Regular Expression Matching | Brute Force | Optimal | Recursion Concepts And Questions

[@21UEE104AngiraDas]

Few days back I was struggling with this problem, thanks a lot for coming up with the solution that too with so much depth , I really needed this.

[@rashid_khan7809]

thanks....1000times (i have tried my times but fails to understand ) but it made easily by you

[@AdityaRaj-xe1vj]

Your teaching skills are just awesome ❤❤

[@Divyansh-n3h]

Itna mazaa aaya na bhai thank you so much!

[@abhiroopmokshagnabheemineni]

Awesome explanation!!!

[@DevOpskagyaan]

How are you so good. I remember I struggled with this problem a lot. The indepth points you share in your videos is a gold mine

[@iamnoob7593]

Crazy good explanation

[@deathigniter2155]

16:40 , It should be true , just like the way a*ab is true , b*ab will also be true.

[@ugcwithaddi]

Thanks a lot

[@Mohit_Yadav_2003]

At 16:40 minutes, answer of "ab" and "b*ab" must be True, because b* can be matched with " " or "b" or "bb" and so on. * means 0 or more of the preceding element, Also 3rd test case of the leetcode of the same problem is "aab" and "c*a*b" and answer for which is true, clearly stating that "c*" means " " or "c" or "cc" and so on.

[@saarthakkhandelwal4256]

Nope, how do you plan on matching the 'a' from "ab" and "b" from "b*ab" i.e the elements at the 0th index.

[@Mohit_Yadav_2003]

@saarthakkhandelwal4256 buddy b* can be matched with "" i.e with empty string. Means we can take 0 times of the preceeding element

[@riyaverma529]

yeah , i stuck at the same point.

[@riyaverma529]

@@saarthakkhandelwal4256 look here is a test case from leetcode : Input s ="aab" and p ="c*a*b" Output=false Expected=true this was 226th test case which i failed due to the same mis-understanding.

[@Mohit_Yadav_2003]

​@saarthakkhandelwal4256 buddy "b*" can be matched with " " i.e empty string, * means 0 or more of its preceeding element means we can take 0 times of b.

[@ronakpatel2826]

super !~

[@gauravbanerjee2898]

Thank you so much ❤❤

[@AmitSaha_15]

best explanation!! thank you so much

[@abhinay.k]

thanks

[@EB-ot8uu]

kya khatarnak parhate ho bhai. legit.

[@healthybestie108]

mja aagya bhaiya

[@aws_handles]

You are the best I was waiting for this. Is the Recursion Concepts playlist complete now ?

[@jatakchatterjee1459]

Aj ka gfg potd - generalized fibonacci number 🤯(hard) ..kahi par acha explanation nehi mila.. Ho sake to iska ek detailed video bana dena pls.

[@JagannathDebGunin]

Thank you dada Can you please do maximum index of gfg please? And comparator video of java?

[@sandiproy6842]

finite automata toc topic

[@user-fm8jx]

Great observation

[@Ramneet04]

How can we check directly the p[1] as it may be out of bound mean what is p length is 1? Im getting confused in it ??

[@wakeUpDoSomething]

bhaiya aap kha se ho

[@mohddanish2553]

bhaiya could u plzz update more question videos

[@nawazthezaifre8870]

plz also share the slide link

[@codestorywithMIK]

Yes already added in the video description. PDF link

[@nawazthezaifre8870]

@@codestorywithMIK ohh I didn't notice. Thanks a lot Sir for today's problem solution also.

[@yashvalecha2860]

Bhaiya please read my comment on previous videos i want to talk to you

[@dayashankarlakhotia4943]

public boolean isMatch(String s,String p){ boolean[][]dp=new boolean[s.length()+1][p.length()+1]; dp[s.length()][p.length()]=true; for(int i=s.length()-1;i>=0;i--) for(int j=p.length()-1;j>=0;j--){ boolean match=(i