Mid-level React Interview

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  • Опубліковано 25 лис 2024

КОМЕНТАРІ • 107

  • @FilhoLouco
    @FilhoLouco 2 місяці тому +5

    Technical Questions and Answers
    React Fundamentals:
    State vs. Context: Explained the difference between state and context in React.
    State Updates: Discussed how to trigger state changes in parent components from child components.
    Redux: Shared thoughts on Redux and its boilerplate nature.
    Component Passing: Explained how to pass components as props to other components.
    Coding Exercise:
    Longest Substring.

  • @Kreshel1
    @Kreshel1 Рік тому +53

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

    • @davronmaxmudov3972
      @davronmaxmudov3972 8 місяців тому +5

      Thanks 😊, you save me 40 minutes)

    • @RealParadox85
      @RealParadox85 4 місяці тому +1

      the type of BS they use in react interviews these days, stupidly enough 😂

  • @licokr
    @licokr Рік тому +12

    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!

  • @mooosafir74
    @mooosafir74 Рік тому +10

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

  • @devzozo
    @devzozo 8 місяців тому +3

    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...".

  • @khalidben9940
    @khalidben9940 11 місяців тому +3

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

  • @michaelyabut5969
    @michaelyabut5969 Рік тому +73

    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

    • @michaelyabut5969
      @michaelyabut5969 Рік тому +5

      Oh just saw the end that it's 2n not n^2 TIL

    • @lowestCommonDenominator
      @lowestCommonDenominator Рік тому +16

      same, i had an optimal solution in my head immediately but in an interview setting i'm pretty sure i'd forget what javascript even was

    • @wlgs
      @wlgs Рік тому +9

      its n^2, subStr.includes(currentChar) is kinda innocent but it's O(n), better to use set/hash map here

    • @MocBocUS
      @MocBocUS Рік тому +1

      @@wlgs my solution is just loop through the string once and keep track of the longest string and current string so it still be O(n)

  • @ahmedAltariqi
    @ahmedAltariqi Рік тому +4

    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! 😂

    • @justinmlawrence
      @justinmlawrence  Рік тому

      Awesome! 👏🏻👏🏻👏🏻

    • @ahmedAltariqi
      @ahmedAltariqi Рік тому

      Thanks!
      Any feedback from you is much appreciated!
      function getLongestNonRepeatingSubstring(str) {
      let currentSubstring = "";
      let longestSubstring = "";
      for (let char of str) {
      if (currentSubstring.includes(char)) {
      currentSubstring = "";
      }
      currentSubstring += char;
      if (longestSubstring.length < currentSubstring.length) {
      longestSubstring = currentSubstring;
      }
      }
      return longestSubstring;
      }

  • @AAmorim1989
    @AAmorim1989 10 місяців тому +10

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

    • @justinmlawrence
      @justinmlawrence  10 місяців тому +2

      Great feedback!

    • @cy3889
      @cy3889 7 місяців тому +2

      i had the same feeling, he didn't answer the basic technical questions at the start. Feels more like junior level

  • @mahmednabil2429
    @mahmednabil2429 Рік тому +10

    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.

    • @alexyan125
      @alexyan125 Рік тому

      Coz the loop starts from first character and only went one time.

  • @ilichpadilla2448
    @ilichpadilla2448 Рік тому +4

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

  • @duyanhcenland2044
    @duyanhcenland2044 Рік тому +4

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

  • @horne1395
    @horne1395 Рік тому +7

    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.

    • @guilhermecruz
      @guilhermecruz Рік тому +2

      yeah, Set would be a better choice performance wise

    • @bennetleff4088
      @bennetleff4088 Рік тому +4

      the trick is sliding window

    • @ASAPKep
      @ASAPKep Рік тому

      I would assume either way the worst case would be O(n) because you still have to iterate through the whole string character by character to append to your map. The implementation here was O(n+n) or O(2n) which is still equivalent to O(n) as the worst case.

  • @preslavgetov1330
    @preslavgetov1330 11 місяців тому +2

    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.

  • @pkorneev5226
    @pkorneev5226 4 місяці тому

    me personally would not even call him a junior at this point, but pretty good interview, questions at the start were more like an internship level max

  • @SuperYoda7
    @SuperYoda7 Рік тому +24

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

  • @PatrykBiegański-f5v
    @PatrykBiegański-f5v Рік тому +3

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

  • @krishnagupta31
    @krishnagupta31 Рік тому +4

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

  • @mahdijafaree5332
    @mahdijafaree5332 Рік тому +1

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

  • @nr7343
    @nr7343 Рік тому +4

    He did not complete the exercise. maxLength=currentVal

  • @Faith_L3S5-ZERO
    @Faith_L3S5-ZERO Рік тому +6

    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

    • @quochoangang6014
      @quochoangang6014 Рік тому +2

      Same. I'm never get confident enough of my skills too.

    • @wykydytron
      @wykydytron Рік тому +2

      It's what happens to most self thought programmers, because we don't have grades given by teachers and often no one to compare too we think we are still not good. Reality is we very well may be overqualified for junior position but are held back by simple lack of experience and by extention lack of confidence. Then there is whole interview stress thing we're i know fir a fact outspoken confident but weak programmers do better then timid shy but very good programmers it's just how it is, so it's important that interviewer is experienced enough to catch wats going on.

  • @gombosmm
    @gombosmm Рік тому +10

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

  • @StivenBallshi
    @StivenBallshi 10 місяців тому +2

    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.

  • @thinhbui4090
    @thinhbui4090 Рік тому +1

    thanks i really need this

  • @jackvu.hustle
    @jackvu.hustle Місяць тому

    1. State and Context

  • @divakarluffy3773
    @divakarluffy3773 2 роки тому +8

    Nice one! hope my interviewer was this kind 😂

    • @rl6382
      @rl6382 Рік тому

      Well was he?

  • @kashmyr8
    @kashmyr8 8 місяців тому +1

    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.

  • @barboradolejsova3502
    @barboradolejsova3502 4 місяці тому

    The music is a little bit distracting and making hard to understand sometimes. :( Othewise awesome job!

  • @yuliasereda5671
    @yuliasereda5671 9 місяців тому +1

    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?

  • @skdtackkskdtackk
    @skdtackkskdtackk 4 місяці тому

    Have you a version without background music?

  • @idanatiya3925
    @idanatiya3925 Рік тому +1

    The interviewer looks like Doctor who

  • @techmoo5595
    @techmoo5595 Рік тому +1

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

  • @user-zv6bv7eu8k
    @user-zv6bv7eu8k Рік тому +1

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

  • @SeRoShadow
    @SeRoShadow Рік тому +2

    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 ?

  • @berndeveloper
    @berndeveloper Рік тому +4

    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)

  • @Madion86
    @Madion86 Рік тому +3

    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.

  • @JS_Skyline
    @JS_Skyline Рік тому +5

    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

    • @nr7343
      @nr7343 Рік тому +4

      Sort? Second loop its better bc time complexty Mr russain

    • @Reubzzzzz
      @Reubzzzzz Рік тому +4

      Why need to sort the elements when you can just take their length and compare that. Its faster that way!

    • @lets_see_777
      @lets_see_777 Рік тому +1

      wow you went O(n) to On(logn) complexity and think its better

  • @andrewwall2730
    @andrewwall2730 11 місяців тому +2

    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.

  • @richardramirez5746
    @richardramirez5746 Рік тому +1

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

    • @justinmlawrence
      @justinmlawrence  Рік тому

      Yeah! Just reach out on our Discord. Always looking for new people! 👍🏻
      discord.gg/rFP29feQ

  • @Amzodt
    @Amzodt 2 місяці тому

    the background music ruins everything smh.

  • @sadique_x_
    @sadique_x_ 11 місяців тому +1

    chale i no reach!

  • @mycraplife
    @mycraplife Рік тому +11

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

    • @cod-the-creator
      @cod-the-creator Рік тому +4

      No joke, the real life correct answer is "dude, Gary, this API is whack, why are you giving me a bunch of bullshit instead of the thing I need?"

    • @maiaklimenko6614
      @maiaklimenko6614 7 місяців тому

      At least it's close to a real interview. Everyone asks questions about data structure

  • @anasouardini
    @anasouardini Рік тому +2

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

  • @21mighty86
    @21mighty86 Рік тому +4

    3:56 no, correct answer is callbacks

    • @techaddictdude
      @techaddictdude Рік тому

      Which question?

    • @21mighty86
      @21mighty86 Рік тому

      Prop drilling state setter does not fit to react flow of data, and it can cause bugs

  • @369-davian
    @369-davian 9 місяців тому

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

  • @Netai19906
    @Netai19906 5 місяців тому

    great content but the music is not helpful

  • @mahmednabil2429
    @mahmednabil2429 Рік тому +1

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

  • @foqsi_
    @foqsi_ Рік тому +1

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

  • @amy_lio
    @amy_lio Рік тому +3

    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;
    }

  • @JS_Skyline
    @JS_Skyline Рік тому +5

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

    • @prompt9416
      @prompt9416 Рік тому

      bruhh

    • @Adidobro
      @Adidobro Рік тому +7

      yeah new born baby level, rotting in trench soldier level, what else?

    • @mikejones8519
      @mikejones8519 Рік тому +7

      In Soviet Ruzzia, code write you!

  • @neok8902
    @neok8902 5 місяців тому +1

    bullshit interview

  • @gooseob
    @gooseob 7 місяців тому +3

    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), "");
    ```

  • @ondrejvrto
    @ondrejvrto 10 місяців тому +1

    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')); //

  • @BraedenSmith
    @BraedenSmith Рік тому +3

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

  • @JadesFitnessBucketList
    @JadesFitnessBucketList 7 місяців тому

    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;
    }, ' ');

  • @anasouardini
    @anasouardini Рік тому +1

    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

  • @elvado1014
    @elvado1014 Рік тому +2

    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;
    }

  • @DevMe-f5v
    @DevMe-f5v 2 місяці тому

    another way
    function getIterateData(str, list) {
    let filterChar = list ?? "ABCDDDEFGHI";
    let filterStr = Array.from(filterChar.split(" ")[0]);
    let pos = filterChar.indexOf(str) + str.length;
    let strVal = [];
    let center = Math.floor((filterChar.length - 1) / 2);
    let uniqueValue;
    if (pos

  • @MegaKaan-p4f
    @MegaKaan-p4f Рік тому +2

    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;
    }

  • @DevMe-f5v
    @DevMe-f5v 2 місяці тому

    this my solution
    function getNoNRepeatedLongChar(str, list) {
    let filterChar = list ?? "ABCDDDEFGHI";
    let pos = filterChar.indexOf(str) + str.length;
    let center = Math.floor((filterChar.length - 1) / 2);
    let uniqueValue;
    if (pos

  • @n_fan329
    @n_fan329 15 годин тому

    let LongestSubString = str.replace(/(\w)\1+/gi,' ').split(' ').sort((a,b)=>b.length - a.length)[0];
    let misingLetter = String.fromCharCode(LongestSubString[0].charCodeAt(0) - 1);
    console.log(misingLetter+LongestSubString)

  • @cagataykaban7592
    @cagataykaban7592 5 місяців тому +1

    const getLongestNonRepeatingSubstring = (s) => {
    let result = ""
    for (let i = 0; i < s.length; i++) {
    let char = s[i];
    let newString = s.substring(0, i)
    if (!newString.includes(char)) {
    result += char
    }
    }
    return result
    }