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.

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++

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// 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;
}

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Java














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// 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 */



















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Python3














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# 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.              



















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Output:


30





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



          
          
          
          

            

    

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