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

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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 <bits/stdc++.h>
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




<?php
// PHP program to find only
// element in an array where
// every element appears twice.
 
// Find non repeating number
// in an array
function 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 array
function 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).

 



Last Updated : 20 Jul, 2022
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