Sum of all odd frequency elements in an array

Last Updated : 27 Mar, 2023

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++

   // 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 array
int 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 Code
int main()
{
    int arr[] = { 10, 20, 20, 10, 40, 40, 10 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << findSum(arr, N);
 
    return 0;
}
 
  

Java

   // Java program to find the sum of all odd
// occurring elements in an array
import java.util.*;
 
class GFG 
{
 
// Function to find the sum of all odd
// occurring elements in an array
static 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 Code
public 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 */
 
  

Python3

   # Function to find sum of all odd 
# occurring elements in an array
import 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 Code
arr = [ 10, 20, 20, 10, 40, 40, 10 ]
 
N = len(arr)
 
print (findsum(arr, N))
             
# This code is contributed
# by HardeepSingh.             
 
  

C#

   // C# program to find the sum of all odd 
// occurring elements in an array
using 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
 
  

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, 10, 40, 40, 10];
let N = arr.length
document.write(findSum(arr, N));
 
// This code is contributed by gfgking.
</script>
 
  

Output

complexity Analysis:

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:

C++

   #include <iostream>
#include <unordered_map>
using namespace std;
 
void sumOdd(int arr[], int n) 
{
 
  // Counting frequency of every element using unordered_map
  unordered_map<int, int> freq;
  for (int i = 0; i < n; i++) {
    freq[arr[i]]++;
  }
 
  // Initializing sum to 0
  int sum = 0;
 
  // Traverse the frequency and sum all elements 
  // with odd frequency multiplied by its frequency
  for (auto it : freq) {
    if (it.second % 2 != 0) {
      sum += it.first * it.second;
    }
  }
 
  cout << sum << endl;
}
 
// Driver code
int main() {
  int arr[] = {10, 20, 30, 40, 40};
  int n = sizeof(arr) / sizeof(arr[0]);
 
  sumOdd(arr, n);
}
 
  

Python3

   # Python3 implementation of the above approach
from 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 code
arr = [10, 20, 30, 40, 40]
n = len(arr)
 
sumOdd(arr, n)
 
# This code is contributed by vikkycirus
 
  

C#

   // C# implementation of the above approach
using System;
using System.Collections.Generic;
using System.Linq;
 
class MainClass {
    // Function to return the sum of elements
    // in an array having odd frequency
    public static void sumOdd(int[] arr, int n)
    {
 
        // Dictionary is used to calculate frequency of
        // elements of array
        Dictionary<int, int> freq
            = arr.GroupBy(x => x).ToDictionary(
                x => x.Key, x => x.Count());
 
        int sum = 0;
 
        // Traverse the dictionary
        foreach(KeyValuePair<int, int> entry in freq)
        {
 
            // Calculate the sum of elements
            // having odd frequency
            // multiplied by its frequency
            if (entry.Value % 2 != 0) {
                sum += entry.Key * entry.Value;
            }
        }
 
        Console.WriteLine(sum);
    }
 
    // Driver code
    public static void Main()
    {
 
        int[] arr = { 10, 20, 30, 40, 40 };
        int n = arr.Length;
 
        sumOdd(arr, n);
    }
}
 
// This code is contributed by phasing17
 
  

Javascript

   function sumOdd(arr, n) {
 
    // Counting frequency of every
    // element using Map
    let freq = new Map();
    for(let i=0; i<n; i++){
        if(freq.has(arr[i])){
            freq.set(arr[i], freq.get(arr[i])+1);
        } else {
            freq.set(arr[i], 1);
        }
    }
 
    // Initializing sum to 0
    let sum = 0;
 
    // Traverse the frequency and sum all elements
    // with odd frequency multiplied by its frequency
    for(let [key, value] of freq){
        if(value % 2 !== 0){
            sum += key*value;
        }
    }
 
    console.log(sum);
}
 
// Driver code
let arr = [10, 20, 30, 40, 40];
let n = arr.length;
 
sumOdd(arr, n);
 
  

Java

   /*package whatever //do not write package name here */
 
import java.util.HashMap;
 
public class GFG {
     
    static void sumOdd(int[] arr, int n) {
         
        // Counting frequency of every element using HashMap
        HashMap<Integer, Integer> freq = new HashMap<Integer, Integer>();
        for (int i = 0; i < n; i++) {
            if (freq.containsKey(arr[i])) {
                freq.put(arr[i], freq.get(arr[i]) + 1);
            } else {
                freq.put(arr[i], 1);
            }
        }
 
        // Initializing sum to 0
        int sum = 0;
 
        // Traverse the frequency and sum all elements 
        // with odd frequency multiplied by its frequency
        for (HashMap.Entry<Integer, Integer> entry : freq.entrySet()) {
            if (entry.getValue() % 2 != 0) {
                sum += entry.getKey() * entry.getValue();
            }
        }
 
        System.out.println(sum);
    }
     
    // Driver code
    public static void main(String[] args) {
        int[] arr = {10, 20, 30, 40, 40};
        int n = arr.length;
        sumOdd(arr, n);
    }
}
 
  

Output

Complexity Analysis:

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

Sum of all even frequency elements in Matrix

barykrg

Improve

Article Tags :

Practice Tags :

Sum of all odd frequency elements in an array

C++

Java

Python3

C#

Javascript

C++

Python3

C#

Javascript

Java

Similar Reads