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this is by far the best explanation for this question

greaaaaaaaaaaaaat video I finally get it

I absolutely loveddd this explanation! Seriously! this is sooo good! I hope you make more such videos!

very clear explanation!!! thanks

Thank you sir.I needed this video at this time!😄

Great explanation. Thank youuu !!

Dude, thank you for this. Embarrassed to say that I've been coming back to this problem for a week while working on Binary Search. LeetCode has this listed as an "easy" problem. I feel as if you just do it recursively then yes, it's easy to solve. But using Binary Search to optimize it turned into a lot of googling and youtubing that made no sense until I came across this video. Thanks again!

explanation is just amazing, thank you!

For those, who like me still didn't understand solution after video, here code in python: ```python from typing import List import logging logging.basicConfig(level=logging.DEBUG) logger = logging.getLogger(__name__) def findCircleNum(isConnected: List[List[int]]) -> int: cnt, visited, l = 0, set(), len(isConnected) def dfs(start: int): visited.add(start) for end in range(l): if isConnected[start][end] and end not in visited: dfs(end) for start in range(l): if start not in visited: cnt += 1 dfs(start) return cnt if __name__ == "__main__": test_cases = [ [[1, 1, 0], [1, 1, 0], [0, 0, 1]], [[1, 0, 0], [0, 1, 0], [0, 0, 1]], [[1, 0, 0, 1], [0, 1, 1, 0], [0, 1, 1, 1], [1, 0, 1, 1]], ] expected = [2, 3, 1] for index, item in enumerate(test_cases): res = findCircleNum(item) logger.debug(f"res == {res}, expected == {expected[index]}") ``` Simply speaking, dfs marks as `visited` all the linked to current item nodes, therefore in the main loop `cnt += 1` called only for remaining nodes

I understand the implementation part, but I am not sure why we are interested in the earliest end time :(. greedy is tough, man.

what pen and setup do you use to draw diagrams?

you are great <3

it took me 45 minutes to use the trie version of this, and I found yours, I am using this version from now on. Thanks!

Thank you so much for giving a great explanation.

bc angrezo thoda dheere bol liya kro banda smaghne aaya hai

This is the simplest version I've come across, can't thank you enough

@babybear4812 - The time complexity of the code will O(N^2) as in the worst case when none of the cities is connected, in that case, the DFS method will called N times, and the outer loop will also be executed n times making it O(N^2).

Your video took this from being pretty incomprehensible to me to being super trivial in 20 minuts. Thank you sooooo much for your very clear explanation. It was incredibly helpful.

Best explanation I have seen. Thank you very much! The solution seems simple but it is not easy to come up with the solution on the spot especially given the limited amount of time in the interviews. I don't think I would be able to fully solve this myself in 20 minutes.

trash solution with comprehensive amount of code even using queue, when problem states that you should use constant space. The coolest approach I have seen, is "needle" approach, when we use each lvl of tree as linked list nodes. it uses O(1)

Excellent video, subbed, keep it up

Thank you for mentioning about aliasing on curPath array when we directly add to result array!

Great explanation !

Request for toonblast leetcode

Amazing!

This problem was so confusing for me to understand, thanks for explaining

Much thanks, quite clearly explained!

Hey, great explaination and video. But I think leetcode has introduced some new test cases and this approach doesn't work anymore. I've followed this and coded up the Go version but it got WA. Directly copied the python version still got WA. 😞

what is the time and space complexity of this code?

yes,bear is right about nonlocal declaration ,we use global var inside function incase we need to update a variable declare/used outside of a function and can be used throughout code .similarly we use non local var inside nested function incase we need to update a variable declared in the outer function.local is when we declare and use variables inside any function.

best best best explanation :)

Nice one, man. Thanks.

Its that simple? 😲😲😲 OMG thanks a lot for this

buena viejo, explicaste chido

cool explanation, thanks :)

goated video

Yesss, the fact that they call top right before bottom left is low key really confusing 😅

I think there is confusion in the complexity, lets say the graph is only directly connected pairwise (like 1-2, 3-4, 5-6 ... etc) in that case the dfs function will be called N/2 time and therefore time complexity is O(N^2) only. We are supposed to track how many times dfs function will be called, not how many times a node gets visited.

Just remove the ANY word from the title. Because this is all the theory etc and even the intuition is discussed everywhere, books, videos etc. But all that still does not help when trying to solve a difficult problem. So I was expecting that you will also introduce the intuition both for recursion and DP with more complex problems so that practice and thought process will help solve ANY problem. But you just discussed one problem and probably the most easiest one and call the video title as ANY. I can guarantee that just by watching this video even you will not be able to the next difficult problem where DP will "improve" the recursion. Ask yourself.

what would be the time complexity with and without using memo ??

it really gave clarity through that dry run!! , i think i mis-read the question though , . . . . .what i thought was -> , it was necessary that we have exgatly k employees selected ? our(your) code can give (maximum )answer even when we dont consider k people , but less than k people , , here is a test case . . . . eff = [100 , 1 , 1] , speed = [100 , 1 , 1 ] k = 3 , so we need 3 people but taking only the first guy gives us the answer as 100*100 , while taking all three of them reduces our answer to 1*(100+1+1)=102 , ,, , this is could be a variant to the original question , small change to our code like start taking maximum only when the size of heap has reached k should just do the thing!! Happy learning !

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How come it’s enough for you to DFS from the curr as START only? How come we don’t have to also consider other STARTS for this (using curr as END) So if for example were iterating on START =2. So yes we check all the [2][0…5]. But how about checking the [0…5][2] ones too?

this is the best explanation out there for word break 2

12:16

Can you define dfs as a nested function and keep graph outside it? Then you don’t need it as a parameter do you? Why in the official solution do they choose to unmark currNode from visited at the end of dfs?

Very nice explanation, easy to follow. Drawings definetely help a lot. Thanks man

Did you ever interview at FAANG? And why are you so happy when doing Leetcode ? :-)

wawaweewa, it's a very nice!

It's great. Your idea is more understandable and helpful than leetcode card for me. I'm lucky today to meet your video. Thank you so much.