# Non-overlapping sum of two sets

Given two arrays A[] and B[] of size n. It is given that both array individually contain distinct elements. We need to find sum of all elements that are not common.

Examples:

Input : A[] = {1, 5, 3, 8}
B[] = {5, 4, 6, 7}
Output : 29
1 + 3 + 4 + 6 + 7 + 8 = 29

Input : A[] = {1, 5, 3, 8}
B[] = {5, 1, 8, 3}
Output : 0
All elements are common.

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

Brute Force Method :
One simple approach is that for each element in A[] check whether it is present in B[], if it is present in then add it to the result. Similarly traverse B[] and for every element that is not present in B, add it to result.
Time Complexity: O(n^2).

Hashing concept :
Create an empty hash and insert elements of both arrays into it. Now traverse hash table and add all those elements whose count is 1. (As per the question, both arrays individually have distinct elements)

Below is the implementation of above approach:

// CPP program to find Non-overlapping sum
#include <bits/stdc++.h>
using namespace std;

// function for calculating
// Non-overlapping sum of two array
int findSum(int A[], int B[], int n)
{
// Insert elements of both arrays
unordered_map<int, int> hash;
for (int i = 0; i < n; i++) {
hash[A[i]]++;
hash[B[i]]++;
}

// calculate non-overlapped sum
int sum = 0;
for (auto x: hash)
if (x.second == 1)
sum += x.first;

return sum;
}

// driver code
int main()
{
int A[] = { 5, 4, 9, 2, 3 };
int B[] = { 2, 8, 7, 6, 3 };

// size of array
int n = sizeof(A) / sizeof(A[0]);

// function call
cout << findSum(A, B, n);
return 0;
}

Output:

39

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