# Lonely Integer

### Problem Statement :

```Given an array of integers, where all elements but one occur twice, find the unique element.

Example
a = [1,2,3,4,3,2,1]
The unique element is 4.

Function Description

Complete the lonelyinteger function in the editor below.

lonelyinteger has the following parameter(s):

int a[n]: an array of integers
Returns

int: the element that occurs only once
Input Format

The first line contains a single integer, n, the number of integers in the array.
The second line contains n space-separated integers that describe the values in a.

Constraints
1 <= n < 100
It is guaranteed that n is an odd number and that there is one unique element.
0 <= a[i] <= 100, where 0 <= i < n.```

### Solution :

```                            ```Solution in C :

In C++ :

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <cstdlib>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;
int lonelyinteger(vector < int > a) {
int ans=0;
vector<int> ::iterator it;
for(it=a.begin();it!=a.end();it++)
{
ans^=(*it);
}
return ans;

}
int main() {
int res;

int _a_size;
cin >> _a_size;
cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n');
vector<int> _a;
int _a_item;
for(int _a_i=0; _a_i<_a_size; _a_i++) {
cin >> _a_item;
_a.push_back(_a_item);
}

res = lonelyinteger(_a);
cout << res;

return 0;
}

In Java :

import java.util.StringTokenizer;
import java.io.IOException;

public class Solution{
public static void main(String[] args) throws IOException {
try {
boolean arr[] = new boolean;
for (int i = 0; i < N; i++) {
int a = Integer.parseInt(st.nextToken());
arr[a] = !arr[a];
}
for (int i = 0; i < 101; i++) {
if(arr[i]) {
System.out.println(i);
break;
}
}
}
catch(Exception e) {
}
}
}

In C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <assert.h>
int lonelyinteger(int a_size, int* a) {
int i=0;
int num=0;
for(i=0;i<a_size;i++){
num=num^a[i];
}

return num;

}
int main() {
int res;

int _a_size, _a_i;
scanf("%d", &_a_size);
int _a[_a_size];
for(_a_i = 0; _a_i < _a_size; _a_i++) {
int _a_item;
scanf("%d", &_a_item);

_a[_a_i] = _a_item;
}

res = lonelyinteger(_a_size, _a);
printf("%d", res);

return 0;
}

In Python3 :

n=int(input())
a=list(map(int,input().split()))
r=a
for i in range(1,n):
r=r^a[i]
print(r)```
```

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