Sum of all even occurring element in an array
Given an array of integers containing duplicate elements. The task is to find the sum of all even occurring elements in the given array. That is the sum of all such elements whose frequency is even in the array.
Examples:
Input : arr[] = {1, 1, 2, 2, 3, 3, 3} Output : 6 The even occurring element are 1 and 2 (both occur two times). Therefore sum of all 1's in the array = 1+1+2+2 = 6. Input : arr[] = {10, 20, 30, 40, 40} Output : 80 Element with even frequency are 40. Sum = 40.
Approach:
- Traverse the array and use a unordered_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 even, if it is even, then add this element to sum.
Below is the implementation of the above approach:
C++
// C++ program to find the sum of all even // occurring elements in an array #include <bits/stdc++.h> using namespace std; // Function to find the sum of all even // occurring elements in an array int findSum( int arr[], int N) { // Map to store frequency 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 // even occurring elements int sum = 0; // loop to iterate through map for ( auto itr = mp.begin(); itr != mp.end(); itr++) { // check if frequency is even if (itr->second % 2 == 0) sum += (itr->first) * (itr->second); } return sum; } // Driver Code int main() { int arr[] = { 10, 20, 20, 40, 40 }; int N = sizeof (arr) / sizeof (arr[0]); cout << findSum(arr, N); return 0; } |
Java
// Java program to find the sum of all even // occurring elements in an array import java.util.HashMap; import java.util.Iterator; import java.util.Map; import java.util.Map.Entry; public class GFG { public static int element( int [] arr, int n) { Map<Integer, Integer> map = new HashMap<Integer, Integer>(); for ( int i = 0 ; i < n; i++) { int count = 0 ; if (map.get(arr[i]) != null ) { count = map.get(arr[i]); } map.put(arr[i], count + 1 ); } int sum = 0 ; for (Entry<Integer, Integer> entry : map.entrySet()) { if (entry.getValue() % 2 == 0 ) { sum += entry.getKey() * entry.getValue(); } } return sum; } public static void main(String[] args) { int arr[] = { 1 , 1 , 2 , 2 , 3 , 3 , 3 }; // sum should be = 1+1+2+2=6; int n = arr.length; System.out.println(element(arr, n)); } } |
Python3
# Python3 program to find the sum # of all even occurring elements # in an array # Function to find the sum of all even # occurring elements in an array def findSum(arr, N): # Map to store frequency of # elements of the array mp = {} for i in range (N): if arr[i] in mp: mp[arr[i]] + = 1 else : mp[arr[i]] = 1 # Variable to store sum of all # even occurring elements Sum = 0 # Loop to iterate through map for first, second in mp.items(): # Check if frequency is even if (second % 2 = = 0 ): Sum + = (first) * (second) return Sum # Driver code arr = [ 10 , 20 , 20 , 40 , 40 ] N = len (arr) print (findSum(arr, N)) # This code is contributed by divyeshrabadiya07 |
C#
// C# program to find the sum of all even // occurring elements in an array using System; using System.Collections.Generic; class GFG { static int element( int [] arr, int n) { Dictionary< int , int > map = new Dictionary< int , int >(); for ( int i = 0; i < n; i++) { if (!map.ContainsKey(arr[i])) { map.Add(arr[i], 1); } else { map[arr[i]] += 1; } } int sum = 0; foreach (KeyValuePair< int , int > entry in map) { if (entry.Value % 2 == 0) { sum += entry.Key * entry.Value; } } return sum; } // Driver code static void Main() { int [] arr = { 10, 20, 20, 40, 40}; int n = arr.Length; Console.WriteLine(element(arr, n)); } } // This code is contributed by divyesh072019 |
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, 40, 40]; let N = arr.length document.write(findSum(arr, N)); // This code is contributed by _saurabh_jaiswal. </script> |
Output:
120
Time Complexity: O(N), where N is the number of elements in the array.
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 even frequency multiplied by its frequency.
Below is the implementation:
Python3
# Python3 implementation from collections import Counter def sumEven(arr, n): # Counting frequency of every # element using Counter freq = Counter(arr) # initializing sum 0 sum = 0 # Traverse the freq and print all # sum all elements with even 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 , 20 , 40 , 40 ] n = len (arr) sumEven(arr, n) # This code is contributed by vikkycirus |
Output:
120