Asymptotic running time in Big Theta notation
Considering the below algorithm,
Loop1 until(i<n^2)
Loop2 until(j<i^2)
....
j=j+4
End Loop2
i=i*3
End Loop1
I think this is Theta(n^2*log(n)). This is correct or is the Big Theta
higher than this?
No comments:
Post a Comment