Algorithms | Analysis of Algorithms (Recurrences)

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) $\theta$ (n)
(B) $\theta$ (nlogn)
(C) $\theta$ (logn)
(D) $\theta$ (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)

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Algorithms | Analysis of Algorithms (Recurrences) | Question 8

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