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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Beatiful countryside landscape.

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 return true }

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

Awesome video! Thank you for making it 😀

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

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

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

Very informative and helpful.

So engaging and interesting thank you sir for this lecture ❤

// 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 if (str[start] != str[last]) { return false } else { start++ last-- } } return true }

Perfect, trainees will be ready!

loved the tutorial by Mukhtar!

Very helpful.

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

Love this

Thanks for your help 🎉🎉🎉🎉🎉🎉

Love this video ❤

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

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

OMG please do this but with Java or Python

@adlen01

Ай бұрын

Ther is ful apk in webstores

Thank you

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

Ай бұрын

var 🤢

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

@vnm_8945

29 күн бұрын

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

Need this in Python!

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

These are way easier than "easy" leetcode problems.

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

Saved

awesome

❤

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

@arzoohashimi519

Ай бұрын

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

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"));

🎉🎉🎉

👍🙂👍

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 }

0:22 He owns a farm

Honestly, any companies could ask questions like these?

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

@feerfeja

Ай бұрын

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

//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; }

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}`); }

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]; }