Skip to content# Lonely Integer code challenge

### Problem

### Example

### Input format

### Returns

### Solution

### Appendix

## Newsletter

— 1 min read

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`

.

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;3
4 for (let i = 0; i < a.length; i++) {5 if (count(a[i]) === 1) {6 return a[i];7 }8 }9}
```

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