🚀 neetcode.io/ - A better way to prepare for Coding Interviews
@Herald50
2 жыл бұрын
Will never forget my first day on the job where a customer requested a feature to find the longest increasing sub-sequence. A totally valid way to test someones competency as a developer
@btKaranDhar
2 жыл бұрын
Your sarcasam is depressive and hence ironic
@kotb2000
2 жыл бұрын
It is important to understand the idea of subsequences, use memory efficiently and understand complexities of exponentially increasing subroutines. A good Programmer is not the one who solves it the first time seeing it. A good Programmer is the one who understands all the dimensions of the problem and Learn as much as he can about underlying intuitions. and remember when Tony Hoare first made Quick Sort He didn't say I am not going to face a situation where I sort a customer's Needs. His attitude as any computer scientist when making a helpful research or an idea was If I am able to think How to sort and even invent a sorting algorithm then I am going to be able to satisfy my Customer needs who were Future Generations that used Quick sort in every Customer related Situation . Good bye.
@francisconovoa6493
2 жыл бұрын
@@kotb2000 gg
@zweitekonto9654
2 жыл бұрын
Which interviewer hurt you bro
@VibeBlind
2 жыл бұрын
@@zweitekonto9654 All of them
@lugiadark21
3 жыл бұрын
Please keep that tone and speed of voice. It really helps to "understand" the solution. All of us are here to "understand" the solution not just for a solution. You will do great my dude.
@shubhamsinghrawat6928
2 жыл бұрын
This how you except a google engineer
@AnupBhatt
8 ай бұрын
You can change the playback speed on KZitem to make it go faster or slower. In case you find a problem that some other youtuber has solved, that Neetcode hasnt solved yet, use that feature.
@dj1984x
Жыл бұрын
"I really doubt your interviewer is going to expect you to get this without a hint; if they do I'd just walk out of the room" Probably worked great up until the layoffs 😥
@MaxFung
6 ай бұрын
yep, now it feels like every other interview problem has been next level. still some easies out there though :(
@srinadhp
3 жыл бұрын
You sir. Are the savior. Your made it so simple - some times you make me guilty why I could not think of it. Keep them coming!
@jagrutitiwari2551
Жыл бұрын
Thank you for letting us know when we should walk out of the room. And what difficulty to expect in the interview :)
@Golipillas
Ай бұрын
You're the best at actually explaining the reasoning behind these problems and all your views are so well deserved. I'm absolutely dreadful at these despite being a senior dev but you don't know how much your channel has helped, thank you!!!
@director8656
3 жыл бұрын
Thanks, great explanation as usual, who needs cracking the coding interview when this exists!
@DanhWasHere
3 жыл бұрын
This explanation went down so smooth -your voice was very easy to follow and your diagrams weren't sloppy and not complex to understand -this might be the cleanest solution I have seen for this problem for beginners to study!
@jerrybao1934
Жыл бұрын
Thank you for the clear explanation! For those wondering why going backwards in dynamic programming, you can actually solve this in forward dynamic programming, start from the beginning, too.
@quocnguyeninh32
3 жыл бұрын
Your explanation is so great. The tone, voice, and the way you say are so clear. Thank you so much
@redomify
3 жыл бұрын
thank you omgosh this really is the best explanation one can find for this question on youtube
@alex-gz7ud
3 жыл бұрын
Clear explanation! I believe you are the rising star in solving leetcode problem.
@NeetCode
3 жыл бұрын
haha, thanks I appreciate the kind words
@christendombaffler
11 ай бұрын
Yeah, I agree that being expected to find the O(nlogn) solution is walkout tier. I came damn close to figuring it out: use an ordered set to keep track of the elements you've inserted so far so that you can easily find the greatest value that's smaller than or equal to your current one. From here, assuming nums[i] is not your maximum thus far, there are two ways, and figuring out either of them is easily upper Hard level: either you actually delete the value you've found (the fact that this works because it means you can just return the size of your set at the end is incredibly unintuitive), or you mess with the way you store everything in the set so that you can still retrieve the index of the value corresponding to your found value (which is awful to implement).
@SunsetofMana
4 ай бұрын
Why would deleting the value you found work? If you have input array [2,3,1,5,6] you cannot delete the value found at 5 when you see the 1, because then you cannot use it for the actual longest subsequence of 2,3,5,6 Tbh the approach of using a heap to store the subsequence length cache is quite reasonable imo… it’s annoying to implement but quite straightforward as an obvious improvement. If you know how to solve heap problems, which are based on the premise that a heap is a priority queue, why not just apply that here when you are searching for the largest element?
@gabrielfonseca1642
14 күн бұрын
@@SunsetofMana Not sure a heap would work, you mean for when you check the max right? You still have to know if nums[i] is less than the value, it's not enough to find the largest subsequence. In that case, it's still O(n^2).
@anushkachakraborty8635
2 жыл бұрын
That was so simple and an epic explanation of how you can start thinking about approaching this problem. Being a beginner at dp, your videos help me understand how to start approaching a problem :) Thankyou!
@thatguy14713
2 жыл бұрын
Easily the best channel for leetcode solutions. So easy to understand and code is always clean and concise. Hats off to you, Neetcode!
@sathyanarayanankulasekaran5928
3 жыл бұрын
this is brilliant, I wonder who can think of this solution for the first time during the interview
@medievalogic
3 жыл бұрын
excellent! This taught be how to choose subarrays recursively, and then the problem is trivial. Thanks a bunch.
@goshikvia
3 жыл бұрын
Optimal Approach O(nlogn) bisect_left is a python function which gives the lower bound of the element in O(logn) time. bisect_left(array, element, start, end) class Solution: def lengthOfLIS(self, arr: List[int]) -> int: subs = [arr[0]] for i in range(1,len(arr)): if arr[i] > subs[-1]: subs.append(arr[i]) else: subs[bisect_left(subs, arr[i], 0, len(subs))] = arr[i] return len(subs)
@davidespinosa1910
2 жыл бұрын
So if arr = [1,3,4,2], then subs = [1,2,4] ? That's not a subsequence. And yet it works. The mystery deepens... :-)
@paulancajima
2 жыл бұрын
@@davidespinosa1910 Yeah, the problem asks for the longest increasing subsequence. So, this will still give you the correct length just not the subsequence itself
@username-zs6dv
Жыл бұрын
noticing the subproblem 'is there an increasing subsequence with length m' is O(n), and m is between 1 and n, we can use binary search and get overall complexity O(nlogn). But it is way neater with DP
@akhtarzaman2189
2 жыл бұрын
"I'd just walk outta the room" you solve your own problem Mr interviewer LOL
@pujasonawane
3 жыл бұрын
I got it crystal clear now. You explained it very well. Thanks a lot.
@The6thProgrammer
10 ай бұрын
Not sure why, but this one felt much easier than the prior 3-4 problems in the Neetcode Dynamic Programming learning path. Got it on my first try, and solved it exactly the way Neet did. Just goes to show the value of the Neetcode Roadmap, and how the patterns start to solidify in your mind over time.
@engineersoftware4327
10 ай бұрын
That's correct, I feel the same way
@rajatsrivastava555
Жыл бұрын
Thanks for the explanation @neetcode , code in java : class Solution { public int lengthOfLIS(int[] nums) { int dp[] = new int[nums.length]; Arrays.fill(dp, 1); for (int i = nums.length - 1; i >= 0; i--) { for (int j = i-1; j >=0; j--) { if (nums[i] > nums[j]) { dp[j] = Math.max(dp[j], 1 + dp[i]); } } } int maxLIS = 0; for (int i = 0; i < nums.length; i++) { maxLIS = Math.max(maxLIS, dp[i]); } return maxLIS; } }
@juliahuanlingtong6757
3 жыл бұрын
Great demostration starting from brute force, work way up to memoization and then leads naturally to dp!!! So Nice and easy it becomes with your approach! 1 question though: Why work from end backwords? How did you get the instinct?Could you please share your thougghts?
@NeetCode
3 жыл бұрын
I'm used to working from the end backwards because it's similar to the recursive approach. But it's possible and maybe more intuitive to start at the beginning. Whatever makes sense for you is the best approach I think.
@eltonlobo8697
2 жыл бұрын
Example: [1,4,2,3], While computing longest common subsequence starting from index 0, the number at index 2, will be used. While computing longest common subsequence starting from index 1, the number at index 2 will be used. What i mean by "Will be used" is i am asking a question: What is the longest common subsequence starting from index 2. So if we had started computing longest common subsquence from backwards, then when we compute longest common subsquence for index 0 and 1, we already have the answer for longest common subsequence starting from index 2 stored.
@TheDeepsz
3 жыл бұрын
Thanks for such a great explanation, I searched for it so many places, but I didn't find anything more than the formula. This video should have more likes.
@BadriBlitz
3 жыл бұрын
Superb Explanation.Anyone having doubt in leetcode can refer this channel.Excellent video bro.I was struggling for this problem you made it clear.Thank you.
@thetrends5670
Жыл бұрын
🎵 this is two, and this is two, so it doesn't really matter, which one we do. 🎵 Music by Neetcode at 13:34
@huseyinbarin1653
2 жыл бұрын
DP always surprises me. What a good approach. Thank you
@amitupadhyay6511
2 жыл бұрын
I came for that nlogn solution. But again, thanks for the tremendous help as usual
@robyc9545
2 жыл бұрын
It is interesting that you calculated DP from right to left. I think it also works if you do from left to right.
@briankarcher8338
Жыл бұрын
Yes it works both directions.
@kevinkkirimii
Жыл бұрын
@@briankarcher8338 i thought so.
@MerlynJohnson
3 жыл бұрын
whether it should be if nums[i] < nums[j] or if nums[j] < nums[i]
@killerthoughts6150
3 жыл бұрын
it would be great to see the n log n approach
@eduardoignacioroblessosa6349
7 ай бұрын
was looking for this comment
@RajasekharReddi
7 ай бұрын
Excellent oration of the logic and the ending is at another level.
@harishsn4866
2 жыл бұрын
since we have already assigned LIS with value 1 for the length of nums, in the first for loop, we can start from len(nums) - 2 instead of len(nums) -1.
@gritcrit4385
4 ай бұрын
DFS: O(2^n) DP: O(n^2) Binary search: O(n logn)
@sameerprajapati8978
Ай бұрын
I got asked to optimize the current dp solution in less than o(n^2)
@themathguy314
5 күн бұрын
"I really doubt your interviewer is going to expect you to get this without a hint; if they do I'd just walk out of the room" LMAO!
@RishinderRana
8 ай бұрын
Agreeing with everything except the walking out part :)
@winterheat
Жыл бұрын
11:34 I think using the name LIS[ ] is not a good choice, as you may think the final solution is LIS[0] this way. The strict definition of this lookup table, let's call it lookup[ ] is this: IF YOU TAKE THAT NUMBER nums[i] into the sequence, then what is the longest you can get. So lookup[i] IS IF YOU MUST INCLUDE nums[i] into that sequence. If you write LIS[i], it sounds like it is the max NO MATTER you include nums[i] or not, which is not the case. So that's why in the code that follows, the final result is not LIS[0], but max(LIS)
@ildar5184
9 ай бұрын
No need to complicate it further by doing reverse looping, from 0 to n works just fine with the same function of max(lis[i], 1+lis[j]) for i=0 to n, j=0 to i if nums[j]
@thetechies2259
Жыл бұрын
last statement : 'walk out of the room' really made me laugh😂😂.. that's the attitude
@rrt19254
2 ай бұрын
Oof this is the first DP Problem I solved on my own! Was so happy that mine looked a lot like yours.
@paul90317
Жыл бұрын
even if it's not the best solution, it's the best tutorial for LIS I've ever seen
@anandkrishnan72
2 жыл бұрын
"I really doubt your interviewer is gonna expect you to get the O(n logn) solution without a hint. If they do, I would personally just walk out of the room." XDDDDD
@noorbasha8725
Жыл бұрын
This happen with me yesterday, he didnt given any hint, result is i failed the interview
@dominikilja
2 жыл бұрын
This was a great explanation! I struggled with this, but I'm happy to learn some new techniques!
@ece-a036nischintasharma5
Жыл бұрын
Was unable to wrap my head around this one. Your explanation was so nice!!
@kineticsquared
9 ай бұрын
Wow, what a great explanation! Thank you for the detailed step-by-step example.
@sachin_yt
3 жыл бұрын
One of the best solutions ever. thank you.
@HC-xh6mh
3 жыл бұрын
The best explanation video I have watched so far!
@alfamatter12
Жыл бұрын
That sarcasm at the end made me laugh like hell😂😂😂! I'm also walking out from this problem
@numberonep5404
2 жыл бұрын
Great explanation as usual!! A repost(?) of the O(nlong(n)) solution, not that hard and really comes in handy for other problems :) class Solution: def lengthOfLIS(self, nums: List[int]) -> int: indices = [None for _ in range(len(nums)+1)] # Mapping of size to the last indice of the subsequence size = 0 def binary(elem, l, r): # a binary search is possible since the sizes are sorted by definition, even if the values in nums are not while l
@bennypham4337
3 жыл бұрын
Really helped me out to understand this question!
@NeetCode
3 жыл бұрын
Thanks, I'm glad it was helpful!
@yahwehagape
3 жыл бұрын
Great explanation. Curious about nlogn solution now.
@NeetCode
3 жыл бұрын
Thanks!
@asdfasyakitori8514
11 ай бұрын
Man 8 lines of code is all it takes, grate solution
@danielmdubois
Жыл бұрын
Great video, much appreciated. However, I didn't understand the logical jump at @10:40 that suggested we were "starting at 3". I would have preferred to see a solution that proceeded front-to-back, because it seemed to me that is what you were doing in the recursive solution.
@hwang1607
7 ай бұрын
leetcode editorial suggests improving time complexity with binary search class Solution: def lengthOfLIS(self, nums: List[int]) -> int: sub = [] for num in nums: i = bisect_left(sub, num) # If num is greater than any element in sub if i == len(sub): sub.append(num) # Otherwise, replace the first element in sub greater than or equal to num else: sub[i] = num return len(sub)
@vietnguyenquoc4948
6 ай бұрын
IDK why but your voice in this video sounds really calming
@testbot6899
2 жыл бұрын
O(nlogn) solution class Solution: def lis(self, A): res = [A[0]] n = len(A) for num in A[1:]: if num>res[-1]: res.append(num) else: res[bisect_left(res,num)] = num return len(res)
@NhanSleeptight
2 жыл бұрын
do you mind sharing the code for the DFS solution? I just want to practice implementing these ideas.
@DavidDLee
Жыл бұрын
def lengthOfLIS2(self, nums: List[int]) -> int: L = len(nums) cache = {L: 0} def dfs(i): if i in cache: return cache[i] n = nums[i] result = 1 for j in range(i + 1, L): if n < nums[j]: result = max(dfs(j) + 1, result) cache[i] = result return result return dfs(0)
@mikemartin6748
3 жыл бұрын
Why do you not cover the best solution: dynamic programming with binary search? That's the one I'm looking for because you need it to solve later problems like the Russian Doll Envelopes.
@CostaKazistov
2 жыл бұрын
AlgoExpert covers it
@handuongdinh9290
2 жыл бұрын
The video made it very easy to understand. Thank you for making this video. Keep up the work. I’m looking forward to view yours next videos.
@jitpatel7692
20 күн бұрын
Great Explanation,Thank a lot
@annabellesun4719
2 жыл бұрын
another day watching neetcode to help me with leetcode. Thank you!
@sarvarjuraev1376
Ай бұрын
Great explanation, thank you
@prempeacefulchannel
3 жыл бұрын
It’s looks easier after your explanation 👏🏻
@NeetCode
3 жыл бұрын
Glad it was helpful!
@theysay6696
3 жыл бұрын
"I would personally just walk out the room" LOL
@ajithkumarspartan
4 ай бұрын
When i feel its so hard to learn DSA problem and crack FAANG like companies my mind tells me "neetcode" and after seeing the video explanation i become calm and motivated to proceed further.
@protyaybanerjee5051
Жыл бұрын
"Personally, I would walk out of the room" - Yeah, man!
@rubenomarpachecosantos7130
2 жыл бұрын
"I would personally just walk out the room" haha
@Sandeepkumar-uv3rp
3 жыл бұрын
Really very helpful, explained in a crystal clear manner👌👌
@Cld136
3 жыл бұрын
Thank you very easy to understand and follow. I had problem understanding the solution on leetcode :)
@dhruvkhurana9235
Жыл бұрын
"I would personally just walk out of the room" I'm dead
@JayPatel12928
2 жыл бұрын
Man this is the best video so far on this problem ✊🏻
@entropy7571
3 жыл бұрын
Make a video about the binary search solution to this problem
@diksha8347
2 жыл бұрын
This explaination is so so good. Thank you.
@amardeepbhowmick3614
3 жыл бұрын
This is GOLD!
@realoctavian
Жыл бұрын
finally a good explanation and solution for this, thanks!
@roxanamoghbel9147
3 жыл бұрын
you're dynamic programming videos are all so well explained and helpful
@irinagrechikhina9538
3 жыл бұрын
Great explanation, as usual, thank you! :)
@draugno7
7 күн бұрын
Haha, after the last sentence I looked at the problem now on Leetcode and sadly read: Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity? :D
@aquere
Жыл бұрын
Brute Force DFS's time complexity is not 2^n. We have a branching factor of n at worst, not 2. So it's gonna be n^n
@apoorvdp
Жыл бұрын
@Neetcode the O(N^2) solution now gives a TLE on Leetcode. Given the popularity of this problem, could you please make a follow-up video or respond to this comment on how to get to the O(N.logN) solution?
@De1n1ol
Жыл бұрын
for me top-down didn't give TLE, the bottom-up did. I use python
@peter0702
9 ай бұрын
It will not in c++, but I guess you can reference to this video kzitem.info/news/bejne/k2ipl6yui6iqZW0 you can see the problem is the O(nlogn) solution is not sth you can figure out but more like applying an algorithm.
@vivekjoshi9073
2 жыл бұрын
Thank you sir best explanation able to do in other programming language easily and concept is clear
@sdsunjay
6 ай бұрын
It feels overly complicated to start from the end of `nums` when we can start from the beginning. The implementation below was *easier* to understand conceptually and ran faster. ``` class Solution: def lengthOfLIS(self, nums: List[int]) -> int: LIS = [1] * len(nums) for i in range(1, len(nums)): sub_problems = [LIS[k] for k in range(i) if nums[k] < nums[i]] LIS[i] = 1 + max(sub_problems, default=0) return max(LIS) ```
@dhanrajbhosale9313
Жыл бұрын
End was epic 😄.. "I'll probably walk out of interview"🙃
@simardeepsinghmudhar7065
Жыл бұрын
very good explanation
@gargichaurasia4103
3 жыл бұрын
You are great.. you explained it very well. Thank you so much!
@thecomputerman1
2 жыл бұрын
You are solving problems like God would solve, I attempted it and couldn't solve it in my first attempt though I knew what Dynamic programming is. Also, the way you explain choices and recursion is far the best way to start attacking problems like this.
@mohit8299
Жыл бұрын
very nicely explained bro thanks a lot
@dcc5244
7 ай бұрын
博主讲的真好!
@shuvo0201
3 жыл бұрын
Awesome! Very clear and thorough explanation 🙂
@srikanthvimjam9753
3 жыл бұрын
Really nice explanation. Your video saved me lot of time.
@NeetCode
3 жыл бұрын
Thanks!
@priyanshupareta9328
3 жыл бұрын
Best explanation out there!! Thank you for your efforts.
@srushtinarnaware4919
Жыл бұрын
Thank You so much
@koraykara6270
Жыл бұрын
Could you solve the question in O(nlog(n)) time complexity please and make a video after that?
@nightmarauder2909
2 жыл бұрын
lol you kept typing LIST great explanation, thanks!
@santoshkadam8431
Жыл бұрын
Wow great explanation!
@kexinliu6105
Жыл бұрын
It does not really make sense when there's an array [9,2,5,3,7] and LIS[0] returns a 1. Doesn't LIS[0] mean the LIS starts from index 0 to len(num) and it should return 3 which is 2,3,7 or 2,5,7
@NickDrian
2 жыл бұрын
No need to iterate through backwards, forwards works just as well.
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