What is the Big O complexity of this algorithm

I can’t seem to figure out the time complexity for this algorithm. I know that the while loop will execute at O(log n) times because n keeps halving. But I’m not super sure how to represent the time complexity of the inner for loop.

while(n>0){
   for(int j = 0; j < n; j++){
        System.out.println("*");
   }
   n = n/2;
}

Answer

Each while loop divides n by 2.

Therefore, the first for loop would run n times, the second for loop would run n/2 times, the third for loop would run n/4 times, and so on.

If we set n to an arbitrary large value, then the complexity would be:

n + n/2 + n/4 + n/8 + n/16 + n/32 + ... = 2n

Of course, n is an integer, and programmatically, the division would result in 0 after enough repetitions.

Therefore, the time-complexity of the algorithm would be O(2N) = O(N)