Video

Best of the best explanation. I think understanding the problem here is the biggest challenge than actually solving it. Row 1: "1" (Base case) Row 2: "11" (Previous has one 1s) Row 3: "21" (Previous has two 1s) Row 4: "1211" (Previous has one 2s and one 1s) Row 5: "111221" (Previous has one 1s and one 2s and two 1s) Row 6: "312211" (Previous had three 1s and two 2s and one 1s)

Thank you.

how does the tree have a right side of one if they all none on the right side??

This was one of the best LeetCode explanations I have ever watched!

im new to this and this problem , although easy, gave me a major headache , cuz i did not know how to run code, and had no idea how to implement it, just to see how it works

nice explanation at the end <3

Visiting your channel after 2-3 years. Hope you are good.

// another solution, I guess faster /** * @param {number} n * @return {string} */ var countAndSay = function (n) { if (n === 1) return "1"; let finalResult = "1" for (let i = 2; i <= n; i++) { let stringInFocus = finalResult; let currentResult = ""; let counter = 1; let prevCharacterIndex = 0; let currentCharacterIndex = 1; while (!!stringInFocus[currentCharacterIndex]) { if ( stringInFocus[currentCharacterIndex] === stringInFocus[prevCharacterIndex] ) { counter++; } else { currentResult = `${currentResult}${counter}${stringInFocus[prevCharacterIndex]}`; prevCharacterIndex = currentCharacterIndex; counter = 1; } currentCharacterIndex++; } currentResult = `${currentResult}${counter}${stringInFocus[prevCharacterIndex]}`; finalResult = currentResult; } return finalResult; };

// following solution is similar to "Level order travelsal ii" var minDepth = function(root) { if(!root) return 0; if(!root.left && !root.right) return 1; let q = [root], depth = 0, isFound = false; while(q.length !== 0 && !isFound){ let nodeCount = q.length; while(nodeCount !== 0){ const current = q.shift(); if(!current.left && !current.right){ isFound = true; break; } if(current.left) q.push(current.left); if(current.right) q.push(current.right); nodeCount--; } depth++ } return depth; }

// following solution is identical to the solution of the "Balanced binary tree" var maxDepth = function(root) { if(!root) return 0; function getHeight(node){ if(!node) return 0; let left = getHeight(node.left) let right = getHeight(node.right) return Math.max(left, right) + 1 } return getHeight(root) };

var isSubsequence = function(s, t) { let pS = 0, pT = 0; while(!!s[pS] && !!t[pT]){ if(s[pS] === t[pT]){ pS++; } pT++; } return (pS - 1) === (s.length - 1) };

Well explained!

took me forever to get this

Hello TW, I have been learning Javascript at a slow pace. I am still outside of the software engineering and development universe (I don't have a job in this field, yet.). I am in my late 50s and have desired to learn code for a while. With regard to my previous attemps of learning programming, I was quite confused by the complexity of writing instructions that a computer can comprehend. Your videos have lessened the fear and confusion that I havepreviously experienced when it comes to leetcode problems. Thank you and I hope to return the favor to you, soon.

THANK YOU FOR SPEAKING NON INDIAN ENGLISH! 🙏

great explanation

didn't pay much attention and thought you coded in java :)) , anyways very good teaching , I finally understood this 1d dp

thanks

This is fantastic and I can't thank you enough for speaking calmly enough to grasp what you are saying without having to repeatedly rewind, as so many tutorials require lol. I will definitely look for more of your videos. (subscribed✔)

Thanks mate

Hmmmm

Dude, this was so helpful with the visuals on the left side breaking down the code! liked and subbed

thank you for making these! Still helpful 2 years later! Your explanations are super helpful!

var lengthOfLastWord = function(s) { return s.trim().split(" ").filter(each => each !== "").slice(-1)[0].length };

Super clear explanation!

var majorityElement = function(nums) { const half = nums.length / 2; let charMap = {} for(let each of nums){ !charMap[each] ? charMap[each] = 1 : charMap[each] += 1 if(charMap[each] > half) return each; } return -1 };

// this beats 74.79% of the leetcode solutions var intersection = function(nums1, nums2) { return Array.from(new Set(nums1.filter(each => nums2.includes(each)))) };

var intersection = function(nums1, nums2) { let finalResult = [] let nums1_uniques = new Set(nums1); for(let each of nums2){ if(!finalResult.includes(each) && nums1_uniques.has(each)){ finalResult.push(each) } } return finalResult };

Here is my regex approach to the isValidCharacter function: _function_ isValidCharacter(char) { *return* char.length === 1 && char.toLowerCase().match(/[a-z0-9]/i); }

The intro song pulls me in every time. Anyone know the name of it?

I think you should rebrand to "Terrific Whiteboard"

Why did you stop making videos?

Thank you. EXCELLENT explanation! Interview prep is such a grind sometimes... When you have questions about a problem and you can watch a 25m video and come away crystal clear.. its invaluable

Have some of the best explaination on CZcams rlly learned a lot, you should post more man would love a watch a whole DS&A playlist by you and the way you explain things with the drawings! cheers.

fantastic explanation

Slight improvement (?): You don't need the break keyword, you can just return lastWordLength and it will break out the loop

This was the best explanation among all others!

simple explanation, made a lot of sense to me

Explain why you used this mid point calculation formula.. getting confused with regular binary search

Dude. Such a great job with the explanation! Thanks so much! Also, see you haven't posted in about 2 years, hope you're doing well.

Super explanation ! Watched many videos, yours is definitely the clearest one. Thank you so much !

I only understood your sol after searching all over, the way u name the var as pointer actually helped me a lot.

the best explanation I’ve ever seen so far. Great work! Thank you

Underrated channel. You are the best. Please upload more videos.

beautiful.

Hi. Why did you stop making videos?

That was really confusing to understand.

i love your explanations and videos. subscribed!

Thank you for great explanation

Thanks Whiteboard... your not that terrible dude hahaha