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by matahwoosh
3283 days ago
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Generally, yes, but I think `O(log n / log log n) == O(log n)` is wrong. log(n) / log(log(n)) = logx(n) (where x = log(n), wasn't sure how to describe logarithm base in a better way).
So you get O(logx(n)). In general the logarithm base doesn't matter for Big-O when it's a constant, but I'm not sure you can apply the same thing to a base of log(n). |
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