LeetCode Explained - Reverse Nodes in K-Groups [HARD]

This question is pretty tricky. Thankfully, if we deconstruct it into a rephrased version of a question we already know the answer to it becomes a lot easier.

Getting ready for your next interview? Coding challenges got you frustrated? LeetCode Explained will help you understand all the popular coding interview questions! First we begin by deconstructing the problem into a statement we are able to understand easily.

This problem primarily tests your knowledge on linked lists in the programming language of your choice. Even though this video is in Java, the solution will definitely transfer to all programming languages including: Python, C++, C#, Javascript, etc.

Then after that we start by writing it out just like how we would on a whiteboard. Next, after we come up with a written solution we begin coding. Throughout coding each step gets explained so you don't get confused.

Check out the problem here:

LeetCode Explained - Reverse Nodes in K Groups [HARD]
0:00 Intro
0:26 Whiteboard Walkthrough
4:01 Coding
6:18 Time Complexity
7:13 Outro

Tags: Java, learning to code, leetcode, algorithms, coding interview, coding challenge, programming, learning java, hard programming question, hacker rank, software engineer, software engineering, day in the life of a software engineer, how to code, median, arrays, binary search

[fbarnea]

This is the best solution I've found. So intuitive, simple and straightforward. I also like that you used the simple reverse list, it makes it so much easier to follow.

[JamesBrodski]

What a great video! Thank you for making it! But I think the background music is kind of distracting.

[kittwwang]

This explanation and code are really the easiest to understand after I tried 4 or 5 different explanations. Thanks a lot!

[adityakrishna]

Thanks. This was very helpful. I hope your YouTube channel makes it big. I'll be sure to check other videos you've uploaded.

[PuneethSrinivas]

Thanks. Which is that background music ?

[dr_]

Thanks for your video. I don't think the time complexity is O(nk). The inner while loop to reverse the k elements has already moved the current node to the next k node, so the outside while loop won't iterate through all nodes one by one. Therefore, I think the time complexity is O(n/k * 2k) = O(n). Correct me if I am wrong.

[uddiptakalita]

Gr8 explanation!
You've earned my subscription.

[vaib]

The most intuitive solution. Thanks Man!

[edwardteach]

Beautiful explanation, I found this explanation better for me to understand and visualize! Thank you!

[tan]

Loved your video, extremely easy to understand. Thank you so much for doing it!

[tan]

Thank you for this video! It was very easy to follow. Loved it!

[satang]

How come root.next become 4 in return statement?
root.next was head which was 1.
I debugged: after the first reverse for (1, 2, 3, 4),
i.e. after prev.next = reverse(tail, K) for (1, 2, 3, 4), root.next becomes 4 from 1
how come it changes root.next from 1 to 4 after the first reverse?

[testaccount-umgj]

Thanks for this awesome video @Coding with Conner. You got a new subscriber.

[hishamjuneidi]

I almost never comment on YouTube, but I just just want to thank you. Keep the good work.

[priyankarrajgupta]

what was the need of dummy root node, cant we have just some tmp pointer variable?

[avikchowdhury]

very good explanation, keep up the good work.

[parikshit]

Great job at explaining the approach! The whiteboard explanation was very clear. Just reduce the background sound next time.

[xiaoyangdong]

It would be much better if there is no BGM.