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I was looking for some proof that greedy works for our currency system for "coin change problem", because in some cases we apply DP and some Greedy. So happy to find this video. Thanks for awesome explaination!

👍🏻👍🏻

Thank you so much for this fantastic explanation

For people trying to implement this, be aware of the following: Let's say you have a set of points where the median in x is 80 for example, but you have multiple points where x is 80. Now on this line: if Y[i].x <= X[v].x then append Y[i] to Yl else append Y[i] to Yr. You will end up adding more to the left side then expected! IN such a way that X_l.size() != Y_l.size() && X_r.size() != Y_r.size().

Wow, very clear!

clear

I'm so grateful...

The residual network is unclear.

Shitty explanation

wow i actually understood immediately! thank you so much!

Hi, I know this is late, At 8:50, doesn't the algorithm return only array of 'unique elements' instead of array of 'unique elements + single instance of element that is repeated?

Are you still alive?

the man stuttered so much i couldnt understand half of it

Thank you sir !

minor mistake, i think on 2:27 the second maximum matching example is wrong, since it only has 3 edges, instead of 4. Thank you for the content, really helped me understand the definition

bullshit

thanks, this is remarkable explaination

Thanks a lot sir

Thanks you very much. Didnt understand though how median is O(1) amortized? can it be maintained while inserting somehow?

THANK YOU SO MUCH!!!!!!!

Can we also use max heap on y coordinate I think that would be more appropriate for this use

This cant be more worse

what happens if you increase the all edges capacity as n+1. Ofcourse we will think that |A| =|B| =n.

that's great! cool too!

thanks but your solution is wrong ! the good upper bond is 2*min{ |L| + |R|} + 1

Incredible, thanks u!

This Sudo codes doesn't work . try a simple input : Coins[2.3,5] and amount = 4 , it gives output 1 , but it should be 2 . He was using greedy concept .

Good one, thank you for the work.

Excellent demonstration! Thank you.

Thank you, thank you!

What if the lines are 2,(1,3) and 1,(2,4) ?

thanks for the explanation! One question though, why is the upper bound |v|+1? You mentioned that v is the number of vertices, so what does the +1 do?

<3

Great detailed explanation!

Good Job! very clear. would you know where can i find information regarding proof of correctness of this algorithm? I mean why does this promise maximum matching?

Hi, I have a question. Minute 12:34 if l==r return false. why are you checking left with right? should you checl the i with the values from the substring?

Thanks. I enjoyed the explanation using the pseudocode, though I'd also like to see the how time complexity of the recurrence relation is calculated.

The solution to this problem is subset of all the solutions to the ramanujan-hardy partition of a positive number. The way to filter the answer is eliminate all solutions from original problem for all coins which are not in given set :)

thank you this was amazing

very nice speech. thanks!

Really nice video, thank you sir!

Extremely interesting and enlightening. I really wish you go on making other videos of that kind, it would really help many people.

Awesome explanation (y)

Thanks for the help!

I used a variation of this for actual software running on production floor.thanks

Thanks! Just had to change i>0 to i>=0 to make it work.

is there any complete algorithm course like this?

Super sir you made to understand it easily thank you

great video, thank you

I have a question about the recursive formula: C(i,a) = min{C(i+1, a), 1 + C(i, a-den[i])}, what is the reason you only consider choosing one den[i] why the recursive formula is not this: C(i,a) = min(C(i+1,a), k + C(i, a - k * den[i])) for all possible k