Longest Common Substring | Recursion and Memoization

liked this video? Click here / @codebix1096
problem :- practice.geeks...
code :- github.com/luc...

[@anktrj]

I saw 10 videos on this . . . every one just filled the dp table but you helped me see the overlapping subproblem. Thanks man.

[@kbhanuprakash9530]

True

[@RahulVerma-x1k]

correct, everywhere they r explaining tabulation

[@ujjwalshrivastava-xs5lx]

Yeah

[@csalgo7345]

You're the Guru of recursion and a very valuable resource to learn from. Please keep up the great work that you're doing!

[@codebix1096]

Thanks, will do!

[@Abhishek-Khelge]

@@codebix1096 Seriously idea is so clear and easy to understand.

[@AnandKumar-kz3ls]

well done bro just sick and tired of youtubers uploading tabular solution and playlist and kids commenting in comment section you're god of dp and this video proves if you know recursion well and never ever heard of dp before you can solved almost any solution

[@imajt5]

Thanks for the explanation, but the cases are like 1. If matched, then we will take the max of three cases - fn(i-1, j-1, count + 1), fn(i-1, j, 0) and fn(i, j-1, 0) Bcz there can be case like "pbax " and "pbpbax" this string, so even though string is matching we cannot ignore the other cases 2. If not match, then we have only two case - fn(i-1, j, 0) and fn(i, j-1, 0)

[@shaulendrakumar6314]

bro can you pls give the correct code of this?

[@shaulendrakumar6314]

pls bro can you send??

[@imajt5]

@Shaulendra bhai de dunga baad me...ab dekhna padega

[@ADNANAHMED-eo5xx]

First video I saw on your channel, what an indepth tutorial

[@MDEMANURRAHAMAN-]

Best Explanations Indeed ❤

[@theuntoldtree]

Guys share this channel, let's support his hardwork , he deserves.

[@adityajoshi8578]

this is the only video that helped me ...! gr8 explanation and content

[@kanishkanand1555]

this is o(n^3) because this problem lacks suboptimal structure , but finding longest common suffix of the prefixes takes just o(n^2) and is far more easy to code and implement.

[@betabias]

can you please elaborate what suboptimal structure is, or any reference link for the same?

[@jatinkumar4410]

Thank you sir...your explanation is very clear!

[@pawanpandey2]

You are pro bro in explaining recursion!!

[@codebix1096]

Thank you. Follow our linkedin page for regular updates www.linkedin.com/company/codebix/?viewAsMember=true

[@judgebot7353]

very nice keep going. This is absolute gold content

[@SouravGhosh-pb5nm]

Thanks for such a crisp and nice explanation.

[@codebix1096]

You are welcome !

[@kushalappaca5324]

Great way of explaining..

[@nishthavan3764]

One Ques though How will string PBAN AND PPBAN Work ?? Longest common is 4 but you it is 3 because P will be removed and BAN and PBAN will be left. I think code is right but explanation is a little wrong there will always be 3 call when we are on current node to account for the problem I just mentioned but there will be 2 calls in the case when not matching but when matching there will be 3 calls.

[@TarunKumar-cn6in]

Crystal clean explanation ..

[@rituraj6056]

tnks a lot sir,u really feeled about recursion tree

[@codebix1096]

Thank you. Follow our linkedin page for regular updates www.linkedin.com/company/codebix/?viewAsMember=true

[@pratikjadon1468]

This solution getting wrong answer in GFG longest common Substring ? Also why we are checking the all three condition even when character of both strings are equal ? plz tell im alot confused

[@mayankbadika3101]

Excellent explanation . Keep it up :)

[@advik8387]

I think while explaining you forgot one thing, for prsq and pprs - even though the first chars are same, it will still go through diff1 and diff2, because there is a possibility that a longer substring could result from that, the code is correct, but the explanation and recursion map misses this. the ans is 3, but according to explanation it's 1.

[@vanshsehgal9475]

Yeah I noticed that thing!!

[@shaulendrakumar6314]

bro can you pls share the corrected code??

[@mtoheed1085]

what the powerful videos...really bomb man

[@anonymous10906]

loved it

[@sanjayjoshi8807]

c++ solution same like yours giving TLE on GFG can you help in this sir !!

[@ankitparashar8730]

Haa🙄

[@AvikNayak_]

only bottom-up approach is getting accepted.

[@varsha4260]

Sir,you should have put a else statement if the characters are equal ,then we don't have to split it into two parts and just return the same till that part.

[@HimanshuKumar-hm4wv]

great explanation sir

[@anktrj]

Also there's one request can you explain how this overlapping subproblem can be visualized in iterative DP technique

[@theuntoldtree]

sir difference vaale case ko else m ku nhi rkha?

[@AnandKumar-kz3ls]

aabc , abc this test case will fail

[@vikramshukla6436]

Key banate waqt beech mein koi bhi string daalna necessary hain kya ?

[@pragneshamadavadi2267]

worth watching content !! Understood

[@nikhilgoyal8340]

Badiya samjhaya bhai

[@pratikjadon1468]

if s[i] == s2[j] then we are checking with 1 + (i-1,j-1) but why the other two recursive condition not in else block . I mean to say if two characters are equal then we use if condtion but we are also executing the (i-1,j) and (i,j-1) even if s[i[ == s2[j] regardless of if condition.

[@vinayjangra1401]

yess, because unlike longest commong subsequence here we are not sure that if the first char matches then it would be our opmtimal answer or not, lets say abcde and afcde, here both first char matches , if we just put that statement in if else it means we are including this 'a' in our final answer but final answer will be cde, cde, which has no 'a'(first char). hope you got it

[@utkarshchauhan5827]

U r Excellent

[@Ankit.yt_885]

Well done!

[@Ajji_stories12]

super

[@anktrj]

And What if some asks me to find out the complexity of the memoization. How to do that? Should I just check the pattern by building the decision tree which can be messy

[@toomuchpoop450]

when its a 2d dp its usually n^2

[@raunakkumar6144]

its 3 d dp

[@priyanshuplays9457]

Bro it is giving tle on gfg :(

[@InderjitSingh-sh9ve]

Cpp solution bhi diya kro please

[@samsingh43]

Wrong s1 = pobox s2 = boxax

[@suyashsingh4338]

Still Gives TLE on GFG

[@anirudh6549]

TLE

[@varsha4260]

Sir your memoized solution is getting time limit exceeded on leetcode

[@soumyadiptabanerjee8256]

memoization here is a n^3 approach. Go for dp.

[@DMDRPBHU]

W aaaaaah

[@codebix1096]

Thank you. Follow our linkedin page for regular updates www.linkedin.com/company/codebix/?viewAsMember=true