# Lonely Integer HackerRank Problem Solution

Lonely Integer HackerRank Problem Solution

In this HackerRank Lonely Integer problem, we have Given an array of integers, where all elements but one occur twice, find the unique element.

Example

The unique element is .

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, , the number of integers in the array.
The second line contains  space-separated integers that describe the values in .

Constraints

• It is guaranteed that  is an odd number and that there is one unique element.
• , where .

## Lonely Integer Hacker Rank Problem Solutions

I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.

### Problem Solution in C#

Dictionary<int, int> dict = new Dictionary<int, int>();
int n = a.Count, count = 0;
for(int i=0;i<n;i++)
{
if(dict.ContainsKey(a[i]))
{
var val = dict[a[i]];
dict.Remove(a[i]);
}
else
{
}
}
foreach(KeyValuePair<int,int> entry in dict)
{
if(entry.Value == 1)
{
count = entry.Key;
}
}
return count;

### Problem Solution in JavaScript

function lonelyinteger(a) {
const findDuplicates = a => a.filter((item, index) => a.indexOf(item) !== index);
let duplicatesArray = findDuplicates(a);
for (let i = 0; i< a.length; i++){
if(!duplicatesArray.includes(a[i])){
return a[i]
}
}
}

### Problem Solution in Python

def lonelyinteger(a):
for i in range(n):
pope=a.pop(i)
if pope in a:
a.insert(i,pope)
else:
return pope

### Problem Solution in Java

public static int lonelyinteger(List<Integer> a) {
boolean n = true;
for(int i=0; i<a.size();i++){
for (int j=0;j<a.size();j++){
if(j != i){
if(Objects.equals(a.get(i), a.get(j))){
n=false;
break;
}
else{
n=true;
}

}
}
if(n){
return a.get(i);
}
}
return 0;

}