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.
- XOR of a number with itself is 0.
- 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 <bits/stdc++.h>using namespace std;// Find non repeating // number in an arrayint findNonRepeating(int arr[], int n){ int res = 0; for (int i = 0; i < n; i++) res = res ^ arr[i]; return res;}// Driver Codeint 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 Codestatic 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 arraydef 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 Codestatic 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
<?php// PHP program to find only // element in an array where // every element appears twice.// Find non repeating number// in an arrayfunction findNonRepeating($arr, $n){ $res = 0; for ($i = 0; $i < $n; $i++) $res = $res ^ $arr[$i]; return $res;}// Driver Code$arr = array( 3, 8, 3, 2, 2, 1, 1 );$n = sizeof($arr);echo(findNonRepeating($arr, $n));// This code is contributed by Ajit.?> |
Javascript
<script>// Javascript program to find only // element in an array where // every element appears twice.// Find non repeating // number in an arrayfunction findNonRepeating(arr, n){ let res = 0; for (let i = 0; i < n; i++) res = res ^ arr[i]; return res;}// Driver Code let arr = [ 3, 8, 3, 2, 2, 1, 1 ]; let n = arr.length; document.write(findNonRepeating(arr, n));</script> |
Output
8
Time Complexity: O(n).
Space Complexity: O(1).
This article is contributed by ASIPU PAWAN KUMAR. If you like neveropen 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 neveropen main page and help other Geeks.
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