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

• Difficulty Level : Basic
• Last Updated : 20 Jul, 2022

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```

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.

Implementation:

## 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[0]);``    ``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).

This article is contributed by ASIPU PAWAN KUMAR. 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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