Longest Common Subsequence - Dynamic Programming - Leetcode 1143

[@NeetCode]

🚀 neetcode.io/ - I created a FREE site to make interview prep a lot easier, hope it helps! ❤

[@Qwantopides]

Absolute legend! Recently I have been realizing how extremely helpful people in the competitive coding community are. Quite shocking to be honest. Like you! How much effort you have put in to help out others is amazing!

[@thecomputerman1]

Damn, I looked at the problem from your list of 75 problems as I am preparing for an interview, the first time I looked at the problem, I went black, I had a 15 minute timer, I just didn't know what to do and how to proceed. not even close. never could comprehend that a 2D matrix could be a way to look at problems like that, You just opened up my thought process. I have more than 10+ years of experience but I never attempted competitive programming (I'm from a systems engineering background (devops/sre sort) and never required these sorts of problems to be solved, neither in real life nor in any interview). Thanks a ton, Dude for all your efforts. Really appreciate it.

[@audioplatform6299]

If the computer man cannot how can lesser mortals like some of us could😢?

[@PhanNghia-fk5rv]

Ty for ur sharing, i have an westion what bring u here to leetcode as u said that u dont need it for job

[@chaitanyaswaroop7669]

@@PhanNghia-fk5rv He said he never had any interviews or required to solve problems in the past. But as said in the first line, he is preparing for an interview now

[@alokesh985]

Great solution, as always, but even if you solve this starting at the first cell of the table, it would still be a bottom-up approach. Tabulation is called bottom-up because we are starting from a smaller problem and we are building up to the larger problem's solution. Memoization is called a top-down approach because we start from the larger problem and break into smaller problems using recursion (something like recursiveFunction(n - 1)). So, basically, tabulation DP, which is what is done here, is bottom-up. Memoization DP is top-down.

[@ashok2089]

Thanks!

[@phantomcorn]

It depends on what the 2d table represents right. I think from this video dp[i][j] represents the lcs of text1[i:] and text2[j:] so in this case starting from the first cell (dp[0][0]) would be a memoization.

[@paulreilley5439]

I didn't understand this concept until you explained it. As an educator I can confidently say fantastic job! I'm working on this concept on Codecademy and it took me a minute to realize that you set the no character columns as bottom and right, whereas Codecademy does them as left and top. Seeing it done both ways and the subtle differences in the code was extremely helpful to fully comprehend the purpose of this exercise. Thank you!

[@sajinkowser7228]

From trivial questions to niche ones, every single video is quality - great stuff man!

[@fatbubble123]

This is the most incredible explanation of 2D Dynamic Programing solutions I have ever seen. Props to NeetCode for that *Fantastic* visual representation

[@amardeepbhowmick3614]

Your channel has the best content on Dynamic Programming, both from the theory as well as the coding perspective!

[@ku-1119]

Thanks for the explanation. It has been exremely helpful as usual!! Note to others: when initialising dp in the beginning, doing dp = [[0] * (len(text2) + 1)] * (len(text1) + 1) does NOT work because this creates "len(text1) + 1" number of REFERENCES, so updating one array will update everything else. You can try and print the dp array and see for yourself. I found this explanation from the LC discuss section, and didn't realise this in my solution and I was stumped on why my answer was incorrect!

[@zactamzhermin1434]

If you want to avoid double for loop list comprehension you can only do that in the inner loop so it should look like this instead: dp = [[0] * (len(text2) + 1) for _ in range(len(text1) + 1)]

[@itachid]

Thank you. Was wondering the same.

[@jahleelmurray2826]

Oh woww thank you man I was so perplexed 🤣

[@quinnlangford6521]

thanks for this

[@srinadhp]

Wow! That is something! Had to watch couple of times to understand the very difficult concept.. but as usual, you made it so simmple with your impeccable explanation. Amazing stuff! Thanks a lot!

[@josecabrera7947]

dude, I never write comments but thank you !!! I hope you know that you have made an impact on several people's lives for the better.

[@jonaskhanwald566]

Bro please upload videos daily. I'm very dependent on you. Your videos are the best to understand.

[@ritikajaiswal3824]

can you tell me whether watching the solutions helped you in doing good in DSA?

[@varunshrivastava2706]

@@ritikajaiswal3824 Try to solve the problem for 30min if you are unable to solve it then look for a video solution. Problem Solving is a skill that will take a lot of time to master!!!!

[@balajivenkatraman2991]

@@ritikajaiswal3824 What an arrogant statement that is, we can do a math in our mind if it comes to algorithm certain formula/approach needed. Certainly his video helped me to improve a lot and able to apply the concept in other problems too.

[@mohammadammech7934]

@@ritikajaiswal3824 it helped me a lot, in both DS and algorithms

[@julesrules1]

I've watched a bunch of videos to understand this question, and this one is by far the most intuitive to me. I can't thank you enough. 👏👏👏👏👏

[@PC-pr8gi]

you make it look so SIMPLE !!! In fact after the visualization/explanation, the code just comes naturally to one's mind.

[@albertd7658]

We could use a full-size 2d grid, but all we really need it 2 rows with the length of the x-axis, in this case, text2. - Time complexity: O(len(text1) * len(text2)) - Space complexity: O(len(text2)) class Solution: def longestCommonSubsequence(self, text1: str, text2: str) -> int: prevRow = [0] * (len(text2) + 1) for r in range(len(text1) - 1, -1, -1): curRow = [0] * (len(text2) + 1) for c in range(len(text2) - 1, -1, -1): if text1[r] == text2[c]: curRow[c] = prevRow[c + 1] + 1 else: curRow[c] = max(curRow[c + 1], prevRow[c]) prevRow = curRow return prevRow[0]

[@kevincai4249]

My favorite channel on youtube!!! Thanks a lot for the great work, it helps me so much! Continue to solve more leetcode problems and help more people to understand it.

[@ukuduck531]

Couldn't appreciate more, really. My daily routine is basically, continue watching your 75 question playlist, try to solve the problem by myself, then watch your brilliant explanation no matter if I pass it or not. I just completed this 2D DP problem, which I previously feared I couldn't think it through, without watching your code. This feels so good! Simply, thank you!!

[@iamdeancoulstock]

Genius solution. Took 3-4 replays to fully grasp, but I feel great about it now. Thanks for your work!

[@kunal_chand]

The best visual explanation for this problem. I finally understand why do we take the 2D DP array. Because every cell in the array holds information to compare every possible subsequence

[@rohitkumaram]

It is still your solution only: but I find normal iteration easier than reverse, it can depend person to person: dp = [[0 for j in range(len(text2)+1)] for i in range(len(text1)+1)] for i in range(1,len(text1)+1): for j in range(1,len(text2)+1): if text1[i-1]==text2[j-1]: dp[i][j] = 1 + dp[i-1][j-1] else: dp[i][j] = max(dp[i-1][j], dp[i][j-1]) return dp[len(text1)][len(text2)]

[@NeetCode]

Yeah, I'll probably do future problems with normal iteration

[@rohitkumaram]

@@NeetCode hey, you replied me thankyou so much would like to connect more. Since I want to become a faang engineer.

[@CostaKazistov]

💯agree - forward iteration is much easier to follow

[@ritvikgaur3444]

Hi Rohit , can you explain that why both for loop are not giving string index out of range error as we are doing (len(text1)+1) which will add extra 1 space at last which cannot be iterated at it is not defined.

[@rohitkumaram]

@@ritvikgaur3444 dp is defined with length +1, and while iterating string we are iterating always i-1, and j-1, so it will always be 1 less than last limit

[@luis-azanza]

I love the way you explain complicated problems like this one. There's just some many different approaches! Thank you so much :)

[@berke4606]

I feel like there is a light at the end of the tunnel! One of the most clear explanation I have ever vitnessed. Thanks mate!

[@celbelle4182]

I am really struggling to wrap my head around coding, but your explanation was so engaging and simple. Thank you so much.

[@kaushiktd]

I have been struggling to get my mind around this until found this video. Thanks 🙏

[@ruthviks]

Hey, buddy. I have been watching your videos for a while and learning a lot. I really have to say thank you for the time and efforts you put on. You are doing a great job.

[@snake1625b]

wasnt expecting the code to be that short. amazing video btw, long your intuition in the beginning

[@il5083]

Using a top-bottom recursive approach will make the code easier to read in my opinion. The drawing walkthrough helped me to get a more concrete understanding, thanks a lot!

[@thankgodukachukwu1168]

There's an important thing to note in solving this problem using bottom up (used here) or top down tabulation approach. Understand that the shorter text has to be the length of the row while the longer text has to be the length of the cols. So in the example, text2 becomes n and text2 m so you can have a nxm mutidimentional array. With this and the excellent explanation here, you can ace this problem in any interview and also show that you understand it making your knowledge of DP better. And also makes you a better programmer.

[@samuelrobert2927]

Do we really need to hold on to the entire matrix? an optimisation can be done for the space such that we compute row by row as the current value only dependence on the previous row and the right of the current row.

[@AlejandroRuiz-pd6ci]

Dude NeetCode you are life savior, keep it up bro

[@danielsun716]

FYI, I did a space compression. The memory complexity is O(n). def longestCommonSubsequence(self, text1: str, text2: str) -> int: dp = [0] * (len(text2) + 1) for i in range(len(text1) - 1, -1, -1): nextDP = [0] * (len(text2) + 1) for j in range(len(text2) - 1, -1, -1): if text1[i] == text2[j]: nextDP[j] = 1 + dp[j + 1] else: nextDP[j] = max(dp[j], nextDP[j + 1]) dp = nextDP return dp[0]

[@castorseasworth8423]

Genius

[@xhenryx14]

Thank you so much, I have had this problem and I never understood the solution until now! They never taught this algorithm in my DS and Algo class

[@luffySenpai-eb3fg]

This is the best explanation of this problem that can ever be made❣

[@anandkulkarni2111]

good one!! one of the best explanations I have heard for why the solution of this problem works!! i.e diagonal (when matching chars ) vs max of side way movements during mismatch

[@sashikshik]

RU: Меня просто угнетает тот факт, что я, не зная алгоритма более двух часов пытался решить эту задачу и не смог EN: I’m simply depressed by the fact that, without knowing the algorithm, I tried to solve this problem for more than two hours and could not

[@issamjm3343]

You shouldn't take that much time with a problem. If you didn't solve it within 45 minutes, it means that there is a trick that you don't know yet. If you want to progress fast, you should spend reasonable time with problems.

[@bprincepandey]

The best explanation of LCS I have seen. Thanks man.

[@itsjustramblings]

Very well explained in a simple and direct way 👍. I initially saw this as inverse of levenshtein algorithm...without the initial values populated in each dimension since we are looking for the max common subsequence instead of no of changes required to turn one string to another. If you change the max comparison to min comparison and start from the beginning you can find the minimum changes required to turn one string to another. This algorithm will work if you iterate from beginning as well. It took me 3-4 hrs to understand levenshtein first time i came across.

[@demaxl732]

you deserve everything this world has to offer

[@avanishgvyas1992]

This is a nice problem, as unlike some interview questions, this one is a real-world problem! Finding the longest common subsequence between two strings is useful for checking the difference between two files (diffing). Git needs to do this when merging branches. It's also used in genetic analysis (combined with other algorithms) as a measure of similarity between two genetic codes.

[@capooti]

great video, thanks as usual. In this case I find it easier to iterate the rows and columns in regular (and not reversed) order.

[@abhilakshmaheshwari9360]

You are a saviour man!!! If you have a solution for a question, I always visit yours first (not to mention I never need to visit any other video/blog after that)

[@tzifudzi]

Thanks!

[@harshithdesai9989]

As usual, its a wonderful explanation of the bottom up approach. Thank you so much!!!

[@dipesh1401]

this is the best video on this problem on yt

[@tnmyk_]

Your explanation made the solution so easy to understand! Thanks a lot for making this video, it was veryyy helpful! Keep up the good work!

[@yogeshkhatri4555]

Man you are a good teacher ! Just found out about your channel and subscribed

[@stringjourney]

Once the logic explanation was done...code looks easy peasy Great work.

[@Dhruvbala]

I find this problem easier to understand in terms of a decision tree / recursive DP, then bottom up. And this can be optimized fairly easily to O(n) memory complexity.

[@hwang1607]

i think this is easy to do after learning the solution but there was no way I would be able to figure out the solution myself

[@divyanshunautiyal4659]

incredibly intuitive explanation, thanks a lot

[@mehershrishtinigam5449]

This was such a simple explanation! Thank you!

[@prachibari731]

You are AWESOME!! I wish I had known about your channel earlier.

[@raminessalat9803]

Hi, Great video. There is one improvement in terms of space complexity. You really don't need to cache all that table. You just need the three values of dp[i+1][j] , dp[i][j+1] and dp[i+1][j+1] you can update these values as you go up and that's it. you dont need more memory. I thought it might be helpful to add it.

[@igx_s2745]

You mean to declare 3 variables, lets say, diagonal = 0, right = 0, down = 0 ?

[@medsabkhi]

thank you for your content Neetcode, it's really helpful, i'm following your 75 leetcode question on a daily basis. I appreciate your work, sending love from North Africa

[@amitvyas7905]

The top down also works. That (top down) was the first thing that came to my mind when I first saw the question and when I saw this video, I got confused but when I saw the leetcode solutions then I found out I was right.

[@AdityaSUnboxings]

thank you so much bro u literally answered so much of my confusion.

[@VSUKUMARBCE]

great explanation Thank You NeetCode I am follow you through out my dsa journey

[@zaidziad1456]

Clear and simple, thanks a lot.

[@city8390]

Great explanation to understand! Thanks.

[@thevagabond85yt]

Since the problem is symmetric (solve start to end or end to start) you can also code a program which starts from 0th index... see this java code : class Solution { public int longestCommonSubsequence(String text1, String text2) { int m= text1.length(), n= text2.length(); int [][] match= new int[m+1][n+1]; for(int i=0; i

[@matthewsponagle5143]

You are great at explaining these problems! thank you.

[@shaggypeach]

This was great really made me understood the concept. Thank you! Is there one for three strings?

[@AsymptoteEducation]

You have quality content, man. Cheers :)

[@Cld136]

Another great tutorial. Much thanks!

[@vantran321]

Holy crap your stuff really helped me out man

[@mdmasudrana7406]

You are the best as always. Keep it up, dear.

[@ventacode]

Why not forward to bottom we can get the last value of the matrix right ? Is there any specific reason for bottom up approach ?

[@SolutionHunterz]

wondering the same since top down approach comes to my mind by default. Is there any specific reason that we are doing bottom up ?

[@ventacode]

@@SolutionHunterz didnt find any but some problems are solved only on that way

[@FluffyBRudy]

may be because we dont need to do array[array.length-1][array[0].length-1] to get the last index

[@ventacode]

@@FluffyBRudy um but as we are adding a row and a column initially it would be array[n][n] where n is the length of the string

[@ventacode]

@@FluffyBRudy understood your point, it would be different if the two strings compared with different length, thanks man !!!

[@meowmaple]

Will the bottom up approach work if you start from the start of the substrings instead of at the end? Great video btw!

[@bereketyisehak5584]

yeap it does, but the length is save at the last element in the array.

[@amankassahunwassie587]

It can be solved in memory complexity of O(n). check my code. row = [0]*(len(text2)+1) for i in range(len(text1)-1, -1,-1): newrow = [0]*(len(text2)+1) for j in range(len(text2)-1, -1, -1): if text1[i]==text2[j]: newrow[j] = 1 + row[j+1] else: newrow[j] = max(row[j], newrow[j+1]) row = newrow return row[0]

[@sickboydroid]

for me the top down approach (or recursive or memoized approach) is more intiuitive. I feel like code is take to me in that way

[@rishabhsharma6193]

superb explanation as always... but can we just do it with 1D dp for last row then update it as we move forward, just like we did in unique paths?

[@yinglll7411]

Thank you so much! learned a lot from your channel

[@mzn102]

Sorry if by any chance this is a dupe comment. Exploiting the symmetry of the problem, it seems, one can pad on left and top and iterate from left to right going from the first row to the last.

[@devonchia5976]

can just maintain 2 rows (the last and 2nd last rows) to optimise for space complexity

[@shankaranarayana6568]

Its possible to optimise this further by calculating the values in the cells we need only right?

[@WdnUlik2no]

That's exactly what I was thinking as well; it seems like it would be possible to only have two rows

[@TheStefanV]

Yes

[@adityakulkarni2764]

Your solution is great but how did u find out that this problem is actually 2D DP problem?

[@Oleg86F]

How it's possible to come up with such elegant solution directly at the interview without being genius and seeing the problem for the first time? )))))))

[@tarunnarayanan1513]

This was so well explained! thank you so much.

[@adityaverma3247]

@NeetCode : Thanks for explanation BUT why and how we directly jump to Tabulation approach ? What's the brute force way to think and solve before jumping to Tabulation aka DP. That's what interviewer will ask, else it will look like I have memorized without showing brute-force approach. let me know your thoughts !!!!! Also what tool are you using for your visuals ? TIA !

[@thatguy14713]

Is it possible to implement a recursive w/ caching solution to this problem, similar to the other DP problems you have solved? If so, do you think you could describe the approach?

[@CharlesGreenberg000]

How would you then use the DP table to obtain the actual best sequence? I suppose you start at the top left. If you find same symbol, move diagonally down. Otherwise take the max of bottom or right. Does that always work?

[@better_dead_than_red]

Can someone help? What if it's "acb" and "abcde"? both 2 alternatives are valid, which one to choose? 7:00

[@kbillore28]

Superb explanation

[@nadeenmoreb1681]

Thank you I finally understand it ❤

[@smtp_yurzx]

Everybody - NeetCode: Subsequence NeetCode: Subsekence

[@NeetCode]

lol thats true

[@0xSuhail]

Great explanation, THANK YOU.

[@prabinlamsal74]

I am just starting leetcode and am at about 15 problems only. Please share how many problems you guys have done and some strategies that might be helpful.

[@vishwaskotegar5248]

blind 75 is the way .. keep at it ...

[@tune6000]

Why does my code not work when I initialize my DP matrix like this: mat = [ [0] * (N2 + 1)] * (N1 + 1) , but it does work when I initialize like this: mat = [[ 0 for j in range(N2 + 1)] for i in range(N1 + 1)] ? N1 is len(text1) and N2 is len(text2)

[@idobooks909]

Better than a Uni course in Algorithms.

[@subratarudra2745]

Nice Explanation🔥

[@tomaparaschiv930]

what are the space and time complexities for this way of solving the problem? thank you

[@binpan3602]

I felt like top down is more intuitive to me in this case?

[@johnpaul4301]

The top down is almost always more intuitive. Especially for 2D dp problems

[@HalfJoked]

Awesome video. We can actually make this O(min(m,n)) space because notice we only have to keep track of 2 rows at any given time. I found the most logical way to write this is to ensure text2 is the smaller variable and base our arrays around that. The reason for that is, we need the bigger string to be the outer loop and the smaller string to be our inner loop in order to make the tabulation work correctly, since we're swapping our rows after each inner loop, and our rows are based around the len(text2). The video solution is perfectly fine, but if you can start with that and get to an ever better space optimization in an interview, it'll look impressive! Passes all test cases. class Solution: def longestCommonSubsequence(self, text1: str, text2: str) -> int: if len(text1) < len(text2): text1, text2 = text2, text1 # text2 is the smaller one, so we base our rows around that cur_row = [0] * (len(text2) + 1) prev_row = [0] * (len(text2) + 1) for i in range(len(text1) - 1, -1, -1): for j in range(len(text2) - 1, -1, -1): if text1[i] == text2[j]: cur_row[j] = 1 + prev_row[j + 1] else: cur_row[j] = max(cur_row[j + 1], prev_row[j]) prev_row, cur_row = cur_row, prev_row return prev_row[0]

[@dohunkim2922]

First thing that came to my mind when I saw this question was using two pointer. I tried so but failed. Is it possible to solve this question using two pointer? Also, do I have to memorize that questions like this requires the use of dp? Did you also memorize it or did it come to you intuitively? Thanks

[@referbaru9368]

i have two question 1. is it posible if size of J is more than size of I ? 2. what the LCS of this quest, if I = [a e d a b] and J = [d e b c a g] ? [e d a b] or [d e b a] ?

[@promethesured]

I dont understand why most solutions have the intuition to start at the end of both strings and work back up to [0, 0]. It works either way, and to me it just made sense to start at the top left and end at the bottom right -- why does everyone intuit the other way?

[@DavidDLee]

It's much easier to explain if you start with the full dfs solution. Otherwise, it's hard to give intuition for the the 2D array.

[@nikhilaradhya4088]

You can bring it down to O(COLS) space complexity

[@anupamarao3783]

would it work if dp is of dimensions [2][j+1] ? because we only need i+1 th row to calculate dp[i][j]! (j is length of second string).

[@TheIcanthinkofaname]

awesome explanation!

[@marchaeldialde]

What application or software you used here? Thank you