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wonder how this video ' made

that was a great explanation, thank you

your video is very very great! I love it. Thank you very much!

Thank you so much. Couldn't be more clear

This is such a great explanation! Thank you so much.

Thank you after 3 years

Thank you so much ❤❤

Should the space complexity not be O(n)? Explaination below. It's true that for each call with make take O(n) space which is the n for that call. And the depth is log n. However, in each next call N is also half. So it's like N + N/2 + N/4 .... 1, which is 2n, so O(n) When we are in a recusrive call, we shouldn't consider space from calls that have finished. And for a given call only it's previous sizes are active. It's similar to how in merge sort (with using extra auxilary array to merge), the space complexity would still be O(n).

Great one, especially the 2δ X δ part!!

An excellent explanation of the problem

Ommggg amazing explaination

4:40 bookmark

好棒好棒🙃

For anyone confused about the "for j=1 to j = 7" part, here is my take at explaining this: For every point p in S, we need only to check points q in distance dist(p, q) <= d. Recall that every pair of points in the left sector has distance at most d. Same thing for the right sector. (If it wasn't that way, we would have a smaller d.😏) Take a paper and draw a circle (a filled circle, not just circumference) of radius d and center point p. Try fitting as many points inside as you can, so that all points in the left half are at least d distance apart from each other. And in the right half, also they are d distance apart. You get a hexagon (6 vertices + 1 center = 7) i.imgur.com/FWN0Okm.png Thus we can iterate through S and check just the next 7 points.

this is amazing. Thank you so much for saving our souls in algorithm class.

hello, i wish you could do more explanation videos like this. I really appreciate your clear explanation!!!

Thank You ma'am.

how did she arrived at the 2nd strip

very very nice explanantion....

Incredible explanation, the best I have found, thank you very much!!!!!!!!!!!!!!!!!!!!!!!!!

牛逼 上课没听懂在你这里听懂了

we definitely need more of these explanations from you

wow this was amazing. thank you so much for this!!

i have a question. does the question assume there are coincident points? if coincident points arent allowed, then when checking in the 2d * d rectangle, can we just check for the next 5 points instead of 7? since coincident points will not happen. thanks.

The only tutorial on the channel had to be the on the topic i was looking for.

Best explanation, bar none

OMG, that was the BEST explanation EVER! You're a lifesaver, thanks a TRILLION!

great explanation, thank you

I think it's enough to iterate j from 1 to 4 because in each side the maximum number of points is 4

Great Explanation but i have not understood clearly

now find the optimal solution for N-dimensional Euclidean space.

For split pairs why we iterating 1 to 7?

That was awesome Keep it going🎉🎉

W vid

感恩

amazing amazing amazing and the best explanation for this problem. Thank you

講的很好！😊

how did you find delta in 6,8 do you brute force the left and the right with out divide it ?

3:55 7 points explanation

so great , that was i am loving version.

Thank you so much , I wanted to bring to your attention a concern I have regarding the time complexity of the closestPair function in this algorithm. Specifically, when calling the function for all elements of Y in the following lines: dl = closestPair(X[1 .. mid], Y); dr = closestPair(X[mid+1 ... n], Y); I believe this approach increases the time complexity. The search for the strip is performed in Y, which has a length of N. Consequently, the work done by subproblems becomes O(N), where N is the number of all the points of the problem. This happens because Y is not getting smaller through the dividing process. I propose a modification where the work by subproblems is O(L), where L is the length of the subproblem. To achieve this, the Y in the call to closestPair(X, Y) should contain the same elements as X, with X being the array that undergoes division. Thank you for considering this suggestion.

Great explanation mam.

Thank you! :D

5:57 Why for j=1 to 7? what's the 7 about?

woah wasn't expecting to find this after reviewing dynamic programming

Pls accept my LinkedIn request 🥺

الله يدخلش الجنة يارب

3:35 2-delta distance 4:08 2d*d O(1) for each point

God Bless you, you are great

thank you