Love your videos! Very well explained. When I tried this code though, I received a time limit exceeded error. I'm sure its on my end, but I'm not sure why because as far as I know its the same code and should work.
so if this answer matters if we write fast != null we will see we have not breaked the linked list in two there is still connection between the 7->2 so that will not be null thats why we write slow != null as slow pointer will point to null 7->null (after the reverse of the list):)
bro but for odd testcase it doesnt work .lets say we have a testcase 1->2->3->2->1->null according to you slow points to middle 3 , now if we reverse it will become 1->2->3->null ,where fast points to starting 1 .now if start comparing values @ fast and slow upto 2 it will be same after that slow points to 3 but fast points to null .so what to do?
Hello, what you can do: Traverse entire L. List. (Let Length of LList come out be n) Just push n/2 elements in the stack. Now you can check if n is odd which is your case. Curr will be pointing to -7. So Curr = Curr-> next; Now start comparing left-out L list with stack elements. If n was even. Eg 2 -3 -3 2 Then We would have done nothing. Coz We have already pushed n/2 elements. And Curr will be pointing to the Third element. Now start comparing left-out L list with stack elements. Hope it helps. 😄
bro trust me u r a god level teacher bro omg
i dont but i get a lot of clarity from your videos , Thanks a lot
Good explanation! Thank you for the great video 😃
thank you for subscribing.. :)
explanation is awesome💥💥👍
Loved the explanation for both implementations! Thanks Nikhil!
your all explanations are awesome
Thank you so much 😀
such a clean explanation, i loved it
Why not using collection linkedlist of java
Why we are making our own linkedlist ?
Why we are making our own tree?
Please answer
Love your videos! Very well explained. When I tried this code though, I received a time limit exceeded error. I'm sure its on my end, but I'm not sure why because as far as I know its the same code and should work.
try the code available on github (link in description)
Not able to understand why i am getting an error when i write
while(fast!=null)
{
if(fast.val!=slow.val)
{
return false;
}
compare to the code I have
so if this answer matters
if we write fast != null we will see we have not breaked the linked list in two there is still connection between the 7->2 so that will not be null thats why we write slow != null as slow pointer will point to null 7->null (after the reverse of the list):)
i am not getting ,what if the no.of nodes is odd ???pls explain
When we are doing fast = head and traversed fast to null then why head is not null. Aren't we dealing with the address?
Your first approach doesn't work in case of 121 or 23732 . Correct me if I am wrong ?
when we do fast= head does it start from for head of ex 237732 this value or is it half
Can you please elaborate on your example?
ok,but what's in case of linkedlist of odd length?
The middle element does not matter.
1 4 x 4 1
The middle element x could be anything
i love it
bro but for odd testcase it doesnt work .lets say we have a testcase 1->2->3->2->1->null according to you slow points to middle 3 ,
now if we reverse it will become 1->2->3->null ,where fast points to starting 1 .now if start comparing values @ fast and slow
upto 2 it will be same after that slow points to 3 but fast points to null .so what to do?
It does work, did you try with such a case?
Y in his videos very less viewers😢
Nice explanation
The stack solution does not work if the pattern is 2 - 3 - 7 - 3 - 2. even tho it is a palindrome
go with the optimized one
Hello, what you can do:
Traverse entire L. List. (Let Length of LList come out be n)
Just push n/2 elements in the stack.
Now you can check if n is odd which is your case. Curr will be pointing to -7.
So Curr = Curr-> next;
Now start comparing left-out L list with stack elements.
If n was even. Eg 2 -3 -3 2
Then We would have done nothing. Coz We have already pushed n/2 elements. And Curr will be pointing to the Third element. Now start comparing left-out L list with stack elements.
Hope it helps. 😄
@@SaumyaSharma007 awesome thanks buddy i will give it ago.. :)
please do a longest palindrome in a linked list
Will add it to my pipeline of video
Please make more explanation videos
Will do..new video every week 😄
Please share as much possible too 😎