Use Master Theorem to solve such recurrence relations.
Let a be an integer greater than or equal to 1 and b be a real number greater than
1. Let c be a positive real number and
d a nonnegative real number. Given a recurrence of the form
then for n a power of b,
- if logb a < c, T (n) = Θ(nc),
- if logb a = c, T (n) = Θ(nc log n),
- if logb a > c, T (n) = Θ(nlogb a).
1) T(n) = 2T(n/2) + 0(1)
In this case
a = b = 2;
logb a = 1; c = 0 (since nc =1 => c= 0)
So Case (3) is applicable. So T(n) = Θ(n)
:)
2) T(n) = T(sqrt(n)) + 0(1)
Let m = log2 n;
=> T(2m) = T( 2m / 2 ) + 0(1)
Now renaming K(m) = T(2m) => K(m) = K(m/2) + 0(1)
Apply Case (2).