Let f be the function with one argument - positive integer n
function f(n)
j = 1
for i=1 to n:
j = 2*j
k = j
while k > 0:
k--
print *
My approach:
- let n = 6
- breakdown the for cycle
- breakdown the while cycle -> when i == 1, we print two *, when i == 2, we print four *, etc.
So I've assumed that it can be $\sum_{i=1}^n 2^i$ what gives me $\Theta(2^n)$.
Am I right or did I something wrong?
