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Online algorithm for range inversions counting

Revision en1, by madn, 2017-08-27 00:33:11

I've been recently thinking about this problem: Given array of n integers and q queries. For each query you need to count number of inversions( i, j pairs such that i < j and a[i] > a[j] ) in subarray from l to r. And you have to answer on queries online(you're not given all queries beforehand).

I was able to come up with a solution that i'd describe below, but I'd like to know if it's possible to solve the problem with better time complexity.

So, my solution: 1) preprocessing Let's divide our array into sqrt(n) blocks, each sized sqrt(n), and merge-sort elements inside these blocks, counting number of inversions in each block. Then, for each pair of blocks, count number of inversions such that first element lies within first block and second lies within second block. We can do this using binary search, because each block is sorted already. After that, for each sqrt-block-range([l, r] such that l is a left border of one block and r is a right border of other block) we'd find number of inversions in it by summing precalculated values( inversions in each block and inversions in pairs ). There are around n pairs so we can do all preprocessing in O(n * sqrt(n) * log(n)). We also need to build merge-sort tree for queries.

2) queries We already know answer for sqrt-block-range that lies within query range. Then we just need to count inversions that include elements outside sqrt-block-range. There no more than 2 * sqrt of them, so we can do this using binary search in merge-sort tree in O(sqrt(n) * log^2(n)), or in O(sqrt(n) * log(n)) if we use partial cascading.

Overall complexity: O((n + q) * sqrt(n) * log(n)) with partial cascading.

So what are your thoughts, can it be solved faster?

P.S. Sorry for my poor English.

Tags inversions, queries, sqrt-decomposition, data structures

History

 
 
 
 
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  Rev. Lang. By When Δ Comment
en2 English madn 2017-08-27 01:02:25 146 Final revision (published)
en1 English madn 2017-08-27 00:33:11 1812 Initial revision (saved to drafts)