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by fmax30
4714 days ago
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Big O is a upper bound , what I meant was that if a piece of code has O(n) then it also has O(n^2).
With emphasis on the upper bound
f(x) = O(g(x)) if f(x) <= cg(x). so my point is if f(N) = O(N) then f(N) is also equal to O(N^2)
as f(N) <= CN^2 . |
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