Question

So I’ve been trying to get a handle on Big Oh calculations. I feel I have the basics down but am stumped on what seems a really easy calculation. So if the calculation below has a big oh of O(n log n) (I really hope I’ve at least got that right) what does changing the order of the loops do to the complexity? Thanks so much in advance for your time.

    int ONLogN(int N) //O(n log n)
    {
        int iIterations = 0;
        for (int i = 0; i < N; ++i)
        {
            ++iIterations;
            for (int j = 1; j < N + 1; j *= 2)
                ++iIterations;
        }
        return iIterations;
    }
    int WhatBigOhIsThis(int N) //???
    {
        int iIterations = 0;
        for (int j = 1; j < N + 1; j *= 2)
        {
            ++iIterations;
            for (int i = 0; i < N; ++i)                
                ++iIterations;
        }
        return iIterations;
    }

Answer 1

The index variables on the two loops are independent, hence the resulting complexity is necessarily the same.