T(n)<=cf(n)
for all n >= k
Inversion - when there is an array a
and a[i] > a[j] && i < j
i, j <= n\2
i, j > n\2
i <= n\2 < j
if n = 1 return 0
else
x = Count (1st half of a, n\2)
y = Count (2nd half of a, n\2)
z = CountSplitInversions (a, n)
return x + y + z
T(n)
- maximum running time of the algorithm