Master Theorem Example

Solve T(n) = 2T(n/2) + nlogn using the Master Theorem !
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[@lalitameena3321]

These videos are really helpful to understand the recursion methods. Please keep on posting for other topics too in datastructure and algorithms. Thank you so much.

[@trafalgarlaw9919]

Thank you for the explanation!

[@patrickkranzpiller6400]

Great video man. Thank you for the help!

[@supernerd10000000000]

You are awesome! Thank you so much for your help. You make everything so simple and easy to read. Wish you were my teacher. Btw, you have a soothing voice ahahhaha

[@chuaselo]

I love your work man please dont ever stop !!!

[@alibaba888]

THANK YOU <3 NOW I UNDERSTOOD HOW TO USE THE MASTER THEOREM

[@Tanessok]

Thanks for your explanations. It helped me :)

[@Gollum1905]

1 question:

[@TheTechnicalman]

Thanks, I believe number 2 if it's base of the log, need to be down ⬇

[@subhankargoswami3521]

awsome explanation

[@kodermanishe]

thanks man a lot, greate video

[@jaybshow1660]

Thumbs up nice work

[@marcus47_]

your description of case 3 converts c into k halfway through i think. ('a*f(n/b) <= k*f(n) where k < 1' should be c not k

[@sonemanibitna3053]

I appreciate you

[@Dan-tf9kb]

This guy reminds me of the episode of My Wife & Kids where Junior became a genius

[@yashthanthratey1942]

Keep it up.. 👍

[@rahulmallah8306]

Isn't n^logba smaller that n log n? Shouldn't case three be used?

[@darkestknight1260]

If 3Tn/2+42?????

[@sqbossh]

Sir I have a feeling you have it wrong... this is CASE 3! You have to compare f(n) with n^log b a and if you compare it it is f(n) = n*log(n) and n^log b a = n ... so "n*log(n)" grows asympt. more quickly then only "n" obviously... so it's case 3 , but it fails in n*log(n) e Omega(n^1+e) because we can find e where this is not true.

[@ejaycarandang5885]

1. T(n) = 4t (n/2) + n 1g n