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Arigato

In the video, you said the TC of building the BIT is O(nlogn). If we use a prefix sum array just for making the tree, we can reduce it to O(n) right?

Not understand

great video

Thanks for a clear explanetion :D

Nicely explained

best explaination on internet

Thankyou so much sir

9:41 if suppose there is a tiny url exising for a long url and you do put, then it will override the existing tiny url.. will it make sense?

Thank you! This is the most thorough video explanation on Morris Traversal that I have been able to find.

there is a problem at 6:45 , after visiting -1 , we got to 2 , then we begin finding predecessor of 2 , Alas , finding predecessor of 2 means going right most in left sub-tree of 2 , here left subtree is -1 , then lets find predecessor ,-1 -> right = 2 , then 2-> right = 5 , 5-> right = 6 , 6-> right = 8 , 8 ->right = 10 , we are moving in totally different direction ....now . Actually this must stop when predecessor is found , but for predecessor , we keep moving right ? Edit: Now I saw the final code , while finding predecessor , you solved the problem by doing , predecessor-> right = current , then break .

Q1. How communication is done b/w servers? Ex. N2 and N3 Q2. How search will be performed on the data present in blob/cold storage?

The best explaination with perfect testcases that I found for this question on Internet. Thank you!

Why do you not make any more videos? These videos are literally gems! I wish you had made more videos on the system design!

Thank you. Very detail and understandable.

Best explanation,I have watched so far!

Thank you great explanation

why 8? i cant undrestand:(

Why do you sfeak and not speak 😂😂

Hero

Won't the counter/all-host/range_based approch generate duplicate tiny urls? With these approaches how do I search if a long url already exists? Won't a DB table [long -> tiny] also be required to avoid duplicate URL generation?

Thank you ,master

i am having a lot of issues with algorithms and logarithms, any thing you can suggest?

We can do this in a 1d array rather than 2d i mean no need to store the whole shit right just compute in a single dimensional array

does the algorithm stays the same when buildings are extended to n-dimension space

Very good explanation!!

Beautiful explanation. You address the actual logic of finding the longest suffix that is a prefix in a substring, which it easier to understand.

Thank you for this video, you help me a lots

you are the best

2024 anyone?

beautiful

thank you so much

great .thanks

Important thing I missed in school: "only one item" 2:39

I understood everything. /s

Coming back to this even after 6 years of college. Big Time Nostalgia !!!

Awesome!

the trick is that If I am interwined by you two strings , my last char must equal to one of your last char. When we deal with each spot in the table , we consider it as the last char of the target string

I love this, great work sir

I don`t understand the O(m+n) part. I don`t understand its purpose.

I have watched Neetcode's and Striver's explanations for this problem, but neither of them come close to this 6-year-old video.

I couldn't understand this earlier. now I'm very clear about this. Such a awesome explanation. Thankyou very much sir.

Man at least go through why we are using this approach ?

2024 hit like

👍

Thank you!! Very helpful video

After 8 years, still so much better than our computer science teachers. Thank you for the video. -Turkey

goat

Thank you so much for the very detailed explanation. I finally understand this.

This is banger. Thank you so much, God bless you.