Just need a confirmation on something real quick.
If an algorithm takes n(n-1)/2 tests to run, is the big oh O(n^2)?
Just need a confirmation on something real quick. If an algorithm takes n(n-1)/2 tests
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Just need a confirmation on something real quick.
If an algorithm takes n(n-1)/2 tests to run, is the big oh O(n^2)?
n(n-1)/2 expands to
(n^2 -n) / 2, that is(n^2/2) - (n/2)(n^2/2)and(n/2)are the two functions components, of whichn^2/2dominates.Therefore, we can ignore the
- (n/2)part.From
n^2/2you can safely remove the /2 part in asymptotic notation analysis.This simplifies to
n^2Therefore yes, it is in O(n^2)