# XOR of Sum of every possible pair of an array

• Difficulty Level : Easy
• Last Updated : 29 Apr, 2021

Given an array A of size n. the task is to generate a new sequence B with size N^2 having elements sum of every pair of array A and find the xor value of the sum of all the pairs formed.
Note: Here (A[i], A[i]), (A[i], A[j]), (A[j], A[i]) all are considered as different pairs.

Examples:

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```Input: arr = {1, 5, 6}
Output: 4
B[3*3] = { 1+1, 1+5, 1+6, 5+1, 5+5, 5+6, 6+1, 6+5, 6+6}
B[9] = { 2, 6, 7, 6, 10, 11, 7, 11, 12}
So, 2 ^ 6 ^ 7 ^ 6 ^ 10 ^ 11 ^ 7 ^ 6 ^ 11 ^ 12 = 4

Input :1, 2
Output :6```

A Naive approach is to run two loops. Consider each and every pair, take their sum, and calculate the xor value of the sum of all the pairs.
An Efficient approach is based upon the fact that xor of the same values is 0.
All the pairs like (a[i], a[j]) and (a[j], a[i]) will have same sum. So, their xor values will be 0. Only the pairs like (a[i], a[i]) will give the different result. So, take the xor of all the elements of the given array and multiply it by 2.

## C++

 `// C++ program to find XOR of``// sum of every possible pairs``// in an array``#include ``using` `namespace` `std;` `// Function to find XOR of sum``// of all pairs``int` `findXor(``int` `arr[], ``int` `n)``{` `    ``// Calculate xor of all the elements``    ``int` `xoR = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``xoR = xoR ^ arr[i];``    ``}` `    ``// Return twice of xor value``    ``return` `xoR * 2;``}` `// Drivers code``int` `main()``{``    ``int` `arr[3] = { 1, 5, 6 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << findXor(arr, n);` `    ``return` `0;``}`

## Java

 `// Java program to find XOR of``// sum of every possible pairs``// in an array``import` `java.io.*;` `class` `GFG {``    ` `    ``// Function to find XOR of sum``    ``// of all pairs``    ``static` `int` `findXor(``int` `arr[], ``int` `n)``    ``{``    ` `        ``// Calculate xor of all the``        ``// elements``        ``int` `xoR = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``xoR = xoR ^ arr[i];``        ``}``    ` `        ``// Return twice of xor value``        ``return` `xoR * ``2``;``    ``}``    ` `    ``// Drivers code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``5``, ``6` `};``        ``int` `n = arr.length;``        ``System.out.println( findXor(arr, n));``    ``}``}` `// This code is contributed by anuj_67.`

## Python3

 `# Python3 program to find``# XOR of sum of every``# possible pairs in an array` `# Function to find XOR``# of sum of all pairs``def` `findXor(arr,n):` `    ``# Calculate xor of``    ``# all the elements``    ``xoR ``=` `0``;``    ``for` `i ``in` `range` `(``0``, n ) :``        ``xoR ``=` `xoR ^ arr[i]``    ` `    ``# Return twice of``    ``# xor value``    ``return` `xoR ``*` `2` `# Driver code``arr ``=` `[ ``1``, ``5``, ``6` `]``n ``=` `len``(arr)``print``(findXor(arr, n))` `# This code is contributed``# by ihritik``   `

## C#

 `// C# program to find XOR of``// sum of every possible pairs``// in an array``using` `System;` `class` `GFG {``    ` `    ``// Function to find XOR of sum``    ``// of all pairs``    ``static` `int` `findXor(``int` `[]arr, ``int` `n)``    ``{``    ` `        ``// Calculate xor of all the``        ``// elements``        ``int` `xoR = 0;``        ``for` `(``int` `i = 0; i < n; i++) {``            ``xoR = xoR ^ arr[i];``        ``}``    ` `        ``// Return twice of xor value``        ``return` `xoR * 2;``    ``}``    ` `    ``// Drivers code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]arr = { 1, 5, 6 };``        ``int` `n = arr.Length;``        ``Console.WriteLine( findXor(arr, n));``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

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## Javascript

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Output:
`4`

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