# Algorithms | Analysis of Algorithms (Recurrences) | Question 8

What is the time complexity of the following recursive function:

## c

 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)

D

(D)

C

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)

Quiz of this Question
Please comment below if you find anything wrong in the above post

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