Sum of all even occurring element in an array
Last Updated :
11 Jul, 2025
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 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 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# 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
<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>
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:
C++
// C++ implementation
#include <iostream>
#include <unordered_map>
using namespace std;
void sumEven(int arr[], int n)
{
// Counting frequency of every element
unordered_map<int, int> freq;
for (int i = 0; i < n; i++) {
freq[arr[i]]++;
}
// initializing sum 0
int sum = 0;
// Traverse the freq and print all
// sum all elements with even frequency
// multiplied by its frequency
for (auto it : freq) {
if (it.second % 2 == 0) {
sum = sum + it.first * it.second;
}
}
cout << sum << endl;
}
// Driver code
int main()
{
int arr[] = { 10, 20, 20, 40, 40 };
int n = sizeof(arr) / sizeof(arr[0]);
sumEven(arr, n);
return 0;
}
// This code is contributed by vikkycirus.
// Java program for the above approach
import java.util.*;
public class Main {
public static void sumEven(int[] arr, int n)
{
// Counting frequency of every element
Map<Integer, Integer> freq = new HashMap<>();
for (int i = 0; i < n; i++) {
freq.put(arr[i],
freq.getOrDefault(arr[i], 0) + 1);
}
// initializing sum 0
int sum = 0;
// Traverse the freq and print all
// sum all elements with even frequency
// multiplied by its frequency
for (Map.Entry<Integer, Integer> entry :
freq.entrySet()) {
int key = entry.getKey();
int value = entry.getValue();
if (value % 2 == 0) {
sum += key * value;
}
}
System.out.println(sum);
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 10, 20, 20, 40, 40 };
int n = arr.length;
sumEven(arr, n);
}
}
# 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
// C# implementation
using System;
using System.Collections.Generic;
using System.Linq;
class GFG {
static void Main(string[] args)
{
int[] arr = { 10, 20, 20, 40, 40 };
int n = arr.Length;
sumEven(arr, n);
}
static void sumEven(int[] arr, int n)
{
// Counting frequency of every
// element using Dictionary
Dictionary<int, int> freq
= arr.GroupBy(x = > x).ToDictionary(
g = > g.Key, g = > g.Count());
// initializing sum 0
int sum = 0;
// Traverse the freq and print all
// sum all elements with even frequency
// multiplied by its frequency
foreach(KeyValuePair<int, int> pair in freq)
{
if (pair.Value % 2 == 0) {
sum += pair.Key * pair.Value;
}
}
Console.WriteLine(sum);
}
}
// This code is contributed by phasing17
function sumEven(arr)
{
// Counting frequency of every element
let freq = {};
for (let i = 0; i < arr.length; i++) {
if (freq[arr[i]] === undefined) {
freq[arr[i]] = 1;
} else {
freq[arr[i]]++;
}
}
// initializing sum 0
let sum = 0;
// Traverse the freq and print all
// sum all elements with even frequency
// multiplied by its frequency
for (let key in freq) {
if (freq[key] % 2 == 0) {
sum = sum + key * freq[key];
}
}
console.log(sum);
}
// Driver code
let arr = [10, 20, 20, 40, 40];
sumEven(arr);
Time complexity: O(n) where n is size of given list "arr"
Auxiliary space: O(1)
Approach#3:using pointers
Algorithm
1. Sort the input array arr.
2. Initialize two pointers i and j to 0.
3. Initialize a variable sum to 0.
4.While j < len(arr):
a. If arr[j] is equal to arr[i], increment j.
b. Otherwise, if (j - i) % 2 == 0, add arr[i] * (j - i) to sum.
c. Set i to j.
5.If (j - i) % 2 == 0, add arr[i] * (j - i) to sum.
6. Return sum.
C++
#include <iostream>
#include <vector>
#include <algorithm> // Required for sorting
using namespace std;
// Function to calculate the sum of elements with even frequency
int sumEvenFreq(vector<int>& arr) {
sort(arr.begin(), arr.end()); // Sort the array in ascending order
int i = 0, j = 0;
int sum = 0;
// Loop to iterate through the array
while (j < arr.size()) {
if (arr[j] == arr[i]) {
j++; // Increment j until the element changes
} else {
// If the frequency of elements between i and j is even
if ((j - i) % 2 == 0) {
sum += arr[i] * (j - i); // Add the element times its frequency to sum
}
i = j; // Move i to the new distinct element
}
}
// Check if the last group of elements has even frequency
if ((j - i) % 2 == 0) {
sum += arr[i] * (j - i); // Add the element times its frequency to sum
}
return sum;
}
// Driver Code
int main() {
vector<int> arr = {10, 20, 30, 40, 40};
cout << "Sum of elements with even frequency: " << sumEvenFreq(arr) << endl;
return 0;
}
import java.util.Arrays;
public class Main {
// Function to calculate the sum of elements with even frequency
static int sumEvenFreq(int[] arr) {
Arrays.sort(arr); // Sort the array in ascending order
int i = 0, j = 0;
int sum = 0;
// Loop to iterate through the array
while (j < arr.length) {
if (arr[j] == arr[i]) {
j++; // Increment j until the element changes
} else {
// If the frequency of elements between i and j is even
if ((j - i) % 2 == 0) {
sum += arr[i] * (j - i); // Add the element times its frequency to sum
}
i = j; // Move i to the new distinct element
}
}
// Check if the last group of elements has even frequency
if ((j - i) % 2 == 0) {
sum += arr[i] * (j - i); // Add the element times its frequency to sum
}
return sum;
}
// Driver Code
public static void main(String[] args) {
int[] arr = {10, 20, 30, 40, 40};
System.out.println("Sum of elements with even frequency: " + sumEvenFreq(arr));
}
}
def sum_even_freq(arr):
arr.sort()
i = j = 0
sum = 0
while j < len(arr):
if arr[j] == arr[i]:
j += 1
else:
if (j - i) % 2 == 0:
sum += arr[i] * (j - i)
i = j
if (j - i) % 2 == 0:
sum += arr[i] * (j - i)
return sum
arr = [ 10, 20, 30, 40, 40 ]
print(sum_even_freq(arr))
using System;
using System.Collections.Generic;
using System.Linq; // Required for sorting
class Program
{
// Function to calculate the sum of elements with even frequency
static int SumEvenFreq(List<int> arr)
{
arr.Sort(); // Sort the list in ascending order
int i = 0, j = 0;
int sum = 0;
// Loop to iterate through the list
while (j < arr.Count)
{
if (arr[j] == arr[i])
{
j++; // Increment j until the element changes
}
else
{
// If the frequency of elements between i and j is even
if ((j - i) % 2 == 0)
{
sum += arr[i] * (j - i); // Add the element times its frequency to sum
}
i = j; // Move i to the new distinct element
}
}
// Check if the last group of elements has even frequency
if ((j - i) % 2 == 0)
{
sum += arr[i] * (j - i); // Add the element times its frequency to sum
}
return sum;
}
static void Main()
{
List<int> arr = new List<int> { 10, 20, 30, 40, 40 };
Console.WriteLine("Sum of elements with even frequency: " + SumEvenFreq(arr));
}
}
function sum_even_freq(arr) {
// Sort the array
arr.sort();
let i = 0;
let j = 0;
let sum = 0;
while (j < arr.length) {
if (arr[j] == arr[i]) {
j += 1;
} else {
// If the frequency of the current element is even, add it to the sum
if ((j - i) % 2 == 0) {
sum += arr[i] * (j - i);
}
i = j;
}
}
// Check the last element
if ((j - i) % 2 == 0) {
sum += arr[i] * (j - i);
}
return sum;
}
let arr = [10, 20, 30, 40, 40];
console.log(sum_even_freq(arr));
Time complexity: O(n log n), where n is the length of the input array arr. The sort() function takes O(n log n) time in the worst case, and the while loop takes O(n) time to iterate through the sorted array. Therefore, the overall time complexity is dominated by the sort() function.
Space complexity: O(1), since we're using a constant amount of extra space to store the pointers i and j, and the variable sum.
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