# Find single in an array of 2n+1 integer elements

• Difficulty Level : Basic
• Last Updated : 30 Apr, 2021

Given an array with 2n+1 integers, n elements appear twice in arbitrary places in the array and a single integer appears only once somewhere inside. Find the lonely integer with O(n) operations and O(1) extra memory.
Examples :

```Input : { 1, 1, 2, 2, 3, 3, 4, 4, 5}
Output : 5

Input : { 7, 9, 6, 8, 3, 7, 8, 6, 9}
Output : 3```

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The idea is to do XOR of all elements. XOR of all elements gives us the result. The idea is based on below XOR properties.

1. XOR of a number with itself is 0.
2. XOR of a number with 0 is the number.

## C++

 `// CPP program to find only``// element in an array where``// every element appears twice.``#include ``using` `namespace` `std;` `// Find non repeating``// number in an array``int` `findNonRepeating(``int` `arr[],``                     ``int` `n)``{``    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``res = res ^ arr[i];``    ``return` `res;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 3, 8, 3, 2, 2, 1, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << findNonRepeating(arr, n);``    ``return` `0;``}`

## Java

 `// Java program to find only element``// in an array where every element``// appears twice.``import` `java.io.*;` `class` `GFG``{` `// Find non repeating``// number in an array``static` `int` `findNonRepeating(``int` `[]arr,``                            ``int` `n)``{``    ``int` `res = ``0``;``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``res = res ^ arr[i];``    ``return` `res;``}` `// Driver Code``static` `public` `void` `main (String[] args)``{``int` `[]arr = {``3``, ``8``, ``3``, ``2``, ``2``, ``1``, ``1``};``int` `n = arr.length;``System.out.println(findNonRepeating(arr, n));``}``}` `// This code is contributed by vt_m.`

## Python

 `# Find non repeating``# number in an array` `def` `findNonRepeating(a, n):``    ``res ``=` `0``    ` `    ``# XOR of all numbers``    ``for` `i ``in` `range``(n):``        ``res ^``=` `a[i]``    ``return` `res``    ` `# Driver code``a ``=` `[ ``3``, ``8``, ``3``, ``2``, ``2``, ``1``, ``1` `]``n ``=` `len``(a)` `print` `findNonRepeating(a, n)` `# This code is contributed``# by 'Striver'.`

## C#

 `// C# program to find only element``// in an array where every element``// appears twice.``using` `System;` `class` `GFG``{` `// Find non repeating number in an array``static` `int` `findNonRepeating(``int` `[]arr,``                            ``int` `n)``{``    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``res = res ^ arr[i];``    ``return` `res;``}` `// Driver Code``static` `public` `void` `Main (String []args)``{``int` `[]arr = {3, 8, 3, 2, 2, 1, 1};``int` `n = arr.Length;``Console.WriteLine(findNonRepeating(arr,``                                   ``n));``}``}` `// This code is contributed by vt_m.`

## PHP

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## Javascript

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Output :

`8`

Time Complexity: O(n).
Space Complexity: O(1).
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