Блог пользователя Arpa

Some method for solving RMQ
Автор Arpa, история, 9 лет назад, По-английски

Hi!

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

It’s very simple to implement and its time complexity is O((n + q)·a(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 an 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
		The answer to this query is a[root of L of this query in DSU].
Code here.

Note that in the above code I used the 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

Update (Feb 2025): The above method is actually . Refer to this blog for algorithm.

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9 лет назад, скрыть # |
 
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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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9 лет назад, скрыть # |
 
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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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    9 лет назад, скрыть # ^ |
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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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      6 лет назад, скрыть # ^ |
       
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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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        6 лет назад, скрыть # ^ |
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        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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9 лет назад, скрыть # |
 
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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
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5 лет назад, скрыть # |
 
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Ok, I found this post, it's too old but anyway. If someone finds it in the future. It's said:

"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)."

But only path-compression works in $$$O(\log n)$$$ time. So if you want it to work in $$$a(n)$$$ time, you need to use both path-compression and size-comparing techniques.

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