Counting inversions in lists algorithmically I want to extract useful info from some data and this makes me think how to do it efficiently.
I will try to explain the problem with math terms.
If we have a sequence of numbers $A=(a_{1}\space a_{2}\dots a_{n})$ and we want to count the number of inversions, ie the number of pairs $(a_{i},a_{j}), 1\leq i < j\leq n$ and $a_{i}>a_{j}$.
I want to find a good algorithm that will solve this problem.
I am looking for an algorithm that is better from the trivial one that solves the problem in polynomial time.
 A: As stated by @joriki, the trivial algorithm that compares all values pairwise has runtime in $\Theta(n^2)$.
A better runtime can be achieved by leveraging balanced binary search trees, e.g. AVL trees. Assume you have AVL trees that store in every node $v$ the number of nodes $t(v)$ in the subtree rooted in $v$; this does not change the runtime characteristics of AVL trees.
Now you take your list and an empty tree and insert the elements one by one. If you have no reversals, every element will end up as the right-most node after inserting. Conversly, the number of reversals an element causes can be read off while inserting; whenever you descend to a left child, add $t(v)$ of the right child you did not go to, and add $1$ if the father is properly larger than the element you insert.
The proposed alorithm has runtime in $\cal{O}(n\log(n))$ as insertion in AVL trees is in $\cal{O}(\log(n))$.
Edit: Apparently you can do better for permutations.
A: The reason you didn't find a solution to this by searching seems to be that the more usual term for what you call a "reversal" is "inversion". Searching for algorithms counting the number of inversions generates lots of hits, including these two questions on stackoverflow with lots of answers:
Counting inversions in an array and How to find the number of inversions in an array.
