test_series

Question

i think answer must be O(1) , as whether n is power of 2 or not,result of if condition is always true

Comments

In the solution part, it is trying to find the next nearest power of 2, for this, we need log(n) time, but for finding whether given value is of the power of 2 or not, we need only O(1) time.

Answer 1

I don,t how will you say if condition is always true. I explain through small example.

#include <stdio.h>
#include<stdbool.h>int main()
{
    printf("Hello World");
    int n =7;
    bool s = !(n & (n-1));
    printf("%s\n",s?"true":"False");

    return 0;
}

 output of !(n & (n-1)) is false if n is odd like 7,9,1 e.t.c  

false && (true/ false) is always false. so if n is odd number then if condition is not satisfied and then enter into while condition that take O(logn) time complexity.

If anything you feel wrong plz..   comment !

Comments

if n =7 means 0111 in binary

n-1 =6 means 0110 in binary

n&(n-1)  = 0110 in binary , means some positive number

now  !(n&(n-1)) means 1001 in binary , means some positive number , MEANS TRUE

if(n&& !(n&(n-1)) ) => if(0111  && TRUE)  => if(TRUE && TRUE ) => condition satisfied

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First, run the above program and clear the concept.

Negation(~)  and not equal to ( ! ) are different things.  Negation are uniary operator.

~ (0110) = 1001

!(0110)  doesn't implies that (1001)

NOTE:

!( non-zero) = False

!(Zero) = True

I hope you get idea!!.

I anything wrong  comment.....!  below

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yes you are correct ,

I forgot C syntax.Did no C programming  since left college 2 years back, only java and SQL