Longest Common Subsequence - Dynamic Programming - Leetcode 1143 

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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.

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.

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!

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!

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

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

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!

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

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

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]

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

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. 👏👏👏👏👏

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.

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)]

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.

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!!

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]

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

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

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!

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

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.

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.

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

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)

Dude NeetCode you are life savior, keep it up bro

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

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

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

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

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.

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

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

you deserve everything this world has to offer

Great explanation to understand! Thanks.

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

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.

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

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?

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!

This was such a simple explanation! Thank you!

Clear and simple, thanks a lot.

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

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.

Another great tutorial. Much thanks!

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?

this is the best video on this problem on yt

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

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)

You are great at explaining these problems! thank you.

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

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]

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

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

How did you come up with the idea of using the 2D grid?

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

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

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

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

Thank you so much! learned a lot from your channel

Holy crap your stuff really helped me out man

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?

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

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

You have quality content, man. Cheers :)

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

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] ?

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

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]

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

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

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.

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.

What application or software you used here? Thank you

Its quite unclear to me, how am i even supposed to understand this is a 2D DP problem during the interview. It would be really helpful if you would go into the solution by following some kind of decision tree you would normally do for other DP problem and explain how that come to be a 2D DP table problem.

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).

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

This was so well explained! thank you so much.

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

if anyone has a lot of trouble understanding still this other video actually explains it easier in my opinion ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-NnD96abizww.html

Lord save us if this is one of the problems we get in an interview and none of the hints make sense. At least when I'm stuck on comparison of two string problem, I always try exploring 2D comparisons

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?

Hi is there is video where you solve this problem but we have to give the actual longest common subsequence?

thanks for great tuts. i have a question. i can understand how algorithm work and follow up with it but after a while i figure out that i could not solve it before watching the the solution. i need to be able to solve it from ground up. with no primer knowledge. and that gives the answer that the first person has solved this problem used his long background on math and DP to solve it. so in general learning these algorithms means memorizing them as without knowing the base science its just a trick.

Awsome explanation!

Great explanation, THANK YOU.

when use dp = [[0]*(len(text2)+1)]*(len(text1)+1) for creating 2d array then in some case it fail , but i am not getting why it fail in some cases.

Hi, how do I know when the problem is needed for a 2D grid dynamic programming? Thanks.

Every problem in 2d dynamic programming set has a different approach. I am finding it hard to solve problem by myself without any hint or looking into the solution. Any suggestion how do i improve it for this?

Superb explanation

what is your drawing software that u use for teach us ?

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? )))))))

Thank you I finally understand it ❤

Here is my bottom up solution: ```python class Solution: def longestCommonSubsequence(self, text1: str, text2: str) -> int: memo = defaultdict(int) for i in range(len(text1)): for j in range(len(text2)): if text1[i] == text2[j]: memo[(i, j)] = memo[(i - 1, j - 1)] + 1 else: memo[(i, j)] = max(memo[(i - 1, j)], memo[(i, j -1)]) return memo[i, j] ```

Better than a Uni course in Algorithms.

Can anyone explain to me why when the two alphabet does not match, we take the max of to the right and 1 cell below?

damn man, it's soo good.

How is this a medium and not a hard 😢

Nicely explained.