CODEWITHSUNDEEP
algorithmloggingbig-o
Sara.a
Sara.a

Reputation: 115

O( log n! ) and O( (log n )! )

To find their relation I substituted log n = x and log n! = n(log n) so with base a , O( log n! ) became a^x(x) and (log n)! became x(x-1)(x-2).... now I think the first one has a higher growing speed. But can you help me to find their relation using big O of n^2

Upvotes: 2

Views: 2206

Answers (1)

Gor
Gor

Reputation: 2908

  1. Actually x(x-1)(x-2).... becomes x^x + ... because you have x scopes. This means that O((log n)!)has a higher growing speed.

  2. Also, if log(n) := x, then n = 2^x and n^2 will become (2^x)^2 = 2^2x which has lower growing speed than x^x

Summary

O(log n!) < O(n^2) < O((log n)!)

Upvotes: 1

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