String Processing in Python: Look-and-Say Sequence

[@caseykongpanickul3426]

I came across this video while doing leetcode problems looking for help. Your explanation here is really great and much appreciated. I'm subscribed now and really looking forward to checking out your other vids!

[@LucidProgramming]

Cheers! If you enjoyed and benefited from my content, please consider liking the video and subscribing to the channel for more content like this. If you would like to support the content creation on this channel please consider unblocking ads when watching my videos as this is how I support my time to make content. I hope to be putting out more similar videos soon!

[@mahmoudtokura]

This playlist has been absolutely brilliant, your explanations are so easy to follow and remember. Thank you so much. I hope you do a playlist on leetcode or codewars questions, that would be awesome. Thanks again

[@LucidProgramming]

Cheers! If you enjoyed and benefited from my content, please consider liking the video and subscribing to the channel for more content like this. If you would like to support the content creation on this channel please consider unblocking ads when watching my videos as this is how I support my time to make content. I hope to be putting out more similar videos soon!

[@debupatra2333]

Thank you for this playlist , really this is clearly the best for python Algo. Your teaching methods are very productive and easy to catch up.

[@LucidProgramming]

Cheers! If you enjoyed and benefited from my content, please consider liking the video and subscribing to the channel for more content like this. I hope to be putting out more similar videos soon!

[@rjedwards158]

Your walk thru of typical python interview questions are very good. Much better than a lot of other users (such as CS Dojo). I think if you were to title your videos "Interview Questions in Python" or something similar rather than "Algorithms in python" like a lot of the other users you may get much higher views (if this is your aim). I think your content is very good but maybe you should bigup your site a bit.

[@vinoddiwan5792]

agree with you CS Dojo is just a big mouth Ex Google Employee

[@KippiExplainsStuff]

This was a fun little exercise :) my approach was slightly different and I would love for you to look at it and see if there is any disadvantage compared to your method (other than having more lines of code): def generate_next_number(num): new_num = "" count = 1 for i in range(len(num) - 1): if num[i] == num[i+1]: count += 1 else: new_num += str(count) + num[i] count = 1 # This takes care of the last digit if count == 1: new_num += "1" + num[-1] else: new_num += str(count) + num[-1] return new_num

[@LucidProgramming]

Cool! From what I can tell, it looks like your approach works. It is kind of a fun problem, and I'm glad you had a good time with it. Cheers, and thanks as always for watching, Amos!

[@sarbjitgahra4667]

That was done so well. Thank you for your knowledge.

[@LucidProgramming]

Thank you very much Sarbjit, I'm thrilled to hear that. :). You're very welcome, and I look forward to continuing to share it. Cheers, and thanks again for watching!

[@Mahdi_B]

Thanks, this was a big help!

[@LucidProgramming]

Thank you! If you like my content, I've been working on some projects during the past couple of months. If you would like to stay up-to-date, please consider subscribing to my mail list. Also, if you haven't already, please consider subscribing! I really appreciate the support, and if you want to support the channel, I do have a PayPal link www.paypal.me/VincentRusso1 for donations that go directly to the creation of content on this channel. I hope that the content I provide there will enhance the videos on my KZbin page. bit.ly/lp_email

[@chiragsoni2946]

Great Explanation. I wonder what is the time complexity of this solution. Is it O(n2), n being the length of the largest string?

[@LucidProgramming]

Thanks, Chirag. One thing I would be mindful of when calculating the runtime is that there is also an outer loop that goes over "n" times and the length of the string (call that "m") as something else. I think your calculation needs a bit of a refinement to fully capture this solution.

[@simonfraile3725]

I have to say, this was really cool and helpful. But I have to ask, do you have a job related to programing or is it just like a hobby??

[@LucidProgramming]

Thanks. And both!

[@simonfraile3725]

@@LucidProgramming What is your job??

[@LucidProgramming]

@@simonfraile3725 You can find out all that info on my website listed in the "about" section of my channel. Cheers!

[@simonfraile3725]

LucidProgramming ok thank you really much

[@LucidProgramming]

@@simonfraile3725 Sure, no problem.

[@janhavibhosle683]

Thank you so much!!

[@LucidProgramming]

Cheers! If you enjoyed and benefited from my content, please consider liking the video and subscribing to the channel for more content like this. If you would like to support the content creation on this channel please consider unblocking ads when watching my videos as this is how I support my time to make content. I hope to be putting out more similar videos soon!

[@niazahmad7823]

basically i did the same question but with better complexity, you can check it out. it is using one loop internally instead of 2 loops and it also uses recursion for n. if any one need help ask i will help. def substr(n, s): if n==0: return(s) l = [] count = 0 cur = None length=len(s) for i in range(length): if s[i] == cur: count += 1 else: if cur is not None: l.append((cur, count)) cur = s[i] count = 1 l.append((cur, count)) an=[] for i in l: an.append(i[1]) an.append(int(i[0])) return substr(n-1,an) s=[1] ans=substr(4,s) print(ans)

[@LucidProgramming]

I don't see how this is a better complexity. In fact, this seems to be worse than what is shown in the video.

[@niazahmad7823]

@@LucidProgramming you are using two loops internally to calculate the next number then you are using one loop for n so you are basically doing it in O(n3). I am using one loop to calculate the next number and using recursion for n. And time wise recursion is O(n). So my code is doing it in O(n2) i may be wrong but thats what my understanding is

[@niazahmad7823]

@@LucidProgramming now I did it using a loop instead of recursion and the complexity of the whole program is O(n2) and one that is shown in the video has the complexity of O(n3) def substr(s): l = [] count = 0 cur = None length=len(s) # 1st pass for i in range(length): if s[i] == cur: count += 1 else: if cur is not None: l.append((cur, count)) cur = s[i] count = 1 l.append((cur, count)) an=[] for i in l: an.append(i[1]) an.append(int(i[0])) return an s=[1] n=4 for i in range(4): s=substr(s) print(s)

[@LucidProgramming]

@@niazahmad7823 How is two loops O(n^3)? Two loops would be O(n^2).

[@niazahmad7823]

@@LucidProgramming sir you are using two loop in function and one loop in the main that makes it O(n3)

[@yt-1161]

this was interesting

[@LucidProgramming]

Cheers! If you enjoyed and benefited from my content, please consider liking the video and subscribing to the channel for more content like this. If you would like to support the content creation on this channel please consider unblocking ads when watching my videos as this is how I support my time to make content. I hope to be putting out more similar videos soon!

[@moeal5110]

How can we do this recursively?

[@LucidProgramming]

Check my recursion playlist.

[@shubhamh7451]

by any chance, can we use count function in this problem? something like, print(list.count[i] , i) and maybe for other number as well and then append all together? did anyone also happen to think this way or is it wrong?

[@LucidProgramming]

You could, although this runtime would be worse. You'd have two loops--one for iterators "i" and "j" and then another factor of "n" for the count function, which would increase the runtime to O(n^3) instead of O(n^2).

[@shubhamh7451]

@@LucidProgramming got it!

[@titouanriot3663]

why do you write Len(s) ? it is an 'int' so I do not understand

[@LucidProgramming]

No, it isn't an int. It's in between quotes and therefore treated as a string.

[@jishanshaikh2003]

What is meant by running time of an algorithm? What are the different types of running time? Like I heard n, log(n), etc. What is meant by this, and how are they identified? Can anyone please explain??

[@LucidProgramming]

It's a general metric to measure the number of computational steps to complete an algorithm. I'm not going to be able to go into detail here, but I suggest looking into "Big Oh" notation on Wikipedia.