entry.Key * entry.Value;
            }
        }

        return sum;
    }

  // Driver code
  static void Main() {
        int[] arr = { 10, 20, 20, 40, 40};

        int n = arr.Length;
        Console.WriteLine(element(arr, n));
  }
}

// This code is contributed by divyesh072019

Javascript

<script>
// Javascript program to find the sum of all odd
// occurring elements in an array

// Function to find the sum of all odd
// occurring elements in an array
function findSum(arr, N)
{

    // Store frequencies of elements
    // of the array
    let mp = new Map();
    for (let i = 0; i < N; i++) {
        if (mp.has(arr[i])) {
            mp.set(arr[i], mp.get(arr[i]) + 1)
        } else {
            mp.set(arr[i], 1)
        }
    }

    // variable to store sum of all
    // odd occurring elements
    let sum = 0;

    // loop to iterate through map
    for (let itr of mp) {

        // check if frequency is odd
        if (itr[1] % 2 == 0)
            sum += (itr[0]) * (itr[1]);
    }

    return sum;
}

// Driver Code

let arr = [10, 20, 20, 40, 40];

let N = arr.length

document.write(findSum(arr, N));

// This code is contributed by _saurabh_jaiswal.
</script>

Output:

120

Time Complexity: O(N), where N is the number of elements in the array.

Auxiliary Space: O(N)

Method 2:Using Built in python functions:

• Count the frequencies of every element using Counter function
• Traverse the frequency dictionary and sum all the elements with occurrence even frequency multiplied by its frequency.

Below is the implementation:

Python3

# Python3 implementation
from collections import Counter


def sumEven(arr, n):

    # Counting frequency of every
    # element using Counter
    freq = Counter(arr)

    # initializing sum 0
    sum = 0

    # Traverse the freq and print all
    # sum all elements with even frequency
    # multiplied by its frequency
    for it in freq:
        if freq[it] % 2 == 0:
            sum = sum + it*freq[it]
    print(sum)


# Driver code
arr = [10, 20, 20, 40, 40]
n = len(arr)

sumEven(arr, n)

# This code is contributed by vikkycirus

Output:

120

Time complexity: O(n) where n is size of given list “arr”

Auxiliary space: O(1)