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Counting inversions in an array

This is an phone interview question: "Find the number of inversions in an array". I guess they mean O(Nlog N) solution. I believe it cannot be better than O(Nlog N) since this is the sorting complexity.

The answers to a similar question can be summarized as follows:

1. Calculate half the distance the elements should be moved to sort the array : copy the array and sort the copy. For each element of the original array a[i] find it’s position j in the sorted copy (binary search) and sum the halves the distances abs(i - j)/2.

2. Modify merge sort : modify merge to count inversions between two sorted arrays and run regular merge sort with that modified merge.

   Does it make sense ? Are there other (maybe simpler) solutions ? Isn’t it too hard for a phone interview ?

Answer 1

It is actually an application of divide-and-conquer algorithm, and if you are familiar with it you can come up with the solution quickly.

Take [1 3 8 5 7 2 4 6] as an example, assume we have sorted array as [1 3 5 8] and [2 4 6 7], and now we need to combine the two arrays and get the number of total inversions.

Since we already have number of inversions in each sub-array, we only need to find out the number of inversions caused by array merging. Each time an element is inserted, for example, 2 inserted into [1 # 3 5 8], you can know how many inversions there are between the first array and the element 2 (3 pairs in this example). Then you can add them up to get the number of inversions caused by merging.