The coding interviews / coding challenges are designed to assess how productive someone can be with the language. In contrast to the algorithm whiteboarding interview, which assesses how thte candidate go about solving a problem, or the architecture design interview, which assesses how thte candidate’s experience in system engineering and product design, the coding interview is is utilized by the company interviewing you to answer one question:

Can you code?

You are usually given the flexibility of choosing the language you are most comfortable with to complete the coding challenge. Thus it’s expected you know how to take advantage of all the language specific features and deal with the language specific idiosyncracies (JavaScript has a lot) to solve problems.

These tests are also designed to gauge your knowledge of computer science fundamentals like various datastructures (e.g., arrays, strings, object/dictionaries), logic building blocks (e.g., loops, if-statements, functions), and problem solving patterns (e.g., recursion, pattern matching, higher order functions).

For coding challenges, there’s always a time limit. The faster you can solve a problem correctly, the better. Thus, this post introduces JavaScript features, best practices, to leverage and common pitfalls to avoid to help you get productive with JavaScript and get the right result quickly.

Table of Contents

For more extensive cheatsheet, check out Let’s Get Productive With JavaScript

Working with Strings


In a nutshell, regular expression, or Regex for short, are patterns you specify to test a string. There are whole books written about Regexes. We are going to focus on the most likely problem we’re going to need to solve with regex during a coding challenge:

Does the string contain this pattern? If so, how many times does this pattern appear? Where does it appear?

Let’s consider this coding challenge:

Given a string, determine if the string is a url.

We can assume we only want to test for websites with the .com and .org top level domains. We don’t care if the website actually exists or not.

Input and expected output:

  • “” -> true
  • “” -> true
  • “google.foobar” -> false
const isStrUrl = str => {
  const matchTld = /(\.com|\.org)$/i
  return matchTld.test(str);

The matchTld is the regex that specifies that we are looking for a string that ends with .com or .org. The \. escapes the The $ at the end of the regex means:

Regex URL Regex URL

You can match an exact sequence in the string using the following regex:

const matchApple = /^apple$/
matchApple.test('apple'); //> true
matchApple.test('orange'); //> false
matchApple.test('apples'); //> false
matchApple.test('Apple'); //> false

The caret (^) and the dollar sign ($) signal “match from the beginning” and “match until the end”, respectively.

As illustrated with the above example, regex is case sensitive.

If you want to match a sequence of characters regardless of whether the letters are upper case or lower case, you could convert every letter in the string to lowercase, then do the pattern matching:

const str = 'Apple'.toLowerCase();
matchApple.test(str); //> true

Other useful regex patterns:

  • whitespace: /^\s$/

String manipulation

Suppose we have a string containing alphanumeric characters sprinkled with some illegal characters, in particular, angle brackets, forward slash, backward slash, and quote. We want to remove the illegal characters and retain just the letters. How would we go about this string manipulation?

Using regex:

const dirtyStr = "<script>window.location=\"\"</script>";
const matchIllegals = /<|>|\\|\/|"/g;
const cleanStr = dirtyStr.replace(matchIllegals, "");
console.log(cleanStr); //> scriptwindow.location=http:evil.comscript

We are using the JavaScript built-in function for string, i.e., replace, to replace every angle bracket, slash, and quote in the string with empty string, effectively removing them from the string.

Get Characters from String

Get One Character

  • str.chartAt(i) - get a character at a index i from str.
let str = 'hello'
str.charAt(0) //> 'h'
str.charAt(str.length-1) //> 'o'
str.charAt(str.length) //> ''

What happens if you do the following?

str.charAt(str.length) //> ''

It returns an empty string! You don’t get any help like the index out of bounds error that people who work with a statically typed language with a compiler like Java would be familiar with. This could be a really nasty bug. A similar thing happens with arrays when you access an index out of bound array element: arr[arr.length] gives you undefined.

Getting Multiple Characters (Substrings)

Say we have “hello world” as our string but we only want “world”. What do we do? Slice recursively. “hello” plus space is 6 characters.

"hello world".slice(6) //> "world"

Using slice works because a string is really just an array of characters.

A more general function for obtaining the substring is appropriately called substring and it takes two arguments: where to start (inclusive) and where to stop (exclusive). Here’s how you use substring:

'hello'.substring(0, 2); //> 'he'

However, using slice and substring for extracting the sub-string requires knowing exactly where is the starting index is. Usually, the substring we want to extract are delimeted by a space or a special character such as a slash. Suppose we have the following problem:

Get the username from the Medium url for user profile.

Usually, webpages for user profile pages on social networking websites have the following form:


What we want is the username at the end. domainName and route could be arbirarily long. So we can’t use slice to solve our problem.

const getUsername = url => {
  return url.split('/').pop().slice(1);

getUsername('') //> xiaoyunyang

getUsername function has a one-liner solution but there are few things going on:

  1. split('/') splits the url string into an array of substrings using teh “/” delimiter. The output of the split('/') operated on our example url becomes [ 'https:', '', '', '@xiaoyunyang' ], which gets piped into the next operation pop()
  2. pop() is a built-in function for arrays we will discuss later. What it does is it returns the last element of the array and in the process, mutating the original array. pop() could get us in hot water (we will discuss later in the arrays section) because it’s mutating the original array but in this case, it’s ok because the array we got from split is an intermediate throw-away datastructure that we are only using for deriving the final result. The pop() operatin gives us @xiaoyunyang, which we pipe into the next operation.
  3. slice(1), as discussed above, returns the substring starting from index 1 until the end of the array. This effectively chops off the @ and gives us xiaoyunyang, which is the username.

Working with Arrays

When we work with arrays in JavaScript, we have a whole suite of built-in functions we can use. This section discusses those functions that returns arrays or mutates the original array. A later section will discuss how to use JavaScript built-in functions for arrays to transform the array to other data types.

Add things to an array and combine arrays


  • array.push(elem) - adds elem to the end of array.
  • array.unshift(elem) - adds elem to the beginning of array.


const mutatingAdd = [1, 2, 3];

mutatingAdd.push(4); //> [1, 2, 3, 4]
mutatingAdd.unshift(0); //> [0, 1, 2, 3, 4]

console.log(mutatingArr); //> [0, 1, 2, 3, 4]


  • array.concat(elem) - adds elem to the front or back of an array without mutating the original array.
  • [...arr1, ...arr2] - ES6 Spread operator to merge two arrays
  • [elem, ...arr2] - ES6 add elem to head of the array
  • [...arr1, elem] - ES6 add elem to tail of the array

Example with concat:

const arr0 = [0];
const arr1 = [1, 2];
const arr2 = [3, 4];

const arr3 = arr1.concat(arr2); //> [1, 2, 3, 4]
const arr4 = arr0.concat(arr3); //> [0, 1, 2]
const arr5 = arr0.concat(arr1, arr2)

Example with spread operator

const arr0 = [0];
const arr1 = [1, 2];
const arr2 = [3, 4];

const arr3 = [...arr1, ...arr2]; //> [1, 2, 3, 4]
const arr4 = [...arr0, ...arr1]; //> [0, 1, 2]
const arr4_altern = [0, ...arr1]; //> [0, 1, 2]
const arr5 = [...arr0, ...arr1, ...arr2]; //> [0, 1, 2, 3, 4]

Extract and remove things from an array

We can access elements in the array using the index. This pattern is commonly used in many other programming elements. This section goes into more details about how to take advantage of JavaScript-specific built-in function to speed up problem solving.

A common thing we need to do with arrays is to remove the last thing in the array. Since the goal is to mutate the original array, one convenient function helps us do exactly that:

  • array.splice(-1) - return the tail of array as an array. mutates array.


const arr = [0, 1, 2];
const tailArr = arr.splice(-1);
console.log(tailArr) //> [2]
console.log(arr) //> [0, 1]

One gotcha with using splice in this way is arr.splice(-1) returns the tail element of arr wrapped in an array. If you just want elem, then you can use ES6 destructuring:

const [tail] = tailArr;
console.log(tail); //> 2

Unfortunately, ES6 destructuring does not support extraction of last element of an array. However, ES6 destructuring and rest operator is very useful for returning the head of the array as elem instead of an array containing an elem without mutating the rest of the array:

const arr = [0, 1, 2];
const [head,] = arr;
console.log(head); //> 0
console.log(rest); //> [1, 2]
console.log(arr); //> [0, 1, 2]

In fact, you can use ES6 destructuring to get extract more than one elements from the array starting from the head of the array:

const arr = [0, 1, 2, 3, 4, 5];
const [first, second, third,] = arr;

If you just want to return the last element of the array without mutating the original array, we can use arr.slice(-1):

const arr = [0, 1, 2];
const tailArr = arr.slice(-1);
console.log(tailArr); //> [2]
console.log(arr); //> [0, 1, 2]

slice and splice are useful for extracting any part of the array, not just the tail. The general way you use these functions is arr.slice(from, until) and arr.splice(from, until) where from and until are indices. The sub-array returned from these operations includes the element at index from until the element that precedes index until. For example:

const numbers = [0, 1, 2, 3, 4];
const lessThanThree = numbers.slice(0, 3); //> [0, 1, 2]
const moreThanTwo = numbers.splice(2, numbers.length); //> [2, 3, 4]
console.log(numbers); //> [0, 1]

Beware! If you reversed the order in which you use slice and splice, in the code above, the code will not have any run-time errors but you’re going to have some nasty logic bugs:

const numbers = [0, 1, 2, 3, 4];
const moreThanTwo = numbers.splice(2, numbers.length); //> [2, 3, 4]
const lessThanThree = numbers.slice(0, 3); //> [0, 1]
console.log(lessThanThree); //> [0, 1] ... not [0, 1, 2] as we expect

Therefore, when you are trying to quickly solve a programming problem correctly, be absolutely disciplined about using immutable functions like slice and ES6 destructuring so you’re not spending 10 minutes of your 30 minute coding challenge time debugging this bug.

Sorting things in Array


const nums = [3,2,7,1,2,0]

nums.sort((a,b) => {
  return a-b

console.log(nums) //> [ 0, 1, 2, 2, 3, 7 ]

nums.sort((a,b) => {
  return b-a

console.log(nums) //> [ 7, 3, 2, 2, 1, 0 ]

It’s important note that the sort function mutates the original array.

If you don’t pass in a function, sort will by default give you the mutated array in ascending order.

const letters = ['c', 'r', 'a', 'b', 'a', 't']
letters //> [ 'a', 'a', 'b', 'c', 'r', 't' ]

However, if you want to get the letters in descending order, we need to use string.prototype.localeCompare.

const letters = ['c', 'r', 'a', 'b', 'a', 't']
letters.sort((a,b) => {return b.localeCompare(a)})
letters //> [ 't', 'r', 'c', 'b', 'a', 'a' ]

Working with Objects

JavaSCript Objects are used to store key-value pairs and can nest other objects as deep as you want. In JavaScript, arrays are actually objects where the keys are numbers. I can make an array using an object notation:

const arr1 = {
    0: 0,
    1: 1,
    2: 2
const arr2 = [0, 1, 2];

How I access the elements are indistinguishable:

arr1[1] //> 1
arr2[1] //> 1

This is how I remember the rules for accessing the value of an object using the key - it looks just like the way you access the element of an array at a given index!

const basket = {
    apple: 1,
    pear: 2
const numApple = basket['apple']; //> 1
const numPear = basket['pear']; //> 2

Size of JS Object

  • Object.keys(dict).length gives you the number of entries in the object called dict, short for dictionary.


let basket = {}
Object.keys(basket).length; // 0
basket = {apple: 2}
Object.keys(basket).length; // 1

Merging two JS objects

  • Use spread operator

There’s a more verbose way of doing it but we are using ES6’s spread operator here:

const foo = {a: 'a', b: 'b'}
const bar = {c: 'c', d: 'd'}
const foobar = {,} //> {a: "a", b: "b", c: "c", d: "d"}

you can also use the spread operator to make a deep copy of the object.

const foobarCpy = {...foobar} //> {a: "a", b: "b", c: "c", d: "d"}

Consistent with the tip above for preferring slice over splice for array operations, we want to err on the side of immutability. We want to make deep copies of things before we start changing things in arrays and objects to avoid nasty bugs and side effects. Assigning objects to another variable does not make a copy of the object, rather, it creates a new reference to the original object:

const basket = {
    apple: 1,
    pear: 2

basket2 = basket;
basketCpy = {...basket}
console.log(basket === basket2) //> true
console.log(basket === basketCpy) //> false

Add and remove things from object

const addToDict = (dict, newKey, newVal) => {
  if (dict[newKey]) return dict;
  const newDict = {...dict};
  newDict[newKey] = newVal;
  return newDict;
const deleteFromDict = (dict, newKey, newVal) => {
  if (!dict[key]) return dict;
  const newDict = {...dict};
  newDict[key] = newVal;
  return newDict;

Compare two objects

What if you have two dictionaries and you want to see if they are equal?

let dict1 = {}
dict1["a"] = 1
dict1["b"] = 2

let dict2 = {}
dict2["a"] = 1
dict2["b"] = 2

JSON.stringify(dict1) === JSON.stringify(dict2) //> true

dict1["a"] = 2
JSON.stringify(dict1) === JSON.stringify(dict2) //> false

Conversion Between Data Types

Conversion Between Different Data Types Conversion Between Different Data Types

We can use functions to transform between these data types as depicted in the graph above.

Data Structures

Linked List

Basic operations are appendToHead, appendToTail. Linked List provides O(1) appendToHead operations but O(N) look up.

function ListNode(val) {
  this.val = val; = null;
ListNode.prototype.appendToTail = function(val) {
  let curr = this
  while(curr) {
    if(! { = new ListNode(val)
      return this
    curr =
  return this
ListNode.prototype.appendToHead = function(val) {
  let head = new ListNode(val) = this
  return head


Basic operations are:

  • push(item) - adds item to top of stack - O(1)
  • pop() - remove the top item from the stack - O(1)
  • peek() - returns top of the stack - O(1)
  • isEmpty() - returns true if stack is empty - O(1)

We can use JavaScript’s array object as a stack.

For example:

const browserHistory = []
const clickLink = link => {
const clickBackBtn = () => {
console.log(browserHistory) //> [ '' ]
console.log(browserHistory) //> [ '', '' ]
console.log(browserHistory) //> [ '' ]
console.log(browserHistory) //> [ '', '' ]

Problem to be solved using a stack is Palindrome

// palindrome (odd number): racecar, dad
// palindrom (even number): abba
// ignore space?
function isPalindrome(str) {
  let seen = [] //> stack

  // construct stack
  for(let i=0; i<str.length; i++) {
    let curr = str.charAt(i)
  let revStr = ''
  // construct reverse string
  while(seen.length > 0) {
    revStr += seen.pop()

    if(revStr!==str.substring(0,revStr.length)) {
      return false

  return str === revStr


  • add(item) - add an item to the end of the list - O(1)
  • remove() - remove the first item in the list - O(1)
  • peek() - return the front of the queue
  • isEmpty() - return true if the queue is empty

How to implement a Queue in JavaScript (See repl)

function DoublyNode(val) {
  this.val = val;
  this.prev = null = null
function Queue() {
  this.front = null
  this.back = null

  this.add = item => {
    let newNode = new DoublyNode(item)
    if(!this.front) {
      this.front = newNode
      this.back = newNode
    } = newNode
    newNode.prev = this.back
    this.back = newNode
  this.printForward = () => {
    let vals = []
    let n = this.front
    while(n) {
      n =
    return vals
  this.printBackward = () => {
    let vals = []
    let n = this.back
    while(n) {
      n = n.prev
    return vals
  this.remove = () => {
    // remove from the front and return the removed node's val
    if(!this.front) return null
    let firstNode = this.front
    this.front =
    this.front.prev = null = null
    return firstNode.val

  this.peek = () => {
    if(!this.front) return null
    return this.front.val


Initialize Array

let arr = Array(10)

Find Max:

const arr = [1, 2, 3]


let a = 'world', b = 'hello'
[a, b] = [b, a]
console.log(a) //> hello
console.log(b) //> world

Generating a sequence

const indices = Array.from(Array(10).keys()) 
console.log(indices); //> (10) [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]