Algorithms | Analysis of Algorithms (Recurrences) | Question 8
What is the time complexity of the following recursive function:
int DoSomething (int n) { if (n <= 2) return 1; else return (DoSomething (floor(sqrt(n))) + n);} |
(A) 
(n)
(B) (nlogn)
(C) (logn)
(D) (loglogn)
(A) A
(B) B
(C) C
(D) D
Answer: (D)
Explanation: Recursive relation for the DoSomething() is
T(n) = T() + C1 if n > 2
We have ignored the floor() part as it doesn’t matter here if it’s a floor or ceiling.
Let n = 2^m, T(n) = T(2^m)
Let T(2^m) = S(m)
From the above two, T(n) = S(m)
S(m) = S(m/2) + C1 /* This is simply binary search recursion*/
S(m) = O(logm)
= O(loglogn) /* Since n = 2^m */
Now, let us go back to the original recursive function T(n)
T(n) = S(m)
= O(LogLogn)
