Поиск Google ничего не нашел

Thinking about it, I realized that the only solutions coming to my mind were those operating in O(n) time. But I found a better solution later. I am going to use the following denotation of sets


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No, O(log n) + O(log n) is not O(n). It is still O(log n). When we have a big-O formula multiplied by a constant, it is equivalent to the value without the multiplied constant. So: O(k * g) = O(g). Where k is a constant, non-zero factor and g is a bounding function.


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I have come to the solution that it belongs to masters theorem case 2 since n lg n is O(n^2). but after referring to the solution manual i noticed this solution that they have. The soluttion says that n lg n = O ( n ^(lg 3 - e)) for e between 0 and 0.58.


Функция f(n) = n2/2 + 3n/2 + 1 возрастает приблизительно как n2/2 (отбрасываем сравнительно медленно растущее слагаемое 3n/2+1). Константный множитель 1/2 также убираем и получаем асимптотическую оценку для алгоритма 1, которая обозначается специальным символом O(n2)...


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