Remove nth node from end of Linked List (LeetCode 19) | A very very easy solution

[@sreeharsharaveendra289]

Best explanation for the solution, neetcode doesnt come close at all.

[@nikoo28]

So glad that you feel that way 😄

[@urvashichaurasia6284]

Sir please start a playlist of hard DSA problems.

[@tundeadebanjo6579]

Wow! Brilliant explanation! You have helped me to finally demystify the fear around coding algorithm.

[@sachinsoma5757]

This channel is a gem. Thanks Nikhil for the wonderful explanation.

[@AadeshKulkarni]

What I like about your videos is how you explain the brute force approach first and then explain the optimised solution. Introducing brute force solutions to noobs like me gives us some confidence and with that confidence + your simplicity of explaining the optimised solution works like charm, baba Beautiful! I also like how the structure is consistent across every video that you upload. Keep rising Nikhil🚀

[@technicalguy.]

Glad to have teachers like you ❤

[@wilberrosales6345]

great explanation bro. Thanks for the video

[@Muhammad_Kabil]

Your way of explanation!!! Thank you so much.

[@jaydeepvasoya6243]

Non forgettable explanation ...! Excellent

[@shayannafees9936]

best explanation ever !!!!

[@sanjeets93]

very nicely explained

[@shivraj940]

Great explanation👍 . For removing nth from start, we can simply iterate (n-1)th node and map n-1 -> n+1. TC : O(n) and SC : O(1)

[@nikoo28]

Very good

[@shivraj940]

@@nikoo28 can you please explain more on the usage of dummy node ? like when we get index out of bound , etc and what’s the problem we face without using dummy node ?

[@nikoo28]

@@shivraj940i can talk about it in a video. But generally, a dummy nodes all those null pointer exceptions. And you can get rid of it anytime you want.

[@atanukundu_10]

Thank you Nikhil bhai, you helped me a lot with linkedlist problems.

[@robertnant1264]

Such amazing content!! These videos are so well done. Thank you

[@Pontiff03]

Make an Adjustment to the code in while moving the secondPtr... set the limits of the for loop to be from i=1 to i

[@uxoyplayz3373]

great explanation you are doing great keep going

[@Aspiringactor02]

Wow explaination

[@121sarthkumar4]

Even without dummy we can do this ??

[@parthmodi2028]

great Explaination

[@lookahead279]

at 9:20 we just have one node, so n should be n=0 . If n=1 then we will have exception . while n=1 , you are removing 8 not 4 in the previous example.

[@nikoo2805]

That is exactly why the dummy node helps. The first node from last is actually the first node that we have to remove

[@srinivas1694]

Thank you for explaining. Can you please start a playlist of solving leetcode 75. It would be very helpfulto us.

[@nikoo28]

once, I complete all the basic concepts...yes, I will start that playlist..otherwise it becomes hard to refer to those explanations

[@tasniatahsin8637]

thanks a lot nikhil you are a gem

[@deepaliyadav6444]

Amazing explanation❤

[@jritzeku]

Can you explain dummy node logic for following case where head is deleted? 1->2->3->4->null , n=4 ....basically removes head node: PSEUDO code: -run first loop to advance 'right; as far as n allows dummy->1->2->3->4(second)->null. //second goes to tail -on second loop , advance 'left' as far as we can dummy(first)->1->2->3->4->null //first stays at dummy .. -our target deletion node ends up being the head node1 first.next =first.next.next dummy->1->2(first.next)->3->4->null HOW does dummy.next give u node 2? Doesn't it point to original head? The only thing that could make sense is firstPtr' serves as an alias to dummy but that logic fails becuase then we should be able to return left.next as final answer. This passes test case for where head node is deleted but not where non head node is deleted.

[@JustMeItsMMN]

Thank you sooooooo much ❤❤

[@nikoo28]

You're welcome 😊

[@sreejasree7427]

thankyou sir

[@Karthik-fl4cq]

Nice explanation sir 👍. Can we store all the pointers in a arraylist and storing its length in one ireration and mapping pointer of that index(len-n) to its next. Is that approach correct sir??

[@nikoo28]

Yes we can, but then you are taking extra space. Ask your interviewer if you are allowed to do so.

[@Karthik-fl4cq]

@@nikoo28 OK sir thanks for your reply

[@ssss22144]

Wow❤

[@naveenkumarkota2090]

@nikhi if not the case of just one traveal can i reverse the linked list and delete it from the first and return the new linkedlist ?

[@nikoo28]

yes, you can do it...but ask your interviewer first

[@THOSHI-cn6hg]

❤

[@ayandutta9758]

In that dry run section in that while loop it'd be SecondPtr instead of SecondPtr.next

[@nikoo28]

what do you mean? it should be secondPtr = secondPtr.next; Can you please clarify?

[@Twintowers9.11]

@@nikoo28 they meant , while(secondPtr != NULL) not while(secondPtr.next != NULL)

[@himanshusingh8693]

// without dummy node class Solution { public: ListNode* removeNthFromEnd(ListNode* head, int n) { ListNode*start=head; ListNode*end=head; while(n--) { end=end->next; } if(end==NULL)return head->next; while(end->next!=NULL) { start=start->next; end=end->next; } start->next=start->next->next; return head; } };

[@yomamasofat413]

man this is a hack I dont know what this teaches. Actually a lot of leetcode solutions are hacks. Like the dp ones