J

#### Johannes Bauer

All the O() tells you is the general shape of the line.

Nitpick: it only gives an *upper bound* for the complexity. Any function

that is within O(n) is also within O(n^2). Usually when people say O()

they actually mean capital Thetha (which is the correct term).

It's perfectly

feasible that for the range of values of n that you care about in a

particular application, there's an O(n^2) algorithm that's way faster

than another O(log(n)) algorithm. [Though that becomes a lot less

likely as n gets large.]

Since O() only gives upper bounds it's also possible for an algorithm

within O(n^2) to always be faster than another algorithm within O(logn).

The O(n^2) algorithm could be Thetha(1).

Regards,

Johannes

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