time complexity

Question

int Dosomething(int n)

{

if(n>=2)

return 1;

else

return (floor(sqrt(n)))+n;

}

time complexity of the program?

Answer 1

The question contains anchor condition IF(N>=2)  RETURN 1;  

Thus if the function DOSOMETHING() is called with any n>=2 it will take constant time and return 1 without involving any recursion.and if is is called with n<2 ... lets say 1. it will return sqrt(1) + 1 = 2... that to without involving any recursion and will  again take constant time O(1).

However if the question would have been  with anchor condition IF(N<=2) RETURN 1 and recursion statement

return DOSOMETHING(FLOOR(SQRT(N))) +N  Then it could be solved using master`s theorem or recursion tree method with Time Complexity being log₂log₂n .