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algorithm - Big Oh notation

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)?

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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 which n^2/2 dominates. Therefore, we can ignore the - (n/2) part.

From n^2/2 you can safely remove the /2 part in asymptotic notation analysis.

This simplifies to n^2

Therefore yes, it is in O(n^2)


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