5 ms·
Agreed! I'm pretty sure they're just using O(x) to mean on the order of x, since big O of any constant is the same (unless that's their point??? :O)
by vaishnavsm 4y ago
Agreed! I'm pretty sure they're just using O(x) to mean on the order of x, since big O of any constant is the same (unless that's their point??? :O)