I like the concept of structuring the interview , very scientific.

As a person who preparing for frontend interviews, it really helped. Many people talk about the question because the question may seem like focused on algorithm. I don't know if this is the best interview for the md-level "react" interview though, the point is that. This video is really helpful for me. I really appreciate it you made this video and uploaded it on youtube. Thank you very much! I'm looking forward to further mock interview videos!

Do more Interviews like this one. Really helpful video. Keep it

Pretty easy to do it faster than n^2 at home, but I would probably freeze up in an interview lol. I could feel that guys anxiety haha

I think a good way to practice would be to copy a hackerrank problems into a react app, display the result(s) in a component, then make an input for the user to see different results from whatever the hackerrank question is asking to do. That seems to be the theme for many of the early-mid level questions. Cody kept saying something along the lines of "Theres probably a better way to do this" and I think that's ok to say, but if you say it too frequently, it makes you seem less confident or qualified. You might want to reword it in a more 'impressive' way by saying something like "I would probably review the documentation to make sure this is the best way to do this...".

thanks i really need this

The use of Array.includes method will affect the time complexity that will be O(n^2), nested loop so to speak.

I remember watching this video when it was released and I didn't understand anything at all. Now, I solved it without watching the video and I feel really happy! 😂

My first thought about the first problem is to have a way to know which characters has already been iterated to. Maybe something like a Set or a Map, array could definitely be used but that would be O(n). then whenever the current iterated character is already in the set or map, that means we have to reset the tracker.

In that getTheLargestSubString problem, u can probably use 'continue' to just skip that repeating char that u don't want in ur array

- He did not crush it, he would have been rejected. - This is not a mid level interview, more like an entry level one.

react interview... 2 react questions and then 90% of the video a leet code exercise

Would this not be a junior/entry interview...? I am on that level, and I feel like I could've done exactly the same

Nice video but I think the candidate overall would have failed in my book. 1. They answered all the React questions in the beginning pretty well. 2. The solution for the coding challenge is pretty much O(n^2) due to the includes nested in the for loop. The solution is not optimal and a pretty messy. 3. The react code was messy. The util method should have been brought in from another file instead of residing in the component. The util function itself should have been "pure" and should not include any React state setter methods. That should be done outside of the util method. Not knocking the interviewee at all though, it's a lot harder under pressure obviously.

ahh.. btw the solution that they agreed on is not correct, consider "ABCDAFG", their code will output "ABCD" but the correct answer is "BCDAFG". also this is an easy two-pointers problem.

Man, I'm struggling to get entry/junior level job and could answer those questions anytime.

guys take it easy, as he said it is a MOCK interview haha

I was wondering about your question smth llike what we get if we pass the component to another component. so, can be the answer that we will get a HOC(high order component) which get and return a component. if we use TS, so we can describe it like React.NodeChild. i'm a junior, so don't judge me please. but can it be the answer?

i won't lie, i did not expect the materials for mid-level position interview would be like this; i understand and can kind of solve the problem by myself which was surprising i always thought that i am more of a junior position guy despite having more than 1.5 yoe because i am never get confident enough of my skills.. but then maybe that's why i am still not good enough for a mid-senior level because i'm still not confident? no telling honestly lol

I would just add (based on my interview experience) that perhaps stating that you have never touched something or don't know it (in the redux/jotai question) does not really add anything to conversation other then name drop. If you don't have any knowledge to share, a simple don't know about it would suffice.

in reality we would use chatGPT for that function and then just build the component.

as a self taugh i feel that its hard to organize what you need to know to get a job

Great video, is there anyway to try this with you?

Nice one! hope my interviewer was this kind 😂

He did not complete the exercise. maxLength=currentVal

Now seriously though, is this interview even junior level. I mean I figured in the first 10s that you should create an array to add chunks then sort it by length. And despite this, when I apply for a position, I dont even get an interview call, not even for an unpaid internship position. It makes me feel bad since I self-studied React 2 Yrs. My portoflio is still in building phase, but the ideas I try to execute require extensive development (Job Platform, Company Portal, ECommerce Everything) Should I copy porfolio projects and then rewrite them as my own (while still working on my own projects) , just to get to the interview ?

The interviewer looks like Doctor who

The solution to this is wrong tho. If we had another string for example str = 'ABCDAEFGH' - the longest string here is 'BCDAEFGH'. The solution cody provided wouldn't work , so you need to check with either the substring method or charAt because you need to see if one of the characters is repeated.

Instead of using second loop you could use sort method to sort the array from biggest to lowest(arr.sort((a,b) => a.length < b.length ? 1 : -1)) and then return the first element of the array

Your solution fails on a string 'ABCDAEFGHI'. Your solution answer = 'AEFGHI', correct answer = 'BCDAEFGHI'. Sorry, but dividing by chunks isnt always enough to solve this problem; so your algorithm is wrong from its idea.

so lost as to how the function knows what the new maxLength is????

I really don't like that much this kind of interviews, because in real life you have an entire world to explore different approaches to hit the best solution. It feels more stressful than a real scenario, if you ask me. Good questions though. (As feedback: Next time don't use music, most of us are listening our own music)

Not sure I understand how the algorithm challenge has anything to do with react. Seems 90% of time not about react. And the code challenge was not well explained. Thought any consecutive same chars should have been eliminated in the results, but guess I was wrong.

chale i no reach!

Often, one's posture when sitting can reveal their level of expertise or competence.

3:56 no, correct answer is callbacks

really loved watching this, i kept pausing after the question to quickly think about what i would say and how I would write the function before Cody answered. My solution was super similar apart from i used **continue ** in my if block in the loop and at the end to find the longest str in the array i did: return strArr.reduce((longest, current) => { return current.length > longest.length ? current : longest; }, ' ');

"Mid-level" is kind of meaningless in this exact interview.

Terrible React interview. It was an algorithms interview, which he will never need to use.

also the real complexity is more than n^2, you should consider the usage of string "+=" operator is O(N) itself

having a b2 level of english, will be deat on the tree first minute 🤮

Cool video, but so difficult for me to listen to people say "uh" or "um" every other word. haha

bthw in Russia this is definitely not a mid-level it’s more like junior or even intern level of knowledge

FWIW: Big O on this solution 100% O(n^2) because of the `.includes`.

Probably smth like that // 'ABCDDDHIGKL' function longestUniqueSeq(str: string) { let maxStr = ""; let bestMaxStr = ""; for (const ch of str) { if (maxStr.includes(ch)) { bestMaxStr = withMaxLength(bestMaxStr, maxStr); maxStr = ch; continue; } maxStr += ch; } return withMaxLength(bestMaxStr, maxStr); } function withMaxLength(str1: string, str2: string) { return str1.length > str2.length ? str1 : str2; }

How is my clever (I guess) solution? #!/bin/env node let longestChunckIndexPair = [0, 0]; let firstIndex = []; const setIfLongest = (lastIndex)=>{ if((lastIndex - firstIndex) > (longestChunckIndexPair[1] - longestChunckIndexPair[0])){ longestChunckIndexPair = [firstIndex, lastIndex]; } } const func = (inputStr)=>{ // O(n-1) for(let i=0; i

interesting interview! I have a short solution to the algorithm question ```js const longestNonRepeatingSequence = (str) => str .match(/.*?(.)(?=\1|$)/g) .reduce((acc, item) => (item.length > acc.length ? item : acc), ""); ```

const getStr = (s) => { let startSlice = 0; let longestUniqString = " "; const compareAndSetString = (endSlice) => { const newSting = s.slice(startSlice, endSlice); if (longestUniqString?.length < newSting.length) { longestUniqString = newSting; } }; for (let i = 0; i < s.length; i++) { const currentLetter = s[i]; const nextIndex = i + 1; const nextLetter = s[nextIndex]; if (currentLetter === nextLetter) { compareAndSetString(nextIndex); startSlice = nextIndex; } compareAndSetString(nextIndex); } return longestUniqString; }

My preety solution with reducer. .... Changing the condition to "actualChar === result[0][0]" gives you the second version. const longestUniqueSequention = (str) => { return [...str] .reduce((result, actualChar) => { actualChar a.length < b.length) .at(0) .reverse() .join(''); } console.log(longestUniqueSequention('ABCDDDDEFGHI')); //

const getLongest = (str) => { var currStr = ""; var tempStr = ""; for(let i = 0; i < str.length; i++){ if( str[i + 1] && (str[i] != str[i + 1])){ tempStr += str[i]; }else{ if(tempStr.length + 1 > currStr.length){ currStr = tempStr + str[i]; tempStr = ""; } } } return currStr; }