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 :



title-img 


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

public class Solution{
    public static void main(String[] args) throws IOException {
    try {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int N = Integer.parseInt(br.readLine());
        boolean arr[] = new boolean[101];
        StringTokenizer st = new StringTokenizer(br.readLine());
        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[0]
for i in range(1,n):
    r=r^a[i]
print(r)








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