if an algorithm takes a total time of T(log n + n^2), is this big O(log n + n^2) or simply big O(n^2) ?
unclepauly 6 Light Poster
Ravalon 62 Posting Whiz in Training
Big O takes the biggest part of the time expression because it has the biggest effect. :) As n gets bigger and bigger, the log n part will get less and less significant as the n^2 part starts to dominate. Since you can toss anything smaller than n^2 in the expression, your big O turns out to be O(n^2).
I hope that helps!
unclepauly 6 Light Poster
awesome - thanks!
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