Video

Thanks for the video, helped me out. keep up the good work! 🌸

I have a doubt why we cannot make prefix sum of size same as nums size, why we initialize prefix sum 0 at 0 index

Thank you bro, this solution really helped me...

Nice explanation!

thank you!

good idea bro thanks

thanks broo

Nyc Explaination Bro

good explanation = thank you so much❤❤❤ please keep doing this kind of videos , because are very helpful for us

Thanks a lot!!!!! <3

code does work when hamsters =".HH.H.H.H.."

promo sm

Thank you so much for the video! It really helped me learn sliding window, such a smart way to solve it, the logic behind it is quite tricky though.

Nice Explanation

Great solution and easy to understand thanks for the whiteboard explanation.

So far. easiest explanation found on youtube ..Thanks !

great solution and explanation!

Can we use vanilla dijkstra instead?

Thanks, good explanation

it is not binary search solution((

Explained it in precise and concise manner ..amazing

your channel is a goldmine got my weekends covered now🙏

Thanks

The time complexity is O(n^4), sorry for that I didn't solve it online, after carefully checking, it should be this: we have s1, s2 with cache, it's n^2, we have for loop, it's now n^3, but we also have s1 == s2, it's another O(n), so the time complexity is O(n^4), if you have any questions, please make a comment.

awesome bro.....thankx

time is O(m + n)

Time is O(n^2), there is for loop inside.

Cool video, thanks!

I think they have added new test cases to this problem, this solution gives TLE now. But my intuition is that you can actually abstract the inner for-loop into its own method (example: def parseSubstring(...){}. If the for-loop in this new method is somehow transformed into a recursive call, then you can add @cache decorator above the new parseSubstring(...) method signature to memoize the results and prevent TLE. I'm too lazy to spend time implementing this though since I'd rather use my time to review other problems for my Amazon on-site lol.....

This really helped me understand this problem better, thanks

Beautiful solution thank you for making a video for this problem

Using python?

Which language is it

Thankyou for these videos great help

hey if possible can u please guide my comp. sci journey , im a first year student

It's the same for res = [] or self.res = [] for python global variable

"Answers within 10^-6 of the actual....." --> nice way to identify a potential Binary Search problem :) This would work for the Sqrt(n) problem too. But unfortunately not for Koko Banana or Aggressive Cows

wait i forgot about this one.Ty so much!!

Nice work man, very clean solution. Your English and explanation is very good 👍

Space complexity O(mn), dfs will call at most m*n times

Hey, I love what you doing. Keep it up. Love the content.

Very good vid

Mistake for analyzing the r value, it should be 2 * 10 ** 8, not 10 ** 16, I made a mistake by 10 ** 8 + 10 ** 8 = 10 ** 16.

Time complexity is O(nlogn) not O(n), heappush and heappop is log(N)

correction: space complexity O(n*k)

fire ❤❤❤❤

i totally understand your logic.. can you give solution in c++

Thank you, very well explained!

A better solution is to use message as a buffer. class Logger: def __init__(self): self.logger_set = set() self.message_queue = deque() def shouldPrintMessage(self, timestamp: int, message: str) -> bool: while self.message_queue: oldest_message, oldest_time = self.message_queue[0] if timestamp - oldest_time >= 10: self.message_queue.popleft() self.logger_set.remove(oldest_message) else: break if message not in self.logger_set: self.logger_set.add(message) self.message_queue.append((message, timestamp)) return True else: return False

Horrible. The map would keep grow until system is out of memory. This kind of implementation should be done using a ring buffer or something.