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Algorithms | Analysis of Algorithms (Recurrences) | Question 8

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

(A)

A

(B)

B

(C)

D

(D)

C



Answer: (C)

Explanation:

Recursive relation for the DoSomething() is 
 

  T(n) =  T(\\sqrt{n}) + 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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Last Updated : 28 Jun, 2021
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