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Amazing explanation, very easy to understand! Great job!

Very clear! I just wish you uploaded more videos. They are a great reference point for me since i mainly code in JAVA

Great job this helped me a lot

I spent hours reading the explanation from Leetcode without understanding until I watched your video. You are my Leetcode savior. Nice explanation.

Hey it shows time limit exceed

hands down, clear code and great explanation

So good!

Nice explanation jassica, Thanks!

very good solution and explenation!

I don't understand line 20 here. if(!dq.isEmpty() && dq.peekFirst() == i - k).... peekFirst() returns the actual deque value, not its index, so why are we comparing it to the index minus the window size (i - k)? I see that it works but I am having trouble following the logic here.

The best I've viewed! Explanation is precise but on point and walking through an example helps so much

i definitely love this video, no bull shit, straight to the example, nice comment, case analyst. And your pronuciation is amazing. OMG

nice

Wow, Jess, this is amazing! Please make more videos like this!

Thank you!

thank you!

Thank you for this great explanation!

Very well explained. Thanks a ton!

vary nice explanation

The best explanation I found so far for this question. Thank you!

讲的很好，但是由于显示器太大，所以字好小

How the hell are you not famous.!?

Really nice solution! Many thanks for posting, & keep posting please :)

Hi again! Can you please do a video where you solve the Basic Calculator III, that comes with all operands and parentheses?

You could have made the size bigger, it's difficult to see your code

Thank you for your explanation! I just stumbled upon your channel and really like it! One thing about your algorithm regarding push n pop during open/close paran is that we should push number before sign, and pop sign before number.

Love your video!! You explained the solution so clear in such a short time!!!

Clearly Understandable!! Thanks!!

Awesome explanation! Please make more videos like this.

Hi Jess, thanks for your videos. Are you planning to make more?

Thanks, great explanation.

If the window is larger than the array, wouldn't we just return the array with the largest element? I think instead the edge case should be if K < 1. Then there's no window, thus returning empty array.

The best explanation of all solutions I've viewed!

Thanks for the explanation! 👍

Wow, very nice! Please do more.

So all this time I was pronouncing "deQewww" in a wrong way. 😅😅

Good one! Thanks!

Thank you so much! Can you please zoom in the web page when taking other video? Because some people use a laptop with only 13-inch screen. (such as me...)

Awesome explanation! :)

if( !dq.isEmpty() && dq.peekFirst() == i - k) { dq.pollFirst(); } what does it really do? Can you please explain. Thanks :)

Have you tried your solution on this test case: [1,3,1,2,0,5] 3

I am a noob in Java. Could you explain what is num=num*10 +s.charAt(i) - '0' do? Especially the +s.charAt(i) - '0' part

Thank you so much, I was having a lot of trouble trying to understand this problem! You should solve more problems

Thank you! It's the only video that's helped me actually understand sliding window with deque

It was very clear, checked other videos with same logic, but yours is pretty neat.

Very clear explanation

great solution + explanation. Thanks for sharing!!

thank you, this helps a lot

Thanks, Very clear and concise. All other solutions of this problem were too hard to follow .

if we were to find the minimum out of the result array then what would be the optimal solution. I thought of using a minHeap instead of an array to store the integers as the window slides and then take to top from minHeap at the end. is there an optimal way to solve this?