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Solving Code Wars Problem in Javascript (ES6) [7th Kyu]

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Stephen G. Vinuya
Dec 06,2018

In this article we are going to solve 3 Codewars' [7th kyu ] problems using Javascript (ES6) and  discuss approaches how problems can be solved .

Codewars is a great site for improving your problem solving skills and it's also great for discovering different approaches of other developers in solving each problems.

Problems at code wars are label by difficulty as kyu, [1st kyu] being the hardest and [8th kyu] being the easiest.

Let's now start with our first problem.

1. Sort array by string length

Codewars link:

The problem is :

Write a function that takes an array of strings as an argument and returns a sorted array containing the same strings, ordered from shortest to longest.

For example, if this array were passed as an argument:

["Telescopes", "Glasses", "Eyes", "Monocles"]

Your function would return the following array:

["Eyes", "Glasses", "Monocles", "Telescopes"]
All of the strings in the array passed to your function will be different lengths, so you will not have to decide how to order multiple strings of the same length.

The solution:

Non ES6 approach: 

function sortByLength (array) {
return array.sort(function(a, b) {
return a.length - b.length; // sort it by ascending order based from string's length


We used Javascript's sort() function here which takes a callback function as an optional parameter for sorting.

Since we are comparing the length of the string as the basis for sorting, we used length property of the string.

a.length - b.length  this will be sorted in ascending order,

if we want it in descending order we just have to reverse the expression b.length - a.length

Would you be believe that our solution above can be reduced into a line of codes?

ES6 approach:

const sortByLength = (array) => array.sort((a, b) => a.length - b.length);

What's new here is ES6's arrow function, click here to learn more about arrow functions.

2. Flatten and sort an array

Codewars link:

The problem is :


Given a two-dimensional array of integers, return the flattened version of the array with all the integers in the sorted (ascending) order.


Given [[3, 2, 1], [4, 6, 5], [], [9, 7, 8]], your function should return [1, 2, 3, 4, 5, 6, 7, 8, 9].


Please, keep in mind, that JavaScript is by default sorting objects alphabetically. For more information, please consult:

Non ES6 approach: 

function flattenAndSort(array) {
array = array.reduce(function (acc, a) {
return acc.concat(a);
}, []);

return array.sort(function (a, b) {
return a - b;

We used reduce() array method to transform the 2d array into 1d array. 

acc stands for accumulator in reduce wherein it is initialized as empty array, concat() the array every iteration.

Again we can reduce the solution above into a single line of code.

ES6 Approach:

const flattenAndSort = array => [].concat(...array).sort((a, b) => a - b);

What we have used new here is this operator "...", which is called Spread Operator

Spread Operator, flattens and spreads the array content. To learn more click here 

3. The Office I - Outed

Codewars link:

The problem is :

Your colleagues have been looking over you shoulder. When you should have been doing your boring real job, you've been using the work computers to smash in endless hours of codewars.

In a team meeting, a terrible, awful person declares to the group that you aren't working. You're in trouble. You quickly have to gauge the feeling in the room to decide whether or not you should gather your things and leave.

Given an object (meet) containing team member names as keys, and their happiness rating out of 10 as the value, you need to assess the overall happiness rating of the group. If <= 5, return 'Get Out Now!'. Else return 'Nice Work Champ!'.

Happiness rating will be total score / number of people in the room.

Note that your boss is in the room (boss), their score is worth double it's face value (but they are still just one person!).

Non ES6 approach:

function outed(meet, boss) {

let avg = Object.keys(meet).reduce(function (acc, k) {

if (k === boss) {
acc += meet[k] * 2;
} else {
acc += meet[k];

return acc;
}, 0) / Object.keys(meet).length;

if (avg <= 5) {
return 'Get Out Now!'
} else {
return 'Nice Work Champ!';

We have used Object.keys() to get meet object keys, which we will need to compare if the person is the boss or not.

If boss multiply it by 2, after accumulating the sum of all team members, divide it by length of keys inside the meet  object to get the average.

After getting the avg check if <= 5 and return the necessary statement.

ES6 Approach:

const outed = (meet, boss) => {

const avg = Object.keys(meet).reduce((acc, k) => (
acc += (k === boss) ? meet[k] * 2 : meet[k]
), 0) / Object.keys(meet).length;

return (avg <= 5) ? 'Get Out Now!' : 'Nice Work Champ!';

We still used the arrow function here and some ternary operator to make the condition much more shorter.

That ends our article, there's a lot of possible code improvements when using ES6.

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