Array - 52: Segregate Odd & Even Numbers in given Array

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  • Опубліковано 5 жов 2024
  • Source Code: thecodingsimpl...
    Solution
    We'll take 2 variable, left & right
    Left = 0 & right = arr.length - 1
    Now we'll start from left & keep on increasing left as far as it's even value as our target is to place even values at first
    Similarly, we'll start from right & keep on decreasing right as far as it's odd value as our target is to place odd values in second part
    Now we swap values
    We do until left is lesser than right
    Time Complexity: O(n), where n is elements of array
    Space Complexity: O(1)
    For more info, please see the video.
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КОМЕНТАРІ • 23

  • @karthikkoppala9664
    @karthikkoppala9664 11 місяців тому +4

    Isn't the time complexity O(n^2)? Since we are using nested loop?

  • @nikhilaggarwal9325
    @nikhilaggarwal9325 3 роки тому +4

    Your content is really good please keep on posting videos. I have learnt a lot from you. Thanks

  • @shivamkumarjha3302
    @shivamkumarjha3302 3 роки тому +1

    Just Keep on. Your content is better than famous youtube channels. Thank you so much.

  • @something_different8968
    @something_different8968 Рік тому

    so nice of you to make this simple. Thanks

  • @a9ocalypse
    @a9ocalypse 2 роки тому +1

    Thanks a lot for this amazing explanation

  • @altheadelavegadewites941
    @altheadelavegadewites941 2 роки тому +2

    Your discussion are good, Would you please make a video on how to separate even and odd number from 1-100, Thankyou!!!

    • @CodingSimplified
      @CodingSimplified  2 роки тому

      Thanks for your feedback. Sure, I'll try to do. Noted your question.

  • @yoshitamahajan353
    @yoshitamahajan353 3 роки тому +3

    Ur channel is so best!

  • @sakshitiwari4367
    @sakshitiwari4367 2 роки тому +2

    Why we swap the value can you please explain

  • @sysybaba420
    @sysybaba420 Рік тому +1

    what if had to maintain order(stability) of all even and odd numbers, then how would the code/login look like?

  • @chandramoulieshn6070
    @chandramoulieshn6070 Рік тому

    To make it effective, why don't we use "Lomuto’s Partition Scheme" --> lesser lines of code and provides the best understanding.

  • @NazmulAlom
    @NazmulAlom 2 роки тому

    Amazing session.

  • @areegahmed3926
    @areegahmed3926 2 роки тому

    what if the odd numbers are more than the even?so the condition of right>left how will it work?

  • @phuongdinh3769
    @phuongdinh3769 3 роки тому +2

    Hi I have 1 question, if you already put (left< right) in the big while loop, why do you still include it in the smaller while loop. Thanks

  • @surajgrandhi6742
    @surajgrandhi6742 2 роки тому

    what to do if they ask in the sorted form of even and odd

  • @its_neel_ok
    @its_neel_ok Рік тому

    how to sort this using in built sort function ----
    sort(a.begin(), a.end(), [](int x, int y) {
    return x % 2 < y % 2;
    });
    i got this code but didn't understand how can we manipulate sort function to our use?

  • @prasenjitmitra8434
    @prasenjitmitra8434 3 роки тому +1

    Sir if we need to maintain the oder of numbers after segregation what we need to do

  • @codewithfaizu1520
    @codewithfaizu1520 2 роки тому

    suver