How to find the number of inversions in an array ? [duplicate]

Answer 1

You could also use a counting sort-like approach, if the array only contains small numbers for example (like if it's a character array):

inversions = 0
let count = array of size array.Length
for i = 0 to array.Length - 1 do
    for j = array[i] + 1 to maxArrayValue do
        inversions = inversions + count[j]

    count[array[i]] = count[array[i]] + 1

Basically, keep a count of how many times each element appears. Then at each step i, the number of inversions the ith element generates is equal to the sum of all the elements bigger than i that come before i, which you can easily compute using the count you're keeping.

This will be O(n*eps) where eps is the domain of the elements in your array.

This is definitely simpler in my opinion. As for efficiency, it's only good if eps is small obviously. If it is, then it should be faster than other approaches since there's no recursion.

Answer 2

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.