Recurrence Relation T(n)=8T(n/2)+n^2 | Substitution Method | GATECSE | DAA

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[@davidrahabi]

Can't understand what he's saying and it's still the most simple and straight forward explanation. Great video!

[@EpicKrushank]

Thanks you saved my 4 marks 😁😁

[@LordSarcasticVlogger]

Very Nice explanation

[@nani-db1rh]

Is iteration nd substi r same method..??? Plz reply sir

[@manojkumarsahoo254]

Sir what will be answer if question starts with T(n)=nT(n/2)+2n^2

[@shirishinamdar7013]

Really Helpful. Thanks.

[@LordSarcasticVlogger]

Thank You Sir

[@swastideep207]

Best

[@eng.mariamalhussainy687]

(like)^1000 ^_^ ... thanks sir sooooooo much

[@shubhamverma-sx4re]

Nyc channel

[@ChandanAkbarpuria]

Tree method se kaise hoga

[@manuelconte2127]

Someone can do this ? T(n) = 4*T(n/2)+n^2

[@avinash1211]

T(n) = 4T(n/2) + n² ....(i) T(n/2) = 4T(n/2²) + n² ....(ii) put eq(ii) in eq(i) T(n) = 4²T(n/2²) + n² + n² similarly after k iterations. . T(n) = (4^k)* T(n/2^k) + k*n² or T(n) = (2^k)² * T(n/2^k) + k*n² ....(iii) assuming T(1) = 1 then n/2^k = 1 n = 2^k k = log k base 2 put values in eq (iii) T(n) = n² + n² logn order of T(n) = Ｏ(n² log(n) ) ........ You can very same by master's theorem. It will form Case 2 i.e., log a base b = k (Hinti: k = 2 )

[@SMART_POOL]

@@avinash1211 Great work bro 👍

[@SMART_POOL]

@@avinash1211 how Bro feels after this - kzbin.info/www/bejne/l2GvZn-gqK2Aj5Ysi=WG_K8dbBnQIIRAbn

[@Haezy_thechef17]

sir 8 thik se likhe