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10^-6n^2 and n^2 both are O(n^2) how?

if we multiply very very small number in n^2 then big O of n^2 and (very very small)*n^2 is n^2,why and how?

asked 21 Oct '18, 00:53

narayansatish's gravatar image

accept rate: 0%

That's literally how big O notation is defined. Any constant does not matter for the asymptotic behavior. Wikipedia defines the term pretty well:

(21 Oct '18, 02:07) algmyr7★

big O is more of "how much slower code gets with larger input" rather than just how slow it is.

(21 Oct '18, 02:13) abdullah7686★
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question asked: 21 Oct '18, 00:53

question was seen: 108 times

last updated: 21 Oct '18, 02:13