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By MR.AC, 13 months ago,

I came across an issue when solving problem C in today's contest. I was trying to find the distance between two pointers (by subtracting) in a set to determine the number of elements between those two pointers. However, I got an error. Is there a way to find this distance?

Submission (under "if(t == 3)")

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 » 13 months ago, # |   -16 let's consider that you have two pointers , the left one in position l and the right one in position r , then if you want the number of elements there (including the pointers) it is r-l+1 , if it was without them it is r-l-1 , and if it was without one of them it is r-l
•  » » 13 months ago, # ^ |   0 but you have to make sure that l
•  » » » 13 months ago, # ^ |   0 Oh my bad I knew that but the issue was not about that, it was about a compilation error.
•  » » » » 13 months ago, # ^ |   0 I thought you said " Is there a way to find this distance?" , so sorry if I wasn't helpful
 » 13 months ago, # | ← Rev. 2 →   0 use ordered set(policy based set)
•  » » 13 months ago, # ^ |   0 That might be a good idea. If I am trying to find elements in range L to R, should I do something like order_of_key(R+1) — order_key(L)? (reminds me of prefix sums)
•  » » » 13 months ago, # ^ |   0 This correctly gives #of rooks between rows L,R, but that's different than "at least one rook in each row in range [L,R]" which is what the problem asks
•  » » » » 13 months ago, # ^ |   0 However, I am using a set combined with a multiset to ensure that only distinct rooks are stored. (Please check my most recent submission) Therefore, in my case, it can be used to check if the condition is satisfied.
•  » » » » » 13 months ago, # ^ |   0 Ah true, my bad
 » 13 months ago, # |   +3 std::distance, works in linear time.
•  » » 13 months ago, # ^ |   0 I knew there had to be something! Thank you.
•  » » » 13 months ago, # ^ |   0 Note that it works in linear time for each call of the function, so if you were to use it in your solution, it still would not pass because the total complexity would be $O(n^2)$. I'm 99% sure that the implementation for std::distance with not random access iterators just increments the smaller one repeatedly until it reaches the bigger iterator...
•  » » » » 13 months ago, # ^ |   +3 Yeah, I tried it and it got TLE on test case 5. But I am sure it will be useful elsewhere!
 » 13 months ago, # |   0 You can't do this with set. You can use ordered_set which requires some installation. Or if you are first inserting then doing queries, i.e You don't insert anything after you make first query, you can use Coordinate Compression and then use Fenwick Tree/Segment tree