Re:Big-O notation explained by a self-taught programmer
by oskarkv on Saturday July 27, 2013 @03:41AM
(#6112151)
Of course O(n) is not equal to O(n^2). O(n) is a set, and O(n^2) is a different set.
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Re:Big-O notation explained by a self-taught programmer
by fmax30 on Saturday July 27, 2013 @07:24AM
(#6112538)
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 .