The complexity of the algorithm with loops

Question

I have algorithm that contains next loops:

for (int i = 0; i < size; ++i) {

    for (int j = i + 1; j < size; ++j) {
        //Do stuff
    }
}

I found that this algorithm has $O(n^2)$ complexity but I can't understand why? I.e. if $N = 4$ then $n^2 = 16$ but my loop has 6 iterations only. Just it's a half of $n^2$ value.

P.S. I understand never how to measure the complexity of the algorithm, I only can understand how to write it in the mathematics.

The point is that $n^2/2 = O(n^2)$. – Yuval Filmus Sep 07 '16 at 02:17 — Yuval Filmus, Sep 07 '16 at 02:17
Do we just reduce a factor? – Denis Steinman Sep 07 '16 at 02:47 — Denis Steinman, Sep 07 '16 at 02:47
@Шах I suggest you check the definition of big-O. – David Richerby Sep 07 '16 at 07:29 — David Richerby, Sep 07 '16 at 07:29

score 1 · Accepted Answer · answered Sep 07 '16 at 06:14

1

Your "stuff" will get executed $N(N - 1)/2 = 0.5N^2 - 0.5N$ times. When analyzing the asymptotic complexity, only the highest order term is kept, and multiplicative constants are removed, leaving you with $O(N^2)$.

It works this way because we're interested in what happens when $N$ goes to infinity (scalability).

answered Sep 07 '16 at 06:14

rcpinto

471
3
15

It seems, I understood. Thank you for explanation! – Denis Steinman Sep 07 '16 at 06:18

The complexity of the algorithm with loops

1 Answers1