Break camelCase

Description

Complete the solution so that the function will break up camel casing, using a space between words.

My answer

function solution(string) {
  return [...string].map(function(letter) {
    return letter === letter.toUpperCase() ? letter = ` ${letter}` : letter
  }).join('')
}

Other solutions

function solution(string) {
  return(string.replace(/([A-Z])/g, ' $1'));
}

function solution(string) {
  return string.replace(/([a-z])([A-Z])/g, "$1 $2");
}

Wrap up

RegExp.$1-$9

$1, ..., $9 properties are static, they are not a property of an individual regular expression object. Instead, you always use them as RegExp.$1, ..., RegExp.$9.
The values of these properties are read-only and modified whenever successful matches are made.
The number of possible parenthesized substrings is unlimited, but the RegExp object can only hold the first nine. You can access all parenthesized substrings through the returned array's indexes.
These properties can be used in the replacement text for the String.replace method. When used this way, do not prepend them with RegExp. The example below illustrates this. When parentheses are not included in the regular expression, the script interprets $n's literally (where n is a positive integer).

Examples

var re = /(\w+)\s(\w+)/;
var str = 'John Smith';
str.replace(re, '$2, $1'); // "Smith, John"
RegExp.$1; // "John"
RegExp.$2; // "Smith"

source: mdn web docs

Multiples of 3 or 5

Description

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Finish the solution so that it returns the sum of all the multiples of 3 or 5 below the number passed in. Additionally, if the number is negative, return 0 (for languages that do have them).
Note: If the number is a multiple of both 3 and 5, only count it once.

My answer

function solution(number){
  let array = [];
  if (number < 0) {
    return 0;
  } else {
    for(let i = 1; i < number; i++) {
      array.push(i);
    }
    const three = array.filter(ele => ele % 3 === 0);
    const five = array.filter(ele => ele % 5 === 0);
    const result = [...new Set([...three, ...five])];
    
    return result.reduce((a,b) => a+b, 0);
  }
}

Other solutions

function solution(number){
  var sum = 0;
  
  for(var i = 1;i< number; i++){
    if(i % 3 == 0 || i % 5 == 0){
      sum += i
    }
  }
  return sum;
}

function solution(number){
  return number < 1 ? 0 : [...new Array(number).keys()].filter(n => n % 3 == 0 || n % 5 == 0).reduce((a, b) => a + b);
}

Wrap up

keys method

The keys() method returns a new Array Iterator object that contains the keys for each index in the array. It returns a new Array iterator object.

array.keys()

Examples

const array1 = ['a', 'b', 'c'];
const iterator = array1.keys();
for (const key of iterator) {
  console.log(key);
}
// expected output: 0
// expected output: 1
// expected output: 2

const arr = ['a', , 'c'];
const sparseKeys = Object.keys(arr);
const denseKeys = [...arr.keys()];
console.log(sparseKeys); // ['0', '2']
console.log(denseKeys);  // [0, 1, 2]

source: mdn web docs

Disemvowel Trolls

Description

Trolls are attacking your comment section!
A common way to deal with this situation is to remove all of the vowels from the trolls' comments, neutralizing the threat.
Your task is to write a function that takes a string and return a new string with all vowels removed.
For example, the string "This website is for losers LOL!" would become "Ths wbst s fr lsrs LL!".
Note: for this kata y isn't considered a vowel.

My answer

function disemvowel(str) {
  return str.replace(/[aeiou]/gi, '');
}

Other solutions

const vowels = 'aeiou';

function disemvowel(str) {
  return str
    .split('')
    .filter(letter => !vowels.includes(letter.toLowerCase()))
    .join('');
}

Wrap up

Regex g, Regex i

g = global, match all instances of the pattern in a string, not just one
i = case-insensitive (so, for example, /a/i will match the string "a" or "A".
source: stackoverflow