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).
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