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Edvins Antonovs

Lonely Integer code challenge

My approach to solving HackerRank’s Lonely Integer code challenge.


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


a = [1, 2, 3, 4, 3, 2, 1]

The unique element is 4.

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.


Returns int value of the element that occurs only once.


First, we start by creating a count function which accepts the value (v) we want to search for. Within the function, we run the Array.filter() method to create a new copy of a given array and populate it with the filtered value(s) (v). Then Array.length returns the exact number of times the value (v) appeared in the given array.

After that, we run a classic for loop where within each iteration, we pass the current array's item to the count() function and see if the result is equal to 1. If so, we return the current value (arr[i]) as it is unique and appears only once.

1function lonelyinteger(a) {
2 const count = (v) => a.filter((i) => i == v).length;
4 for (let i = 0; i < a.length; i++) {
5 if (count(a[i]) === 1) {
6 return a[i];
7 }
8 }


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