Time complexity n^2

Question

Is this O(N^2) or O(nlogn). Isnt it n^2 when there are nested loops?

int a[], N;
int f1(){ int i, j, sum=0;
    for (i=1;; i=2*i)
    {
        If (i>=N)  return sum;
        for (j=1; j<2*i;j++) sum+=a[i];
    }

Answer 1

This is O(N log N) as the outer loop is doubling the value of i in every iteration. So the complexity of outer loop is O(log N) instead of O(N) .

If you had i++ or similar instead of i=2*i then the time complexity of two loops would have been O(n^2) .

Edit: this is a simplified analysis. Please see the answer from R Sahu for more rigorous analysis.

Answer 2

Is this O(N^2) or O(nlogn).

It is neither.

Isnt it n^2 when there are nested loops?

That is true when you iterate over the items linearly. That is not true in your case.

In your case ...

The values of i are: 1 2 4 8 16 ... N

The inner loop is executed 2 + 4 + 8 + 16 + 32 ... N times.

That is a geometric series. The sum of a geometric series is a(1 - r^n)/(1 - r) .

In your case, a is 2 , r is 2 , and n is log2(N) (log with base 2). Hence, the sum, after some simplification, is 2*2^(log2(N)) , which is same as 2*N .

ie your algorithmic complexity is O(N) .

Thanks are due to @LedHead for correcting the error in the initial post.