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 :)

Auto comment: topic has been updated by bensonlzl (previous revision, new revision, compare).Assuming the changes revert after each query, then:

You don't need to modify any values; you can solve this subproblem of counting greater elements than a given value on some subarray in $$$O(logN)$$$. You can reduce them to prefix queries and then solve those with a persistent segment tree:

The current problem I'm working on has persistent changes, so this won't work, but I was also thinking about the case where the swaps revert after each query as a side problem. Thanks for the idea!

ngmh pls help me ^^ :| :<

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