Top 10 Javascript Algorithms to Prepare for Coding Interviews 

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

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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 }

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Awesome video! Thank you for making it 😀

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

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

Very informative and helpful.

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('') }

So engaging and interesting thank you sir for this lecture ❤

Perfect, trainees will be ready!

loved the tutorial by Mukhtar!

Very helpful.

// 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 }

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

Love this video ❤

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Love this

We want more videos on these String and Array manipulation methods and programming questions

We are learning well from these type of Videos But please zoom in something more so that we can see the screen properly.

Thank you

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'))

Need this in Python!

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

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.

OMG please do this but with Java or Python

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

awesome

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

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

Saved

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

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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"));

👍🙂👍

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

0:22 He owns a farm

Honestly, any companies could ask questions like these?

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 }

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

//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,' '); }

//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; }

for(var i = 1; i { console.log(i); }, 1200); } Output is 4 4 4 & for(let i = 1; i { console.log(i); }, 1200); } Output is 1 2 3 Why??

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]; }

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.

i'm sure no one understand this

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

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}`); }

These are way easier than "easy" leetcode problems.

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()}`);

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]; }