Can T(n) = 2T(n/2) + nlogn use master theorem?

Yes, T(n) = 2T(n/2) + nlogn can be solved using the master theorem.

First, we identify the values of a, b, and f(n) in the recurrence relation:

a = 2 (the number of recursive calls)
b = 2 (the size of the subproblems)
f(n) = nlogn (the cost of dividing the problem into subproblems and combining their solutions)

Next, we compute the value of logb(a):

log2(2) = 1

Since f(n) = nlogn is asymptotically larger than n^(log2(2)) = n, we can apply case 2 of the master theorem:

If f(n) is Θ(n^c) for some constant c and a >= b^c, then T(n) is Θ(n^c logn).

In this case, c = 1 because f(n) = nlogn.

Since a = 2 and b^c = 2^1 = 2, we have a > b^c and can apply case 2:

T(n) is Θ(nlogn).

Therefore, the solution to the recurrence relation T(n) = 2T(n/2) + nlogn is T(n) = Θ(nlogn).