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

Yes, we can use the master theorem to solve this recurrence relation.

The master theorem states that for a recurrence relation of the form T(n) = aT(n/b) + f(n), where a ≥ 1, b > 1, and f(n) is asymptotically positive, we have:

- If f(n) = O(nᵣ) for some r < logb(a), then T(n) = Θ(nᵣ).
- If f(n) = Θ(nᵣ logⁱ n) for some r = logb(a) and i ≥ 0, then T(n) = Θ(nᵣ logⁱ⁺¹ n).
- If f(n) = Ω(nᵣ₊ₑ) for some e > 0 and r > logb(a), and if af(n/b) ≤ cf(n) for some constant c < 1 and all sufficiently large n, then T(n) = Θ(f(n)).

In this case, we have a = 2, b = 2, and f(n) = nlogn.

We can compute logb(a) = log₂(2) = 1 and r = 1.

Since nlogn = Θ(n¹ log n), which matches the form nᵣ logⁱ n with r = 1 and i = 0, we have case 2. Therefore, T(n) = Θ(nlog₂ n).