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 is 1 and 2, and it's number
of occurrence is 2. 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:
// CPP 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 stroe sum of all // even occurring elements int sum = 0; // loop to iteratre 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; } |
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Output:
120
Time Complexity: O(N), where N is the number of elements in the array.
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