How to find nth node from the end of a Singly Linked List in Java?

[@itsdineshvaryani]

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[@shubhamagarwal1434]

Best series for DS & Algo. I am also 10 yrs java exp guy. Was looking for DS & Algo free course over KZbin with java implementation and found this. Hats Off To You Man...Excellent Work. GOD BLESS YOU :)

[@itsdineshvaryani]

Thanks !!!

[@AshwaniKumar-fp3nk]

Hands down, that's the best explanation ever.

[@itsdineshvaryani]

Thanks !!!

[@kritisingh3194]

My approach( O(n) ): int getNthFromLast(Node head, int n) { Node current = head; int len = 0; while (current!=null) { current=current.next; len++; } if(head == null || n>len) { return -1; } int count = 0; current = head; while (count < (len - n)) { current = current.next; count++; } return current.data; }

[@itsdineshvaryani]

cool !!!

[@codingc3684]

bhut vdia yr 🙃

[@keerthigeetha4182]

your teaching style is excellent make more videos on different subjects the best you tube channel I have ever seen 🎉🎉

[@itsdineshvaryani]

Thanks

[@arpitamandal6469]

simplest explanation .. thanks for making it so simple

[@itsdineshvaryani]

Welcome !!!

[@danielsenik2432]

You can also use your code from previous video(find nth node), just pass as parameter list.size - n

[@itsdineshvaryani]

didnt get you !!!

[@sathishkumarr7106]

@@itsdineshvaryani First find the length of the List. Just traverse from head to till (length - n).

[@arkojyotibhattacharya9302]

Absolutely Phenomenal. Thank you, sir🙏

[@itsdineshvaryani]

Welcome !!

[@meetshah4432]

Hello sir, love your videos, they teach me a lot. But I have a question, why don't you ever use for loop? In most of the algorithms that I saw in this series, for loop makes the code way shorter. So can you please tell me why do you avoid using for loop?

[@itsdineshvaryani]

I have used it ... there is no such preference ...

[@namankabadi2792]

Sir your videos are very nice.. Sir it will be best if u could add time complexity for the algorithm you are explaining..

[@itsdineshvaryani]

Will try

[@mitigateddrip6129]

If n is equal to the number of nodes in linked list then it will give the right answer but it will be throwing exception at same time

[@itsdineshvaryani]

You can correct code by adding that edge case ... Ok ???

[@mpbe]

sir you are best...

[@itsdineshvaryani]

thanks !!!

[@poojithagampa4315]

is this code work for even length of list sir

[@surajmahato2557]

Wow what an explaination👏

[@itsdineshvaryani]

Thanks !!!

[@garimakhare15]

Thank yo so much for uploading this video

[@itsdineshvaryani]

Welcome !!! There is complete playlist in the description of video !!!

[@juliemoore728]

Excellent!

[@itsdineshvaryani]

Julie Moore Thanks !!!

[@shashikantmaurya1038]

sir there is issue with this algorithm suppose last three node value is need to be addressed but there is only two element in linkedlist so the count method to take the reference pointer ahead will give nullpointer Exception.

[@nupursurbhi3987]

yes that corner case needs to be addressed.

[@rajarishabhsingh8296]

Tussi great ho sir ji!!

[@itsdineshvaryani]

Thanks !!!

[@leoking2386]

for n = 0 code gives exception. while (count < position) { count++; refPointer = refPointer.next; } while (refPointer.next != null) { refPointer = refPointer.next; mainPointer = mainPointer.next; } if (mainPointer.next == null) { return mainPointer; } return mainPointer.next; }

[@itsdineshvaryani]

ok

[@ananthukkumar987]

Wow, Intelligent algorithm

[@marals1061]

Thank you

[@vedantmahajan558]

awesome content

[@itsdineshvaryani]

Thanks. I request you to share it with ur friends and colleagues.

[@Ankit-we8ym]

why this work ,sir please explain anyone?

[@deliveringIdeas]

O(n^2) solution..? Not good, right?

[@itsdineshvaryani]

O(n)

[@itsdineshvaryani]

O(n)

[@deliveringIdeas]

@@itsdineshvaryani two while-loops would be O(n^2) right? (Two traversals)

[@itsdineshvaryani]

@@deliveringIdeas wrong ... Its O(n)

[@nagabhushanr5599]

@@deliveringIdeas loops are not within one another so its not O(n^2)

[@arijitmohapatra147]

Sir can't we get the source code ?

[@itsdineshvaryani]

Its in description of video !!! Pls check !!!

[@arijitmohapatra147]

@@itsdineshvaryani ty sir

[@davidfield2030]

The video should be renamed: How to find the nth node from the START of a Singly Linked List

[@itsdineshvaryani]

nope !!!

[@vikasbalani6378]

Nope. He has created two pointers actually. One (refPtr) started early and from head node to count 'n' ,then another pointer (mainPtr) will start along with refPtr. So once refPtr ends its journey then mainPtr would be nth position before the end of linkedlist.

[@itsdineshvaryani]

@@vikasbalani6378 thanks !!!

[@wakeelmalik8387]

if newNode.data=0 , then your code throwing NullPointerException