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It's still useful for comparing potential performance of algorithms with different O(x). e.g. an O(n) algorithm will most likely outperform an O(n^2) algorithm
by jonny_eh 4y ago
It's still useful for comparing potential performance of algorithms with different O(x).
e.g. an O(n) algorithm will most likely outperform an O(n^2) algorithm, but two different O(n) algorithms can still perform quite differently.
But you're right, it's meant for comparing scaling.