# Lonely Integer Hacker Rank Problem solution Lonely Integer Hacker Rank Problem solution

HackerRank Lonely Integer problem solution in java python c++ c and javascript programming with practical program code example

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 ## HackerRank Lonely Integer  SOLUTIONS:

You are given an array of integers and every integer except one occurs more than once. The integer that doesn’t have a repeating value inside the array is considered unique. The goal of the function is to output that unique number.

let a = [1,1,2,2,6];

Looking at our array input above, values 1 and 2 have repeating values but 6 is the only number that doesn’t. The function will return 6 because it is a unique number.

Let’s turn this problem into code.

### HackerRank Lonely Integer Solution in JavaScript:

function lonelyinteger(a) {
let loneValue=a;
for(let i=1; i<a.length; i++)
loneValue ^= a[i];
return loneValue;
}


### Lonely Integer Solution in Python:

def lonelyinteger(a):
unique=0
for i in a:
unique^= i
return unique


### Lonely Integer Solution in Java:

public static int lonelyinteger(List<Integer> a) {
int unique=a.get(0);
for(int i=1;i<a.size();i++){
unique^=a.get(i);
}
return unique;
}


### 3 Possible Approches of Solving Lonely Integer :

   # Naive Approach
for i in range(len(a)):
single = True
for j in range(len(a)):
if i!=j and a[i] == a[j]:
single = False
break
if single:
return a[i]

# Using Hash map
hash_map = {}
for i in a:
if i not in hash_map:
hash_map[i] = 1
else:
hash_map[i] += 1
for key, value in hash_map.items():
if value == 1:
return key

# Using Xor
result = 0
for i in a:
result ^= i
return result