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A comment on The true cost of linked lists. Parent comment
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.
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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.