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)
(B)
(C)
(D)
A
(B)B
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)
Quiz of this Question
Please comment below if you find anything wrong in the above post