Top 10 Javascript Algorithms to Prepare for Coding Interviews

Build a solid foundation and prepare you for Leetcode-style coding challenges. Learn the top 10 must-know Javascript algorithms interview questions to help you ace your coding interviews. This course will help you build a strong foundation in Javascript algorithms and tackle Leetcode problems with confidence.
💻 Code: github.com/codingmoney/javasc...
✏️ Course developed by @CodingMoney
⌨️ (0:00:00) Introduction
⌨️ (0:01:00) Reverse String & Integers
⌨️ (0:11:29) Palindrome
⌨️ (0:15:58) Max Char
⌨️ (0:33:43) Array Chunking
⌨️ (0:41:56) Title Case
⌨️ (0:49:31) Anagrams
⌨️ (1:07:54) Count Vowels
⌨️ (1:15:21) Fizz Buzz
⌨️ (1:20:02) Steps String Pattern
⌨️ (1:30:52) Pyramid String Pattern
⌨️ (1:39:24) Bonus - Spiral Matrix
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[@CodingMoney]

1:00 Reverse String & Integers 11:29 Palindrome 15:58 Max Char 33:43 Array Chunking 41:56 Title Case 49:31 Anagrams 1:07:54 Count Vowels 1:15:21 Fizz Buzz 1:20:02 Steps String Pattern 1:30:52 Pyramid String Pattern 1:39:24 Bonus - Spiral Matrix

[@augustoeduardo209]

Beatiful countryside landscape.

[@sayedrezaomid1305]

I am very happy that i saw my Afghan and ex colleaque trainings at this channel.

[@daisy_haniii]

Palindrome using two pointers technique: function isPalindrome(str) { let min = 0 let max = str.length - 1 do { if(str[min] == str[max]) { min++ max-- } else { return false } }while(min < max) return true }

[@MyCodingDiarie]

You're changing lives with your content. Keep inspiring!

[@orangeprogrammer]

Awesome video! Thank you for making it 😀

[@WhitneyChakara]

This is excellent and able to be completed as it's short and straight to the point.

[@marioerichsen1977]

function palindrome(str) { //with two pointer technique var stringArr = str.split('') const reverseArray = (arr) => { let i = 0; let j = arr.length-1; while (i < j) { let tmp = arr[i]; arr[i++] = arr[j]; arr[j--] = tmp; }; return arr }; const reversed = reverseArray(stringArr) return str === stringArr.join('') }

[@khanhphamthanh5012]

amazing spiral Matrix. It really so hard for me to find out the way. Thank you for your code

[@atultrp1]

// every method function palindrome(str) { return str.split('').every((ele, i) => { return ele == str[str.length - i - 1] }) } // two pointer technique function palindrome2(str) { let start = 0 let last = str.length - 1 while (start < last) { if (str[start] != str[last]) { return false } else { start++ last-- } } return true }

[@arzoohashimi519]

Very informative and helpful.

[@sergeynazarov2473]

Perfect, trainees will be ready!

[@user-dy8si5bf8t]

So engaging and interesting thank you sir for this lecture ❤

[@shulamohammad2275]

loved the tutorial by Mukhtar!

[@LokeshKumar-tk7ri]

Thank you Free code Camp for providing DSA in Javascript and also need more content on JS DSA

[@Adolfofilipe7369]

Thanks for your help 🎉🎉🎉🎉🎉🎉

[@murshidalam1982]

Very helpful.

[@edwarddk9007]

my solution for the Pyramid String Pattern exercise: function pyramid(n, level, body, space) { if(level

[@user-lp1jw1fq7d]

Love this video ❤

[@lazarokabira2945]

Love this

[@savchenkoilliya9131]

In my opinion I can give better and more javascript-style solutions for at least couple tasks. Maybe it's a little bit harder to read, but it works better when we are talking about js: var chunk = (array, size) => (array.length (str.split(" ") .map((word)=> (word.charAt(0).toUpperCase() + word.slice(1)).join(" "))) P.S. Good practice show O(n) complexity for each solution, because it's a think that can be asked on any interview.

[@deha9210]

var 🤢

[@user-cu7tw3vb2k]

Q1: function reverseIt(str){ let newArr = ''; for(let i = 0; i< str.length ; i ++) { newArr += str[str.length -i-1]; } return newArr; } console.log(reverseIt('hello'))

[@MS-gi3hc]

Thank you

[@dudekool12]

Need this in Python!

[@narekmeliksetyan9262]

it was very useful. but I will suggest the following solution for problem 9. function foo(n){ for(i=1;i

[@vnm_8945]

You could just use Array(i) instead of Array.from({length:i}), nice solution!

[@user-ko6zp7rt6b]

function isPolindrom(str) { let check = str.split('').reverse().join(''); return str === check ? true : false }

[@DavidMorenoH]

OMG please do this but with Java or Python

[@adlen01]

Ther is ful apk in webstores

[@mr_aha4713]

12:23 function pal(str){ console.log(str) if (str === str.split('').reverse().join('')){ return true }else { return false } } console.log(pal(""))

[@alikh5601]

Saved

[@-.nisa.-]

1:39:20 This is such an overcomplication... One 'for' loop is more than enough here: for (let i = 1; i

[@arzoohashimi519]

Big o notation for repeat function is o(n) which is similar to a for loop. So ur solution is also the same.

[@m0slah]

❤

[@ITRebels]

awesome

[@SaeedITSpot]

🎉🎉🎉

[@ajmalshahanpm938]

function count(str){ let counter=0; // for(let i=0;ifi===str[i])) { // counter++; // } // } for(single of str){ if (['a','i','o','e','u'].some((fi)=>fi===single)) { counter++; } } return counter; } console.log(count("aj code"));

[@moshtabamorsali]

I did each of them in a different way but still working in the same performance😶‍🌫

[@rishiraj2548]

👍🙂👍

[@vijayakabajo]

0:22 He owns a farm

[@user-ko6zp7rt6b]

function checkVowels(str) { let arr = [] const vowelsArray = ['a', 'e', 'i', 'o', 'u'] let arrFrStr = str.toLowerCase().split('') for (let i of vowelsArray) { for (let y of arrFrStr) { if (i === y) { arr.push(i) } } } return arr.length }

[@g0nt411]

Honestly, any companies could ask questions like these?

[@HazimTulumovic]

Max chars : function maxChar(str) { const usedChars = {}; for (const char of str) { usedChars[char] = usedChars[char] ? usedChars[char] + 1 : 1; } return Object.keys(usedChars).sort((a, b) => usedChars[b] - usedChars[a])[0]; }

[@subashstino899]

Is this MERN Stack developer needs to Study DSA ???… And I want to know Any best book for DSA( Data structure and algorithms )……????

[@feerfeja]

this video course is the best for learning algorithms in Javascript :p

[@nicolasmanns7249]

Palindrome with some kind of two pointers made with a for loop (also removes caps and spaces) function palindrome(str) { strToTest = str.toLowerCase().split(" ").join(""); let isPalindrome = true; for (let i = 0; i

[@phyapanchakpram]

//alternate newbie solution function capitilizeFirstLetter(str) { let arr = str.split(' '); let result = []; for(let s of arr) { result.push(s.charAt(0).toUpperCase() + s.substr(1)) } return result.toString().replace(/,/g,' '); }

[@phyapanchakpram]

//To return all char if max occurence has more than one time function maxOccurence(str) { let charMap = {}; for(let char of str) { charMap[char] = (charMap[char] || 0) + 1; } let max = 0; let maxChar = []; for(let key in charMap) { if(charMap[key] > max){ max = charMap[key]; } } for(let char in charMap) { if(charMap[char] === max) { maxChar.push(char); } } return maxChar; }

[@dogukankarabeyin3059]

Guys these questions and answers are identical to Stephen Grider's Coding Interview Bootcamp udemy course. I get these are common questions but it just seemed too similar.

[@OthmaneAHMARLAHYA]

i'm sure no one understand this

[@DavidDroddy]

For String and Int reversal and Palindrome -- .every and two pointer techniques: // STRING.SPLIT INTO ARRAY //const reverseString = (theString) => theString.split('').reverse().join(''); // SPREAD STRING(ARRAY-OF-CHARACTERS) INTO ARRAY // const reverseString = (theString) => [...theString].reverse().join(''); // ARRAY.EVERY(element, index, array) const reverseString = (theString) => { let reversedString = ''; const reversiosa = (character, i, charArray) => { // using index and array (all tests pass) // let lastElement = charArray[charArray.length - ++i]; // return reversedString += lastElement; const j = charArray.length - (++i) return reversedString += charArray[j]; // using element (all tests pass) // return reversedString = character + reversedString; }; [...theString].every(reversiosa) return reversedString; } // "TWO POINTER" SOLUTION // const reverseString = (theString) => { // let i = 0; // let j = theString.length-1; // let stringArray = [...theString]; // while(i parseInt(reverseString(theNumber.toString())) * Math.sign(theNumber); const isPalindrome = (theString) => reverseString(theString) === theString; // TESTS const stringToReverse = 'jar'; const expectedReversedString = 'raj'; const reversedString = reverseString(stringToReverse); let expectedAndReceived = `expected: ${expectedReversedString}; received: ${reversedString}`; if(reversedString === expectedReversedString) { console.log(`pass - reverseString('${stringToReverse}') ${expectedAndReceived}`); } else { console.error(`FAIL - reverseString('${stringToReverse}') ${expectedAndReceived}`); } let intToReverse = 53; let expectedReversedInt = 35; let reversedInt = reverseInt(intToReverse); expectedAndReceived = `expected: ${expectedReversedInt}; received: ${reversedInt};`; if(reversedInt === expectedReversedInt){ console.log(`pass - reverse a POSITIVE integer reverseInt(${intToReverse}) ${expectedAndReceived}`); } else { console.error(`FAIL - reverse a POSITIVE integer reverseInt(${intToReverse}) ${expectedAndReceived}`); } intToReverse = -54; expectedReversedInt = -45; reversedInt = reverseInt(intToReverse); expectedAndReceived = `expected: ${expectedReversedInt}; received: ${reversedInt};`; if(reversedInt === expectedReversedInt){ console.log(`pass - reverse a NEGATIVE integer reverseInt(${intToReverse}) ${expectedAndReceived}`); } else { console.error(`FAIL - reverse a NEGATIVE integer reverseInt(${intToReverse}) ${expectedAndReceived}`); }

[@salaminipples965]

These are way easier than "easy" leetcode problems.

[@petehotchkiss4830]

Does the capitalise method need to split strings out? Would it not be more performant to use a RegEx for this? eg const capitalise = (str): string => str.replace(/\w\S*/g, (s) => `${s.charAt(0).toUpperCase()}${s.slice(1).toLowerCase()}`);

[@HazimTulumovic]

Max chars: function maxChar(str) { const usedChars = {}; for (const char of str) { usedChars[char] = usedChars[char] ? usedChars[char] + 1 : 1; } return Object.keys(usedChars).sort((a, b) => usedChars[b] - usedChars[a])[0]; }