Inversion counting with swap queries in O(log N)

Revision en2, by bensonlzl, 2020-02-15 11:21:50

I am currently working on the following problem:

  • Given a permutation $$$A$$$ of length $$$N$$$, you have $$$Q$$$ queries specifying 2 numbers $$$X$$$ and $$$Y$$$ that swap the elements at indices $$$X$$$ and $$$Y$$$. After every query, output the number of inversions in the new permutation. All queries must be answered online.

My current solution can process every query in $$$\log^2$$$ $$$N$$$ per query and $$$N$$$ $$$\log$$$ $$$N$$$ precomputation, by first precomputing the number of inversions in the initial permutation and using a segment tree with a BBST in every node. Each BBST stores all of the elements in the range covered by the segment tree node.

We perform a range query on the value at index $$$X$$$ and the value at index $$$Y$$$ to determine the number of numbers smaller than either number in the segment between them, and then compute the change in inversions. After that, we update the 2 indices in the segment tree. Each of the $$$\log$$$ $$$N$$$ nodes takes $$$\log$$$ $$$N$$$ time to update, which gives an overall complexity of $$$\log^2$$$ $$$N$$$.

My question is: Is it possible to compute the change in inversions more quickly than $$$\log^2$$$ $$$N$$$? eg. computing the change in $$$\log$$$ $$$N$$$ time. I suspect that it is possible with either a modification to the segment tree, or using a completely different data structure all together. Any ideas are welcome :)

Tags inversions, data structure


  Rev. Lang. By When Δ Comment
en3 English bensonlzl 2020-02-15 11:22:06 0 (published)
en2 English bensonlzl 2020-02-15 11:21:50 361 Tiny change: ' query in $O(log^2 N)$ per query' -> ' query in log^2 N per query'
en1 English bensonlzl 2020-02-15 11:18:53 1019 Initial revision (saved to drafts)