Shovels Of Code  # Common JavaScript Interview Coding Problems

October 15, 2019

Todays post is about the dreaded interview coding challenges we tend to face as developers on the prowl for an awesome place to work. I’m back on the prowl for a job myself and I wanted to share some solutions to common interview coding problems that I’ve failed myself. Failing is painful but it’s usually the most memorable teacher. All the solutions below are my own. If you are new to coding some of this might be tough. Try and figure it out anyway. Set breakpoints in the debugger and see how things change as it runs line by line.

For several of the challenges I solve the problems multiple ways. This is a good practice to get into and will teach you a lot. That being said interviews are a performance art where programming is not. I have no answers to this problem but the dichotomy exists and must be coped with.

## .css-1fcz27o{margin-left:-20px;padding-right:4px;color:#663399;}1. Remove the Duplicates in an Array

Here is the function signature

`function removeDups(arr: any[]): any[]`

`let sampleArr = [1,2,5,5,8,19,18,16, 14,14, 'spiderman', 'spiderman'];function removeDups(arr) {const removeDupsMap = {};arr.forEach((item) => {    removeDupsMap[item] = item;});`

Object.keys returns an array of strings that represent the key names of the dictionary. I then map over the values and check if the item at each index is a string or number so that the function supports strings and numbers

`return Object.keys(removeDupsMap).map((item) => {    return isNaN(parseInt(item, 10)) ? item : +item;});}`

## 2. Compare two strings and determine whether they are anagrams.

The signature of the function looks like this

`function anagram(word1: string, word2: string): boolean`

`function anagram(word1, word2) {     word1 =  word1.toLowerCase().trim();     word2 =  word2.toLowerCase().trim();    const word1Map = {};    const word2Map = {};`

Now it’s time to count up all hits for each letter and store them inside a simple dictionary (hash map) so they can be compared against.

`for (var i = 0; i < word1.length; i++) {  word1Map[word1[i]] = word1Map[word1[i]] + 1 || 1}for (var j = 0; j < word2.length; j++) {  word2Map[word2[j]] = word2Map[word2[j]] + 1 || 1}`

Because objects do not have a length property use Object.keys which converts all existing keys to an array of strings. Therefore because an array has a length property we can easily compare the length of both hashmaps.

Check that both hash maps are of the same length otherwise there is no sense in going further.

`if (Object.keys(word1Map).length !== Object.keys(word2Map).length) {  return false}`

if maps are of same length then loop through and compare the key values

`for(key in word1Map) {   if(word1Map[key] !== word2Map[key]) {       return false;   }}return true;}`

## 3. Find all the vowels in a word

The signature of the function

`function findVowels(word: string): number`

This one is pretty easy using a quick hashmap / dictionary of the vowels. First split the word into an array. Then you can simply loop over the word one letter at a time. When a match returns truthy add one to the increment variable called vowelsInWord When finished return the result.

`function findVowels(word) {  let vowelsInWord = 0  const vowels = { a: "a", e: "e", i: "i", o: "o", u: "u" }  word.split("").forEach(l => {    if (vowels[l.toLowerCase()]) {      vowelsInWord += 1    }  })  return vowelsInWord}`

## 4. FizzBuzz

This one is a classic and it tests to see if you understand how to use modulo operator otherwise known as the remainder of division operator.

The challenge is usually stated something like this:

• console logs the numbers from 1 to n, where n is the integer the function takes as its parameter
• logs fizz instead of the number for multiples of 3
• logs buzz instead of the number for multiples of 5
• logs fizzbuzz for numbers that are multiples of both 3 and 5
`// solution 1function fizzBuzz(nums) {  let result = ""  nums.forEach(num => {    if (num % 5 === 0 && num % 3 === 0) {      result += "FizzBuzz" + "\n"    } else if (num % 3 === 0) {      result += "Fizz" + "\n"    } else if (num % 5 === 0) {      result += "Buzz" + "\n"    }  })  return result}// a test arr generatorlet sampleArr = []let i = 1while (i < 100) {  i += 1  sampleArr.push(Math.floor(Math.random() * 100 + 1))}console.log(fizzBuzz(sampleArr))`

Just a slightly different approach for the heck of it.

`// solution 2function divisibleBy(num, div) {  return num % div === 0}function fuzzBizz(num) {  if (divisibleBy(num, 3) && divisibleBy(num, 5)) {    return "FizzBuzz \n"  } else if (divisibleBy(num, 3)) {    return "Fizz \n"  } else if (divisibleBy(num, 5)) {    return "Buzz \n"  } else {    return ""  }}function fuzzBizzIt(sampleArr) {  let result = ""  sampleArr.forEach(number => {    result += fuzzBizz(number)  })  return result}console.log(fuzzBizzIt(sampleArr))`

# 5. Palindrome

A palindrome is a word, sentence or other type of character sequence which reads the same backward as forward.

Here is the function signature

`function isPalindrome(word: string): boolean`

a simple way to solve it goes like this:

`function isPalindrome(word) {  return word    .split("")    .reverse()    .join("") === word    ? true    : false}`

What if they ask you to solve it without using reverse? Oh snap it’s game time!

`function isPalindrome(word) {  const reversedWord = []  word.split("").forEach(letter => {    reversedWord.unshift(letter)  })  return reversedWord.join("") === word ? true : false}`

`function isPalindrome(word) {  word = word.split("")  const reversedWord = []  let j = 0  for (let i = word.length - 1; i >= 0; i--) {    reversedWord[j] = word[i]    j++  }  return reversedWord.join("") === word.join("") ? true : false}`

look ma! no array manipulation methods used! How does this work? Well glad you asked! Since I’m not using `Array.prototype.push() or Array.prototype.unshift()` I have to resort to a old fashioned for loop. I have set it to decrement down starting at `word.length - 1` and stopping at the zero index. The j variable is set above the loop at zero as well and increments while the i variable decrements.

If you get this question and they ask you to do it without the reverse method what they really want to know is if you know how to reverse a string by hand. Learn all 3 ways.

# 6. Check for values within a set range [50…99] inclusive

This one I literally just failed and it was so simple. Nerves get my heart pumping so quickly I could barely think. Don’t let this one get you. Logic is the first thing to get very hard when your entire body is in fight or flight mode. Never assume something is easy when you are in the hot seat. The easiest things become much harder.

`// Both values are within range of 50 ... 99function withinRange(x, y) {  if ((x >= 50 && x <= 99) && (y >= 50 && y <= 99)) {    return true  }  return false}console.log(withinRange(45, 100)) // returns falseconsole.log(withinRange(22, 101)) // returns falseconsole.log(withinRange(45, 155)) // returns falseconsole.log(withinRange(60, 88)) // returns trueconsole.log(withinRange(50, 99)) // returns trueconsole.log(withinRange(45, 99))  // returns falseconsole.log(withinRange(50, 30))  // returns falseconsole.log(withinRange(455, 234)) // returns false```---`