Rotate List - Linked List - Leetcode 61 - Python

[@CST1992]

6:49 "k modded -" Rude interruption there

[@floatingfortress721]

Clear explanation with clean code :) Step 1: Compute length of list, storing a reference to tail node Step 2: Find pivot node, which becomes new tail node Step 3: Rotate list by reconnecting both parts

[@JameS00989]

Best channel on Internet for DS and Algo 🎉 Thanks NeetCode ❤

[@fazliddinfayziev-qg1vg]

I always watch your videos, when I cannot solve algorithms by myself. You are good at explaining. Thank you bro.

[@MrACrazyHobo]

I love you man, thanks for getting me a job!

[@licokr]

When I solve coding problems and use itereation, somehow, I think of iteration of all elements I think it's because many problems need to iterate. I think this is really important to break what I have known and think about the solution fit the best to the problem I currently encounter. My first approach was making a list then iterate them. Time: O(n) Space: O(n). After your explanation, I only replaced with two points. and I thought, okay it looks nice. Then I saw your solution. It iterates one more though, you don't use any space I mean O(1). Hoo, great. I really learned a lot from your videos. Your eyes to see problems is absolutely stand out

[@louiswoolley987]

great explanation！ could you make a video on leetcode 105 Construct Binary Tree from Preorder and Inorder Traversal ?

[@ronpb3943]

I wrote a recursive solution, it went good until the cases where the k number was huge appeared

[@meghanapg7194]

Best teacher ever! Thank you :)

[@salomeoumarouwate2024]

NeetCode is GREAT!!!

[@digestable_bits]

very clear explanation, thanks as usual :)

[@MrLeyt1125]

Hey why dont you use 2-pointers method from Leetcode 19?

[@ComputerScienceSimplified]

Great video, keep up the incredible work! :)

[@user-jn5se1kb2q]

hello guys sorry i'm a beginner, when we assign the cur = head is it create a copy of the linked list or referencing the same linked list?

[@garcj1499]

it means cur is referencing the head of the same list!

[@theysay6696]

Great drawing! Thanks!

[@rohitkumarsinha876]

great explanation bro and thanks alot

[@juheehong7445]

would a fast/slow pointer to get to our desired point work here? how exactly would we implement that, though?

[@prateeksharma2354]

ya even i thought the same

[@anchitbhushan6172]

I think that would work too, but in worst case for reaching at k-1 th node will have O(n) complexity, but in case of for loop implemented here the worst case is O(k)

[@Prince-gt5ju]

even i thought same but if k is larger than len of linked list then it wont work

[@derrickagyemangduah2353]

@@Prince-gt5ju just find the k mod len

[@Ishaana3290]

Why we are doing k=k% length

[@AshokKota-p3w]

because it reduces the time even for k=2000000000 .we will get the answer by doing only two rotations.

[@AshokKota-p3w]

Thank you Brother

[@Ishaana3290]

And what exactly happening in for loop New head = cur.next ie 4 Cur.next = none means 3 -> none What is the value of tail.next ??

[@archanayogi7856]

i didnt understand the tail.next last line part after nhead become 45 how tail = head(12345) it should be added to the nhead , i am beginner pls help me

[@dpsingh_287]

we already broke the connection between 123 and 45 when we set cur.next=None after exiting the loop therefore, head would now be (1->2->3). Now we essentially put this head at the end of the tail (which is 4->5). This gives us the desired rotated list which is 4->5->1->2->3

[@sabyasachisahoo8975]

I love your way of explaining .

[@triscuit5103]

I knoooow, right? Nobody else can do it like he does, I absolutely love the way he explains things

[@maamounhajnajeeb209]

I am the one who like this video for the 500 time

[@shaantalk]

you should like it off number if times 😂..otherwise the like would go away

[@rudranshsharma2074]

k = k % len(linked_list).....right?

[@codeduel]

yes

[@edwardteach2]

U a God

[@shrimpo6416]

clean & neet

[@halahmilksheikh]

I don't understand, isn't the k = k%len part just an optimization? Without it, it should just be slow but should still work. But if you comment it out, there are test cases that fail. Why is that? Edit: Oh I get it now. The formula for finding the break is length - k - 1. It's possible to have a huge K which sets it to negative, so you need the % to keep k smaller than length.

[@qazaqempire2446]

if k > total len of linked list, then we will do more rotations than usual but the result will be same. so we only need to do actual remainder of k shifts, which is k % len

[@vijayakumareyunni6010]

Good explanation. But, some time can be saved by limiting the explanation of the problem.

[@ngneerin]

In short: 2 pointer

[@ximengzhao2168]

Great video, line 26: cur.next = tail.next works as well :)

[@codeduel]

It does but for simplicity, None is preferred.

[@davidmbugua6994]

you can also the method of converting in into a list, then using the rotate list method!

[@saran703]

can you please make a tutorial about python data struture !! That would me more helpful

[@infiniteloop5449]

That is literally his whole channel

[@ary_21]

why cant we use reversals algorithm on this question?

[@colonelondghoskots7759]

I THINK THE code is not working

[@jimwang8]

two passes

[@davidlee588]

the skip length is 5 - 2 = 3; while we need i < 3 - 1, which means i starts from 0, and 1, it only moves one step, not two steps, that's why i DON'T understand the length - k - 1part; Could anyone explain more?

[@qazaqempire2446]

it's k = 2 but index starts from 0 so hence minus