this video is amazing

Can you suggest a similar algorithm that is more GPU friendly? I don't think I can span a fixed size grid across my scene, because the scene is too big.

That's IN SHA ALLAH really made me happy, to see a Muslim man making these great videos with easy and visual explanations! 🔥 Kudos to you 💯 keep doing the hard work, you will reach 1M some day IN SHA ALLAH!

thanks man, you explained the thing really well. I had fun taking the class

The linear time solution isn't truly same as the first two because the last element in the array is always moved away. However, it should have a small chance to stay in its original position in the final result. I think we should add an additional swap to fix this.

very clear after watching this.

🤍

This video is one of the best I've seen on the topic. Simply brilliant! Keep up the good work!

Perfect explanation

How about case y1 = y2

@7:03 isn't it like this? slow traverses c3l+z and fast traverses x?

Beautiful visualisation! and explanation of time complexities and algorithm, love the video so much!!❤

at 4:21, maybe something is wrong with the formula? There are two ">"?

cast a ray to "any" direction? I can easily make a point outside to cast a ray crossing the polygon 3 times!

Thanks a lot.

i want to ask how did you do the graphic effects and all stuff in the video which tools did you use? thanks!

Shouldn't you start with right cell search to optimize memory access though ?

😅

Thanks

Awesome. I watch searching for an explanation on ray casting and yours was perfect. Now, I have a different problem to solve. I want to generate the shape of a city starting from a rectangle and want to morph semi-randomly in a way that it contains some Point of Interest (POI) while it excludes Points of Avoidance (POA). Any Idea as to how to do this smartly; without checking each time if every single deformation left some POI out or let some POA in?

I have come up with a solution but it seems like brute force, and if so, someone help me optimize it: static int maxVowelsNum(String s, int k) { var list = List.of('a', 'e', 'i', 'o', 'u'); var p1 = 0; var p2 = k + 1; var max = 0; var count = 0; while (p2 < s.length()) { var chars = s.substring(p1, p2); for (var ch : chars.toCharArray()) { if (list.contains(ch)) { count++; } } max = Math.max(max, count); p1++; p2++; count = 0; } return max; } Thank you.

Hi, why cant we take d/2 from both side of mid line?

is there a way to modify this to get all pairs starting from the one of the 2 closest points and ending with the pair of the two furthest apart, i.e. get all pairs sorted by distance

Do in C++ too please !!

Isn't this just post-order DFS?

these animations are bomb! Great job!

Do in C++ too

I am confused, in Knapsack code, we did not use knapsack(values, weights, k), (we are not skipping the current value and moving on to i+1 right), what am I missing here? Could you tell me the intuition?

Thanks for the video! In 4:56 you say, that there is no alternating path ending in B4, but there is one between a4 and b4, isn't that so?

last example is useless other than in this toy example, almost never values translate to indices in a array(in bound).

Sorry, but using x as a constant just doesn't feel okay 13:20

Thank you. I've been struggling to wrap my mind around ChatGPT's explanation of sliding window technique for a few days now, this is so much simpler. def count_vowels_efficientcode(word:str, k:int) -> int: VOWELS = set('aeiou') max_vowels:int = 0 n:int = len(word) if n <= k: for c in word: max_vowels += c in VOWELS return max_vowels for c in word[:k]: max_vowels += c in VOWELS cur_vowels:int = max_vowels for i in range(k,n): cur_vowels -= word[i-k] in VOWELS cur_vowels += word[i] in VOWELS max_vowels = max(max_vowels, cur_vowels) return max_vowels

absolutely smashed ... no words about your explanation

WOW. I always expected making a Sudoku solver to be insanely complicated, but clearly not. I actually suck at solving Sudokus, but I'm planning on making a Sudoku solver for a school project, and this is perfect. I didn't expect to understand anything at all when I clicked on this video. I'm really surprised how easy this seems. I hope it is how it seems 🤞

I love these kind of explanations

Hello my friend, allah bless you, I could not find the course cause the link is broken

This was amazing! I actually understood the sudoku backtracking algorithm finally.

You can alsow think in terms of relative velocity From the point of view of slow pointer, fast pointer is moving 1 node ahead at a time So if finally fast pointer reaches the slow one, then definitely there is a cycle I think, using the similar approach if you move slow pointer n times forward and the fast one (n+1) times forward, that should also work

Thanks!

I think, In the end code slow and fast should point to arr[0] instead of just 0

Superb explanation , Thanks.

I wasted a lot of time understanding it, thanks to this amazing video I finally understood :).

Help full video thank you❤

Best video

Really liked the tree visualtion, made it so much easier to understand.

Thanks a lot. We need more non-indian math/coding youtubers.

Hey. Could you please explain how to calculate the time complexity?

Giving the mathematical explanation for getting the entry point of cycle was the best part !

Crazy explanation 🔥😎

Amazing video