Counting inversions in an array

Question

I\'m designing an algorithm to do the following: Given array A[1... n], for every i < j, find all inversion pairs such that A[i] > A[j]

Answer 1

The number of inversions can be found by analyzing the merge process in merge sort :

When copying a element from the second array to the merge array (the 9 in this exemple), it keeps its place relatively to other elements. When copying a element from the first array to the merge array (the 5 here) it is inverted with all the elements staying in the second array (2 inversions with the 3 and the 4). So a little modification of merge sort can solve the problem in O(n ln n).
For exemple, just uncomment the two # lines in the mergesort python code below to have the count.

def merge(l1,l2):
    l = []
    # global count
    while l1 and l2:
        if l1[-1] <= l2[-1]:
            l.append(l2.pop())
        else:
            l.append(l1.pop())
            # count += len(l2)
    l.reverse()
    return l1 + l2 + l

def sort(l): 
    t = len(l) // 2
    return merge(sort(l[:t]), sort(l[t:])) if t > 0 else l

count=0
print(sort([5,1,2,4,9,3]), count)
# [1, 2, 3, 4, 5, 9] 6

EDIT 1

The same task can be achieved with a stable version of quick sort, known to be slightly faster :

def part(l):
    pivot=l[-1]
    small,big = [],[]
    count = big_count = 0
    for x in l:
        if x <= pivot:
            small.append(x)
            count += big_count
        else:
            big.append(x)
            big_count += 1
    return count,small,big

def quick_count(l):
    if len(l)<2 : return 0
    count,small,big = part(l)
    small.pop()
    return count + quick_count(small) + quick_count(big)

Choosing pivot as the last element, inversions are well counted, and execution time 40% better than merge one above.

EDIT 2

For performance in python, a numpy & numba version :

First the numpy part, which use argsort O (n ln n) :

def count_inversions(a):
    n = a.size
    counts = np.arange(n) & -np.arange(n)  # The BIT
    ags = a.argsort(kind='mergesort')    
    return  BIT(ags,counts,n)

And the numba part for the efficient BIT approach :

@numba.njit
def BIT(ags,counts,n):
    res = 0        
    for x in ags :
        i = x
        while i:
            res += counts[i]
            i -= i & -i
        i = x+1
        while i < n:
            counts[i] -= 1
            i += i & -i
    return  res