3 ms·
Sure. Trivially true. I meant to overloaded O() to mean "order of magnitude" when used for a constant factor, but you're right that in computational complexit
by kjhughes 16y ago
Sure. Trivially true. I meant to overloaded O() to mean "order of magnitude" when used for a constant factor, but you're right that in computational complexity's big O notation, O(k) = O(1) for all constants k.