# Non-overlapping sum of two sets

• Difficulty Level : Basic
• Last Updated : 27 Jul, 2022

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 ``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);` `    ``// 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 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

 ``

Output

`39`

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

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