Check if the singly Linked List is a Palindrome

[@vatsalanand9417]

Your videos are of great help and the language you use is so simple and understandable.Thanks.

[@farazromani]

These are great series of videos! I'm glad I came across them. One suggestion -- at the end of the videos, could you also discuss the space and time complexity for your algorithms? Thanks and keep 'em coming!

[@vivekanandkhyade]

yes sure...

[@anweshvalleshetti3499]

It throws nullpointer exception in the even case, because p is already null and your trying to access p.next()

[@avinashkhetri9575]

yeah, just reverse the order of the if conditions though

[@Ashish.Shukla9]

Yeah just put if(!p.null && p.next=null) it will work

[@ezrahn]

Just check for the even condition first, and then the odd.

[@sergiojimenez3445]

I came here looking for cooper and I found gold

[@SnehasishGhoshSg]

Why were you looking for Cooper ? what would cooper Fix ?

[@AbhishekKumar-qr5ql]

kuch bhi matlub

[@siddhartharaja9413]

@@AbhishekKumar-qr5ql it's good and logical

[@saketkumar7335]

you mean copper??

[@ProgrammingTutorials1M]

We can also use stack to determine whether it's a palindrome or not.

[@TechTanka]

When p is null, p->next will throw null pointer at this line on even length check :- if(p ->next == NULL)

[@jamesqiu6715]

exactly! should add a bound check

[@seekngeek]

Check if(p == NULL) first and then check if(p->next == NULL). Just swap the checks. That should solve it.

[@debanjankuri5274]

@@seekngeek Thank you so much

[@VanTran-zf7mg]

Would you please make an example of copy constructor in the Linked List in C++? Thanks

[@pankushmahajan4598]

Mr. Vivekananda ji , try to make videos on hashing

[@ProgrammingArena]

Your program will "Crash" in the second scenario (for Even Number of Elements of Linked List) --> At the last Iteration when "P" becomes a NULL, and your try to access the data from P->Next --> Result in Crash as there is no any address of P. You need to modify the if statement as follows: Option1: if(p->next == NULL) -- Modify to --> if(p && p->next == NULL) Option2: swap the positions of both the "if" statements and the second if statement would be an "else if" rather than just "if". Thanks!

[@soumilyade1057]

perhaps, the conditions can be put in an if..else if statements.. if(p==NULL) //p is not null in the first case. so this condition is skipped// ..... else if(p->next==NULL)// valid for the first case; if the if part executes, the else if part is skipped and the execution continues unhindered!!??// I did not get your first suggestion though, could you please clarify it? what I understood is that if(p!=NULL && p->next==NULL) can be put as the first if condition.

[@ProgrammingArena]

yes, exactly you understood my first condition correctly. The first condiition if(p->next == NULL) will crash the whole program if in case the pointer "p" goes NULL. Hence, to avoid such instances, we must check if p is not NULL. just to clarify you: if (p) and if (p != NULL) both are same !

[@soumilyade1057]

yes !! I understand that now xD Thanks...

[@rohankrishna156]

It's a Null pointer exception for this test case: [1,0,1] . The when we have to assign secondStart = q.next.next, it will point to Null

[@koeber99]

Or you could add all the elements to stack and then check half the LinkedList: time is O(n).... but space is O(n)

[@vivekanandkhyade]

Yes koeber99 , that trick would work.

[@satvikkhare1844]

This was asked in my Nvidia experienced software interview. Please make more videos.

[@SK-qn5ry]

Oh nvidia, great

[@rishabhsahu5257]

Amazing !!Effort for us .Huge respect bro

[@cemyasar4858]

The point is finding middle node. The solution is creating a race condition. One pointer's velocity must be two times bigger than other pointer's velocity. When faster pointer reaches the end, other pointer will be in the middle. I prefer recursive way. I will find the middle by recursively then start comparing from middle nodes(even or odd problem also here) then by returning next node to upper recursive iteration and comparing prev node and returned next node and so on..

[@shailbenq]

The first IF condition in while loop "if(p.next == null)" should be "if(p != null && p.next == null)"

[@talhacoder5084]

dude this one condition T_T Thank you :)

[@r3fl3xplays36]

sir could you please make a video on XOR linked list , SKIP list , UNROLLED list..

[@informative_idiot-x6e]

Thanks for the wonderful video.... Please take a look into the first "if" condition, as many mentioned in the comments.... Thanks!!!! :)

[@HarshKumar-nh6be]

Your videos are very great can you please upload videos on dynamic programming problems like gold mining, tiling problem

[@palagatideepthireddy3659]

in even linked list if p is nul lthe statement p->next will give segmentation fault because pointer pointing to non existing locaton

[@uzairkhurshid3036]

seriously dude this is really good .appreciate it

[@mdanishossain026]

sir , can you make please ad_hoc algorithm playlist

[@PrashantNigam]

Can you please make a video on Single Pair All shortest paths problem? Given a source and a destination find all the shortest between them (there can be more than 1 path with the same length, hence all the shortest paths).

[@rajatbais6388]

SIR YOUR WAY OF TEACHING IS GOOD, But It will be a good practice if you explain the approach first means what actually you are going to tell for this or any problem then go for the algorithm or dry run of the code. Once we know the approach we can understand the video very easily and quickly.

[@parthsinghparihar1904]

Why can't we reverse that linked list and compare with the original value at each node , If any of the respective nodes aren't equal to both , then return 0 ?

[@shubhammalik9418]

I don't see the full working code on the given link. There are only few codes, not all. Can you please share the full code for this problem?

[@alifruslan2728]

Okay, I wanna add, as we are reversing after splitting up, can't we just reverse the linkedlist at the very beginning and check if both are same? This doesn't require splitting the list. Thanks in advance

[@dankyt9249]

We can do it but it is not feasible as it needs traversing whole list and then comparing it. In this we are just doing half comparisons and also running loop for half. So for very long linked list it will reduce to half time complexity.

[@AddieInGermany]

Sir please discuss how to recursion and when and where ? Especially with linked list, tree etc

[@devotion_surya3741]

Sir, what an explanation, amazing💥

[@chetanshrivastava3762]

Nice video.God bless you..

[@rohan5713]

but we also use the reverse of linked list approach also while(i

[@nitinjangra5164]

we cannot back traverse linked list

[@bharathik6479]

Awesome Explanation. I am new to algorithm yet understood this very clearly. Thank you so much , You are saving my grades and career. One small request , Can you please add the gib hub link in the description .

[@kalpatarupodder708]

sir, we can also do it like taking two pointers at two end and traversing in the opposite direction and matching the elements,if all elements matches then its palindrom

[@SOURABHGOLCHHABCE]

bro it is singly linked list and in the case of singly linked list there is no means of going in the backward direction as it is not the doubly linked list thanks !!!

[@shwetayadav3288]

Very clear explanation.

[@viewgood]

Thank you a good explanation, your voice is the same as the tutor in the KudVenkat channel, are you the same ?? just curious !

[@adi_sekar]

No they are different

[@ganeshjaggineni4097]

NICE SUPER EXCELLENT MOTIVATED

[@shloch2007]

very good with explanations..........always checkout this channel, dope

[@prashantshukla7203]

you are doing great job sir!!

[@GeekXpert]

why cant we just reverse the entire linked list and then check?

[@nitinjangra5164]

if you are creating a new reversed linked list that would increase the space complexity...acc to me what we can do is traverse linklst and store in a number then compare accordingly using modulo operator O(1)---space and O(n)---time

[@parambharti7095]

Very nice explanation.

[@ArpitDhamija]

you can rather do it by recursion, it can be more simple and small.....

[@ArpitDhamija]

@Monkey D. Luffy sure , I will make it , I have my competitive programming channel , my exams are going on, after that , I will consistently upload videos there

[@ShivaniSingh-mq4pv]

Amazing vedio

[@16avnisharma]

very useful videos.. keep posting sir

[@harshtulsyan9894]

Hello sir....i need a video to find theLength of longest palindrome list in a linked list using O(1) extra space.

[@mohammedsadiq1567]

Why not reverse the list first?? And then compare?

[@priyanshu4016]

public static void main(String[] args) { System.out.println("entier string to check palindrome :"); char[] input = new Scanner(System.in).next().toCharArray(); int forword=0; int backword=input.length-1; int mid=input.length/2; for(int i =0;i

[@SK-qn5ry]

I think you write for array, But question is singly linked list Though iam not sure whether yoy written for singly linked list or not

[@deepakborah8335]

Really your videos are excellent

[@sallyzhangt]

Thank you. This is very clear.

[@brijyotchawla3247]

What would be space complexity of this solution ?

[@gunslingerarthur5865]

I dont think space complexity will be an issue here as all you require is just three extra pointers p,q and start

[@ridimamittal4064]

amazing explanation , thank you so much :)

[@srafayal]

big an sir love you from odisha❤❤❤❤

[@hibritto]

Genius.

[@suriyapriyajayavel1176]

hello sir....I need program for reverse a linked list from position n to m... n=2 and m=5 LL 1->2->3->4->5->6 output becomes 1 5 4 3 2 6

[@gurpeetsingh9152]

sir can u provide a link to a source code...i am unable to find it

[@pkl520]

explain very clear! Big thanks!

[@VishalYadav-ss4qv]

Excellent 🙏🙏

[@divyagupta5195]

how will this work for a b c c b ?

[@sushmitanigam4979]

awesome explanation. Thanx a lot sir.

[@vivekanandkhyade]

Thanks Sushmita..!

[@jeezradz]

great video!!!

[@ZeeshanKhan-mo9mu]

plzz give me a function that check linked list is sorted or not? plzz give me code

[@pawanbasu6777]

didn't find the program in github!

[@hardikrana2210]

check this: github.com/vivekanand44/codes-KZbin-videos/blob/master/check%20if%20a%20string%20is%20a%20palindrome.cpp

[@vaibesb4u]

Thank you sir. Very useful video

[@vivekanandkhyade]

welcome vaibes.!

[@shaikshabbir4479]

Nice video sir. One query ? Does it work for string "aba" ?

[@bhavyanayyer3391]

sir pls provide with the code???? not available on github

[@vivekanandkhyade]

I will give the link

[@AddieInGermany]

Perfect explanation sir. Thanks :)

[@vivekanandkhyade]

Thanks Aditya..!

[@ayoubaoumrani3200]

Thank you sir

[@vivekgr3001]

Amazing!

[@HarshaVardhan-jf9sd]

there is a bug in the algo

[@ghoshdipan]

This approch is ok but is not correct for all cases

[@webTVfiction]

thank you brother

[@gurmeetchawla8362]

please put the lilnk to github

[@hardikrana2210]

github.com/vivekanand44/codes-KZbin-videos

[@ashishsinha8893]

ye bro can you make the videos on bucket sort plZ

[@kapztcs]

This is flawed , try to run the code against 1-2-1-2-1-2-1

[@aka_gym_lad]

sir tree ke question v dalo

[@imaneorj512]

Thanks ❤you save me

[@ashishsinha8893]

hello brother plz share github link

[@md.abdullahal-alamin8059]

thanks

[@saranyasaravanan7780]

Thank u sir for your explanation. I could not find your source code in github.com/vivekanda44. Can u pls give the correct link?

[@vivekanandkhyade]

github.com/vivekanand44/codes-KZbin-videos/blob/master/check%20if%20a%20string%20is%20a%20palindrome.cpp

[@saranyasaravanan7780]

Thank u so much sir..