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
Answer: (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)
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