I am currently working on the following problem:↵
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- 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.↵
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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. ↵
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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 $\log^2$ $N$.↵
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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 :)↵
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↵
↵
- 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 :)↵
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