วีดีโอ

Really clear and concise explanation. This video helped a ton.

bro... come back ur vids r so good 😭

Wer sieht das auch für's HAW Studium?

Thank you! This was very helpful

Python code: # WITH Path Compression def find(self, parent, x): # finds root or representative of disjoint-set/union if parent[x] != x: parent[x] = self.find(parent, parent[x]) # path compression step return parent[x] def union(self, parent, x, y): root_x = self.find(parent, x) root_y = self.find(parent, y) parent[root_x] = root_y # WITHOUT Path Compression def find(self, parent, x): # finds root or representative of disjoint-set/union if parent[x] != x: return self.find(parent, parent[x]) return parent[x] def union(self, parent, x, y): root_x = self.find(parent, x) root_y = self.find(parent, y) parent[root_x] = root_y

Solved my hours long quandary in 5 minutes. Thank you!

Great explaination with beautiful graph!!! Love it with your DSA explaination.

Great explanation

is there any other way to do it?

very clear! Thanks!

To optimize it, the find function should set the parent of it's parent to find(x). (This will probably help a lot) function find(x): if Parent[x] != x: parent[x] = find(parent[x]) return parent[x] If I'm wrong please tell me.

Correction, this does not run in O(logn), but in O(n). To get O(logn) optimization, you need to use union by rank or by size.

This video is so gud but looks like the channel ain't active anymkre 🥺💔 I really loved the explanation n quality so thankuu n hope u r doing well ✨

But why are you doing log K I think for every level you are just doing one operation so the time complexity is just O(N)

This was actually insanely helpful in understanding the concept. Thanks!

It so simple and straight forward. Thanks you so much

beauty

That was so easy i spent an hour on this😭😭😭

The Big-Theta runtime for find() would be Θ(n).

You got a video on path compression?

Really great explanation. Couldn't understand it earlier but now got it. Thank you.

good video

Great explanation. Was able to recall what I had learnt a while back. Thank you!

Bro this topic is confusing AF, but at least this vid makes it somewhat understandable

Thanks Mr. Tater.

Thank you very much. Nice video.

Amazing! It's very clear. Thank you

Best explanation in this question

This is O(n) time complexity the modulo is a constant operation, also since were only modding 2 here we can just check the first bit in k to see if its a 1 or 0

This video was great! definitely not something that's very clearly covered in alot of other sources

Perfectly explained. Thanks

Short and clear, thanks

wonderful explanation! Thanks mate!

Still don’t understand shit

Thanks for the concise and easy-to-understand explanation!

really kool I have same approach in mind but wasn't able to solve thanks for the video

Thank you so much this is great!

Great explain, helpful visuals. Thnaks

Thank you so much man

Thanks

awesome

fantastic video! thanks!!!

Beautiful demonstration with clear mind

best explanation every ! :D

Why do we need to take ceil for k/2 i.e. k/2+k%2? Why just k/2 does not work, can someone explain, please? Thank you.

Really love your explanation and your simplistic code! This is by far the best video content on the topic I've seen so far, many thanks!

thanks a lot for explaining in few minutes.

Amazing video, very easy to follow, thank you! 🎉

There are two issues here: 1. When performing union, you set parent of the smaller set to the representative of the larger set. 2. When finding a path, you compress the path on your way back, improving O(logn) ops to O(log* n) in average.

dood this is amazing please do more videos man , like dp problems