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Asked: 2026-06-15T10:06:36+00:00

Is time complexity O(n^2) or O (n(logn)^2) better? I know that when we simplify

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Is time complexity O(n^2) or O (n(logn)^2) better?

I know that when we simplify it, it becomes

O(n) vs O((logn)^2)

and logn < n, but what about logn^2?

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  1. Editorial Team

    n is only less than (log n)2 for values of n less than 0.49…

    So in general (log n)2 is better for large n…

    But since these O(something)-notations always leave out constant factors, in your case it might not be possible to say for sure which algorithm is better…

    Here’s a graph:

    enter image description here

    (The blue line is n and the green line is (log n)2)

    Notice, how the difference for small values of n isn’t so big and might easily be dwarfed by the constant factors not included in the Big-O notation.

    But for large n, (log n)2 wins hands down:

    enter image description here

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