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

Lonely Integer Hacker Rank Problem solution 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[0];
    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) {
    // Write your code here
        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

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