What complexity would be?

#1

int count = 0;
for (int i = N; i > 0; i /= 2)
for (int j = 0; j < i; j++)
count++;

#2

For a input integer n, the innermost statement is executed following times.

n + n/2 + n/4 + … 1

So time complexity T(n) can be written as

T(n) = O(n + n/2 + n/4 + … 1) = O(n)

The value of count is also n + n/2 + n/4 + … + 1

#3

I hope I’ve cleared your doubt. I ask you to please rate your experience here
Your feedback is very important. It helps us improve our platform and hence provide you
the learning experience you deserve.

On the off chance, you still have some questions or not find the answers satisfactory, you may reopen
the doubt.