Arpa's blog

By Arpa, history, 4 years ago, In English

(This blog has been restored as Errichto and some other users had wanted.)

Hi !

Here is some implementation for solving RMQ (Tarjan’s algorithm) (Range Maximum / Minimum Query).

It’s very simple to implement and it’s time complexity is O((n + qa(n)), a() stands for Akerman inverse function used in DSU.

Problem : Given array a of n integers, and q queries, for each query print the maximum value in range [L, R].

Solution: We need a array of vectors, called assigned. assigned[r] contains queries that their R is r. When getting queries, push each query in assigned[R]. We need a dsu, first pari is i. We need a stack, named st.

For i from 0 to n, do:
	While st is not empty and a[st.top] <= a[i]
		Set i parent of st.top in dsu and pop this element from st.
	Push i to st
	For each query assigned to i
		Answer of this query is a[root of L of this query in DSU].
Code here.

Note that in above code I used path-compression technique for dsu only, size-comparing technique can be used too (but it has lower performance).

It’s obviously true, because each time for any j ≤ i, a[root(j)] is the greatest value in range [j, i].

Performance test

This method (known as Arpas trick)
Vector + Binary search
Sparse table
O(n) method
generator

Here is the result:

Method\Time(milliseconds)Strictly increasing arrayStrictly decreasing arrayRandom
This method (known as Arpa's trick) 294328902946
Sparse table 361235953807
Vector + Binary search 310161303153
O(n) method 378839203610
 
 
 
 
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4 years ago, # |
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I know this technique as "Arpa's trick". Are you sure you've got the right name?

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    4 years ago, # ^ |
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    I discovered this technique myself, but it seems this technique existed before (as Zlobober said). So I can't name this technique "Arpa's trick", but I call it "Arpa's trick", you can call it "Arpa's trick" as well.

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    4 years ago, # ^ |
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    At least he's helping others ... You can also write tutorials and name techniques using your fake name . Are you sure you wrote "lucian bicsi" correctly ?

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      4 years ago, # ^ |
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      Fake name ?! In fact, "Arpa" is compressed version of my name, my real name is "AmirReza PoorAkhavan", my friends call me Arpa at school, chat, etc. I'm more familiar with "Arpa" instead of my real name ("PoorAkhavan" or "AmirReza").

      There are much persons, knowing me as "Arpa", and don't know my real name even.

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        4 years ago, # ^ |
          Vote: I like it +8 Vote: I do not like it

        I was talking about I_love_Retrograd

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          4 years ago, # ^ |
          Rev. 2   Vote: I like it 0 Vote: I do not like it

          Oh right ! But your comment prepared a opportunity for me to talk about my nick name :D

          Edit: sorry anyway.

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4 years ago, # |
  Vote: I like it +6 Vote: I do not like it

You could also binary search on the stack instead of using DSU, which might be easier to code depending on the implementation, and still achieve a reasonable performance (binary searches are pretty light).

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    4 years ago, # ^ |
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    I'll add some statistics about what method is faster soon, please wait for several hours.

    Edit: Done.

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      4 months ago, # ^ |
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      Can anyone please explain the vector+binary search approach for this problem or provide any source where i can learn the approach. Thank you.

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        4 months ago, # ^ |
        Rev. 2   Vote: I like it +5 Vote: I do not like it

        Iterate over the elements of the array, and answer all queries that end in the current element. And you maintain a decreasing vector with the elements (and indices), on which you can binary search to find the answer.

        Think of it this way, if you fixate the right boarder and look at the RMQs [0, R], [1, R], [2, R], ..., [R, R], then you can notice that the RMQs are monotone decreasing. I.e. RMQ(0, R) >= RMQ(1, R) >= RMQ(2, R) >= ... >= RMQ(R, R). In the vector you maintain the list of all values that appear in that RMQ list, together with their most-left index.

        If you add a new element (so you go from R to R+1), you just need to look at how the RMQs change, since now you want the list to reflect the RMQs for [0, R+1], ..., [R+1, R+1]. Which means if the new element in the array is smaller than the previous minimum, then you can just add the element to the vector, as all the previous R+1 RMQs don't change, and the last one will have the RMQ A[R+1]. But if the element is bigger, then you need to delete a few of the previous RMQs from the list, since by adding that element their RMQ will get bigger.

        The example code from Arpa should explain the details.

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4 years ago, # |
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You can outline header of the table like this:

A/B 1 2 3 4
Arpa's trick 1 3 6 0
Sparse table 6 3 1 0
O(n) 3 1 6 0

Code:


A/B | 1 | 2 | 3 | 4 - - Arpa's trick | 1 | 3 | 6 | 0 Sparse table | 6 | 3 | 1 | 0 O(n) | 3 | 1 | 6 | 0