Sum of all odd frequency elements in an array

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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 cde 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 

Output:

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

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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