L-2.6: Recurrence Relation [ T(n)= 8T(n/2) + n^2 ] | Master Theorem | Example#1 | Algorithm

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0:00 - Master Theorem
3:56 - Question

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amg
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ni
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mayurkalsekar
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Thanks sir for making these videos for us.It takes a lot of effort to make these videos in a busy schedule.

salilshukla
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Sir you are just too good. You have simplified the concepts so well. Thankyou so much sir.

aditiagarwal
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This is wonderful. Sir how simply you explain everything. My professor explains this topic in such a pathetic manner that i feel like I can never solve any master theorem question. But today I watched your video and everything became so easy. Thank you sir. I'm so grateful to you 🙏

vishakasaxena
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The next day is my exam and I don't have any idea of any the subject...
But after seeing his lectures is giving me next level confidence 🙂

karthiksm
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for those who are asking N3 (cube) kaise aaya.... its actually n raise to(log base b a ) that is log base2 8 = 3 ... only log base b a value is substituted nothing else.. Hope its clear ;)

shubhamsukum
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this is the probably the best video to remember master theorem

no-body
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outstanding way to make us understand these difficult problem, in a easy way.
thank you dada ❣️

Soumik
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Wah bade vaiya mauj kardi 🙏🙏 Itna aasan aur engaging tarikese concept koi nahi samjhaya
Thank You❤️❤️

subid.majumdar
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payalsonawane
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VAISHALIBEDI-hm
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aryanpatle
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This is wonderful. Sir how simply you explain everything

ANSARIGANSARIG-ri
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Thanks for this video to find the solution using master theorem.It is very useful for competitive exam.Thanks a lot.

dhasvino
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Sir u said that substitution method can solve any recurrence relation but here we are not given base case ?

ShubhamSharma-rptu
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salmanmanzoor
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Sir, thank you soooo much !! What my teacher could not explain in 2 hours, you explained here in 6 minutes. Please upload more problems with video solutions for practice also. Aur sir, aise hi har topic ko explain karna ! Good Job Sir!!

nannubedi
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n3 issliye hua kyuki Log 8 with base 2 == 2x2x2 = 8, 3 times 2 to get 8

jeetyadav