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Non-overlapping sum of two sets

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  • Difficulty Level : Basic
  • Last Updated : 27 Jul, 2022
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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(n2).

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 += x
     
    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

C#




// C# program to find Non-overlapping sum
using System;
using System.Collections.Generic;
     
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
        Dictionary<int, int> hash = new Dictionary<int, int>();
        for (int i = 0; i < n; i++)
        {
            if (hash.ContainsKey(A[i]))
            {
                var v = hash[A[i]];
                hash.Remove(A[i]);
                hash.Add(A[i], 1 + v);
            }
            else
                hash.Add(A[i], 1);
 
            if (hash.ContainsKey(B[i]))
            {
                var v = hash[B[i]];
                hash.Remove(B[i]);
                hash.Add(B[i], 1 + v);
            }
            else
                hash.Add(B[i], 1);
        }
 
        // calculate non-overlapped sum
        int sum = 0;
        foreach(KeyValuePair<int, int> entry in hash)
        {
            if ((entry.Value) == 1)
                sum += entry.Key;
        }
 
        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
        Console.WriteLine(findSum(A, B, n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript program to find Non-overlapping sum
 
// function for calculating
// Non-overlapping sum of two array
function findSum(A, B, n) {
    // Insert elements of both arrays
    let hash = new Map();
    for (let i = 0; i < n; i++) {
        if (hash.has(A[i]))
            hash.set(A[i], 1 + hash.get(A[i]));
        else
            hash.set(A[i], 1);
 
        if (hash.has(B[i]))
            hash.set(B[i], 1 + hash.get(B[i]));
        else
            hash.set(B[i], 1);
    }
 
    // calculate non-overlapped sum
    let sum = 0;
    for (let entry of hash) {
        if (parseInt((entry[1]).toString()) == 1)
            sum += parseInt((entry[0]).toString());
    }
 
    return sum;
 
}
 
// Driver code
 
let A = [5, 4, 9, 2, 3];
let B = [2, 8, 7, 6, 3];
 
// size of array
let n = A.length;
 
// function call
document.write(findSum(A, B, n));
 
// This code is contributed by gfgking
 
</script>

Output

39

Time Complexity: O(n), since inserting in an unordered map is amortized constant.
Auxiliary Space: O(n).

This article is contributed by Aarti_Rathi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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