Algorithms - Solving Recurrence Relations By Substitution

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In this video I talk about what recurrence relations are and how to solve them using the substitution method. We analyze two popular recurrences and derive their respective time complexities.

[khanhchung]

Thank you for your explanation of this technique. This is the first video about DSA that I could watch entirely without feeling boring. I hopefully you can make more videos about the techniques that are frequently used in interview problems.

[zetrox]

What 5 websites, 3 other tutorial videos and my prof couldnt explain well enough for me to get, you have managed to explain perfectly clearly how to do this. Thank you.

[rohanwadhavkar]

That was incredible. You explained this much better than my professors or the TAs ever did. Subscribed!

[christianremwood]

This one video just clarified 4 weeks of confusion for me. Thank you so much!

[شوقالرجيعي]

THANKS i almost cried all other algorithms explanation videos sucked so bad its like they want me to fail the course (bad instructor bad videos) but im saved by you

[Amandep]

You've explained this better than any other video on youtube. Finally, I understand this stuff now.. Thanks for posting this

[amruthsar]

Thank you so much.. I hadn't understood this for years and this one lecture changed it..

[brianstorm]

Amazing presentation man. I wish my university professor was as clear as you were in the video. Kudos!

[IsaacC]

I found this video very helpful but there needs to be clarification at @12:52. The linear term, n*c, didn't fall from the sky. It comes from the 'merging' operation of elements from two sorted containers (each containing a subset of elements sorted from a more recent recursive call). The merging operation (a sorting algorithm in its own right, I suppose) involves comparing 2 elements at a time, one from each container, until either or both containers are exhausted. The larger (or smaller, depending on whether you sort high-low or low-high) of the two element is inserted into another container to be used in the next recursive step. The comparison and insertion constitute n*c complexity.

[adhwaaalanazi]

you are the BEST, you solve the biggest problem for me in the

[Tylermania]

you know there is a very high pitched screech in your into.

[JinayShah]

Thank you for making the video, you've explained it really well!

[Furighteous]

You made this very easy to understand, thank you!

[bulbul]

Loved the presentation. 😍 Which software do use to make such amazing stuff?

[millermeares]

18:00. Much simpler just to say 2^k = n as you already had that instead of doing the complicated log simplifying thing

[rajdevlive]

Exactly what I was looking for. :-) Thanks.

[icosmini]

The time complexity of the first algorithm is NOT 2^N.
It is N: to obtain the tenth Fibonacci number you just need to compute the previous nine numbers. You definitely don't need 2^10 (one million) computations.

[janssenkuhn]

These are great videos. Too bad they don’t show up in Google results and instead I get stuff from people who are better at SEO than SE.

[eXtrems]

Very nice it all. BTW which software do you use to write?

[fleal]

In case of recurrences like T(n) = T(2n/3) + T(n/3), this method doesn’t work. What kind of methods could we then use?