First example is a job for Master Theorem: a=2, b=3, f=n/log^2(n)
. log_b(a) < 1
, so it is the third case, because f(n) grows (significantly) faster than n^(log_b(a))*log^k(n)
for any k. This means that main job is done outside recursion and T(n)=O(n/log^2(n))
.
Second example can be solved this way:
T(n)
= T(n - 10) + log(n)
= ...
= log(n) + log(n - 10) + ...
= log(n * (n-10) * (n-20) * ...)
= [n = 10k]
= log(10^k) + log(k!)
= k*log(10) + k*log(k) - k + O(log(k))
= O(k*log(k))
= O(n*log(n))