Sum of all odd frequency elements in an array

16 Nov, 2021
Given an array of integers containing duplicate elements. The task is to find the sum of all odd occurring elements in the given array. That is the sum of all such elements whose frequency is odd in the array.
Examples: 
Input : arr[] = {1, 1, 2, 2, 3, 3, 3}
Output : 9
The odd occurring element is 3, and it's number
of occurrence is 3. Therefore sum of all 3's in the 
array = 9.

Input : arr[] = {10, 20, 30, 40, 40}
Output : 60
Elements with odd frequency are 10, 20 and 30.
Sum = 60.
Approach:  
Traverse the array and use a 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 odd, if it is odd, then add this element to sum.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// CPP program to find the sum of all odd// occurring elements in an array #include <bits/stdc++.h>using namespace std; // Function to find the sum of all odd// occurring elements in an arrayint findSum(int arr[], int N){    // Store frequencies 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    // odd occurring elements    int sum = 0;     // loop to iterate through map    for (auto itr = mp.begin(); itr != mp.end(); itr++) {         // check if frequency is odd        if (itr->second % 2 != 0)            sum += (itr->first) * (itr->second);    }     return sum;} // Driver Codeint main(){    int arr[] = { 10, 20, 20, 10, 40, 40, 10 };     int N = sizeof(arr) / sizeof(arr[0]);     cout << findSum(arr, N);     return 0;}
// Java program to find the sum of all odd// occurring elements in an arrayimport java.util.*; class GFG{ // Function to find the sum of all odd// occurring elements in an arraystatic int findSum(int arr[], int N){    // Store frequencies of elements    // of the array    Map<Integer,Integer> mp = new HashMap<>();    for (int i = 0; i < N; i++)        mp.put(arr[i],mp.get(arr[i])==null?1:mp.get(arr[i])+1);     // variable to store sum of all    // odd occurring elements    int sum = 0;     // loop to iterate through map    for (Map.Entry<Integer,Integer> entry : mp.entrySet())    {         // check if frequency is odd        if (entry.getValue() % 2 != 0)            sum += (entry.getKey()) * (entry.getValue());    }     return sum;} // Driver Codepublic static void main(String args[]){    int arr[] = { 10, 20, 20, 10, 40, 40, 10 };     int N = arr.length;     System.out.println(findSum(arr, N));}} /* This code is contributed by PrinciRaj1992 */
# Function to find sum of all odd# occurring elements in an arrayimport collections def findsum(arr, N):         # Store frequencies of elements    # of an array in dictionary    mp = collections.defaultdict(int)         for i in range(N):        mp[arr[i]] += 1         # Variable to store sum of all    # odd occurring elements    sum = 0         # loop to iterate through dictionary    for i in mp:                 # Check if frequency is odd        if (mp[i] % 2 != 0):            sum += (i * mp[i])    return sum     # Driver Codearr = [ 10, 20, 20, 10, 40, 40, 10 ] N = len(arr) print (findsum(arr, N))             # This cde is contributed# by HardeepSingh.            
// C# program to find the sum of all odd// occurring elements in an arrayusing System;using System.Collections.Generic; class GFG{    // Function to find the sum of all odd    // occurring elements in an array    public static int findSum(int[] arr, int N)    {         // Store frequencies of elements        // of the array        Dictionary<int,                   int> mp = new Dictionary<int,                                                   int>();        for (int i = 0; i < N; i++)        {            if (mp.ContainsKey(arr[i]))                mp[arr[i]]++;            else                mp.Add(arr[i], 1);        }         // variable to store sum of all        // odd occurring elements        int sum = 0;         // loop to iterate through map        foreach (KeyValuePair<int,                              int> entry in mp)        {             // check if frequency is odd            if (entry.Value % 2 != 0)                sum += entry.Key * entry.Value;        }         return sum;    }     // Driver code    public static void Main(String[] args)    {        int[] arr = { 10, 20, 20, 10, 40, 40, 10 };        int n = arr.Length;        Console.WriteLine(findSum(arr, n));    }} // This code is contributed by// sanjeev2552
<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 arrayfunction 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 Codelet arr = [10, 20, 20, 10, 40, 40, 10];let N = arr.lengthdocument.write(findSum(arr, N)); // This code is contributed by gfgking.</script>
30
 
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 odd frequency multiplied by its frequency.
Below is the implementation:
Python3
# Python3 implementation of the above approachfrom collections import Counter def sumOdd(arr, n):     # Counting frequency of every    # element using Counter    freq = Counter(arr)         # Initializing sum 0    sum = 0         # Traverse the frequency and print all    # sum all elements with odd frequency    # multiplied by its frequency    for it in freq:        if freq[it] % 2 != 0:            sum = sum + it*freq[it]    print(sum)  # Driver codearr = [10, 20, 30, 40, 40]n = len(arr) sumOdd(arr, n) # This code is contributed by vikkycirus
Output:
60
Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(N)
HardeepSingh
princiraj1992
princi singh
sanjeev2552
vikkycirus
gfgking
simranarora5sos
rohan07
cpp-unordered_map
frequency-counting
Arrays
Hash
Arrays
Hash
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.