Hi,
Inversions are defined as when i<j then a[i]>a[j]

Hi,

Inversions are defined as when i<j then a[i]>a[j]

dedicated invest is a={2,3,8,6,1}

hence reversals are

2,1 3,1 8,6 8,1 6,1 b. the most reversals appear when the invest is in descending adjust, i.e { n,n-1...1} now, the reversals earn be any span elements fine from it which is n(n-1)/2 c.Implantation nature basically reduces the reversals by 1 in each interation, hence compute of reversals is instantly proportional to the interval of implantation nature, so hence interval of reversal nature is dedicated by O(n+f(n)) where f(n) is compute of reversals, hence if compute of reversals are n then implantation natures operate in O(n) yet if its n^2, then it operates in O(n^2) here we want to apprehend control complete n where

8n^{2} < 64nlogn

=n < 8logn

=n/8 < logn

i.e n-8logn>0

on solving the level we gain, n=43

hence control n<=43, implantation nature is better