# 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(n2).
Auxiliary Space: O(1), As constant extra space is used.

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 the 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[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 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).

Another method: Using set data structure

• Insert elements of Array A in the set data structure and add into sum
• Check if B’s elements are there in set if exist then remove current element from set, otherwise add current element to sum
• Finally, return sum

Below is the implementation of the 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)``{``    ``int` `sum = 0;` `    ``// Insert elements of Array A in set``    ``// and add into sum``    ``set<``int``> st;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``st.insert(A[i]);``        ``sum += A[i];``    ``}` `    ``// Check if B's element are there in set``    ``// if exist then remove current element from``    ``// set, otherwise add current element into sum``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(st.find(B[i]) == st.end()) {``            ``sum += B[i];``        ``}``        ``else` `{``            ``sum -= B[i];``        ``}``    ``}` `    ``// Finally, return sum``    ``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;``}` `// This code is contributed by hkdass001`

## 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``    ``public` `static` `int` `findSum(``int``[] A, ``int``[] B, ``int` `n) {``        ``int` `sum = ``0``;` `        ``// Insert elements of Array A in set``        ``// and add into sum``        ``Set st = ``new` `HashSet<>();``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``st.add(A[i]);``            ``sum += A[i];``        ``}` `        ``// Check if B's element are there in set``        ``// if exist then remove current element from``        ``// set, otherwise add current element into sum``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(!st.contains(B[i])) {``                ``sum += B[i];``            ``}``            ``else` `{``                ``sum -= B[i];``            ``}``        ``}` `        ``// Finally, return sum``        ``return` `sum;``    ``}``  ` `    ``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 lokesh.`

## Python3

 `# python program to find Non-overlapping sum` `# function for calculating``# Non-overlapping sum of two array``def` `findSum(A, B, n):``    ``sum` `=` `0``;` `    ``# Insert elements of Array A in set``    ``# and add into sum``    ``st ``=` `set``();``    ``for` `i ``in` `range``(``0``,n): ``        ``st.add(A[i]);``        ``sum` `+``=` `A[i];``    ` `    ``# Check if B's element are there in set``    ``# if exist then remove current element from``    ``# set, otherwise add current element into sum``    ``for` `i ``in` `range` `(``0``, n):``        ``if` `(B[i] ``in` `st):``            ``sum` `-``=` `B[i];``        ``else` `:``            ``sum` `+``=` `B[i];` `    ``# Finally, return sum``    ``return` `sum``;` `# Driver code``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));`

## C#

## Javascript

 `
``// Javascript program to find Non-overlapping sum

// function for calculating
// Non-overlapping sum of two array
function findSum(A, B, n)
{
let sum = 0;

// Insert elements of Array A in set
let st = new Set();
for (let i = 0; i < n; i++) {
sum += A[i];
}

// Check if B's element are there in set
// if exist then remove current element from
// set, otherwise add current element into sum
for (let i = 0; i < n; i++) {
if (!st.has(B[i])) {
sum += B[i];
}
else {
sum -= B[i];
}
}

// Finally, return sum
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));
`

Output
`39`

Time Complexity: O(n*log n)
Auxiliary Space: O(n)

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