Question 10668 – Time-Complexity
When n = 22k for some k ≥ 0, the recurrence relation
T(n) = √(2)T(n/2) + √n, T(1) = 1
evaluates to :
Correct Answer: A
Question 30 Explanation:
√n (log n + 1)
√n log n
√n log √n
n log √n