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Non-overlapping sum of two sets

Given two arrays A[] and B[] of size n. It is given that both array individually contains distinct elements. We need to find the 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.

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:

C++

// 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;
}

Java

// Java program to find Non-overlapping sum 
import java.io.*;
import java.util.*;
  
class GFG 
{
  
    // function for calculating 
    // Non-overlapping sum of two array 
    static int findSum(int[] A, int[] B, int n)
    {
        // Insert elements of both arrays
        HashMap<Integer, Integer> hash = new HashMap<>();
        for (int i = 0; i < n; i++)
        {
            if (hash.containsKey(A[i]))
                hash.put(A[i], 1 + hash.get(A[i]));
            else
                hash.put(A[i], 1);
  
            if (hash.containsKey(B[i]))
                hash.put(B[i], 1 + hash.get(B[i]));
            else
                hash.put(B[i], 1);
        }
  
        // calculate non-overlapped sum 
        int sum = 0;
        for (Map.Entry entry : hash.entrySet())
        {
            if (Integer.parseInt((entry.getValue()).toString()) == 1)
                sum += Integer.parseInt((entry.getKey()).toString());
        }
  
        return sum;
  
    }
  
    // Driver code
    public static void main(String args[])
    {
        int[] A = { 5, 4, 9, 2, 3 }; 
        int[] B = { 2, 8, 7, 6, 3 }; 
  
        // size of array 
        int n = A.length;
  
        // function call 
        System.out.println(findSum(A, B, n));
    }
}
  
// This code is contributed by rachana soma

Python3

# Python3 program to find Non-overlapping sum 
from collections import defaultdict
  
# Function for calculating 
# Non-overlapping sum of two array 
def findSum(A, B, n): 
  
    # Insert elements of both arrays 
    Hash = defaultdict(lambda:0)
    for i in range(0, n): 
        Hash[A[i]] += 1
        Hash[B[i]] += 1
  
    # calculate non-overlapped sum 
    Sum = 0
    for x in Hash
        if Hash[x] == 1
            Sum +=
      
    return Sum
  
# Driver code 
if __name__ == "__main__"
  
    A = [5, 4, 9, 2, 3
    B = [2, 8, 7, 6, 3
      
    # size of array 
    n = len(A) 
  
    # Function call 
    print(findSum(A, B, n)) 
      
# This code is contributed 
# by Rituraj Jain

Output:

39


This article is attributed to GeeksforGeeks.org

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