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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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.			 





Output:


30





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



          
          
          
          

            

    

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