Recurrence Relation T(n)=2T(n/2)+nlogn | Substitution Method | GATECSE | DAA

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[@LordSarcasticVlogger]

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[@ashakiranpratihari4463]

no youtube channel provides with such type of explanation and these many question .thanks for helping me a lot

[@uroobashahid8037]

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[@m3ssiah575]

That was really helpful sir . Thank you

[@prasoongautam4097]

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[@njcoder2641]

Such a good explanation sir

[@batuhanbayr7613]

He is the god of matematic

[@davidrahabi]

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[@prasoongautam4097]

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[@zoltatwarz1956]

Thank you sir.

[@Emperor723]

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نحن بحاجة الي معلمين متلك بجامعتنا والله 😞

[@Duda-kf1pd]

Bom vídeo, estava difícil encontrar esse exercício. Mas uma coisa, o resultado ficou incorreto quando tirou o denominador 2, fiz a prova por indução sem retirá-lo e estava correto. Só de fazer o teste para n=2 já dá para perceber que está errado. Fiquei um tempo tentando entender isso antes de testar, então é um aviso para quem ficou com a mesma dúvida.

[@swopnilmaharjan8630]

wait, i understand the sum of natural number to log(n) but there is substraction too?? for each term of log containing n?
why are we just ignoring that?

[@mastershifu8908]

Why at the final part the denominator 2 is cancelled?

[@LordSarcasticVlogger]

Sir aapse padhne me mazaa aa rhaaha h

[@barkasafeer4832]

agr isko master theorem sy kryn tu b Big O k sath he aye ga ans ya phr Theta k sath?

[@tanmoybera3673]

Is there any difference between Substitution Method and Iteration Method

[@yrkkhcreationworld786]

Badhiya ji baki sab to thek hai but jo marker ka sound hai wo bahut bekar lag rha tha or disturb bhe kar raha tha

[@moonthoughts8822]

is question ko master theorem se bhi solve karien

[@abhinavprakash5267]

sir you look like Dino james
nice explanation

[@aboveallff7698]

Isko recurrence se kar sakte ha

[@mmusic.9507]

This Is a bit wrong from 2:12 when we will multiply n/4 it'll be multiply with 1/4 not 1/2 will be T(n/16) At third Not T(n/8)

[@LARAIBFATIMA-qe4in]

please solve this question using substitution method.
12T(n/2)+1/5n^3
please solve this one

[@jbm5195]

Is there no condition that n must be powers of 2?

[@THEGATEHUB]

I don't understand your doubt... Could you please explain in detail ..for further communication you can contact me at instagram

[@G.T828]

there must be some value of k for which 2^k >= n, we just take 2^k = n for simplification because the time is still constant, even if n/2^k < 1 and we can take it to be 1 unit of time

[@alien77333]

answer should be O(n^2(logn)).

[@aksacts]

Yess

[@prajjwaljaiswal2832]

please explain and how we resolve log n/2

[@saurabhkumar-cj5ff]

log(n-1) or logn- 1

[@THEGATEHUB]

logn-1

[@saurabhkumar-cj5ff]

log(n/2)=logn-log2 =logn - 1. is it right sir

[@THEGATEHUB]

Yes.. I mentioned in Vedio also..

[@saurabhkumar-cj5ff]

ok thanku sir

[@THEGATEHUB]

For further communication you can contact me at instagram..