What is the running time of an O(n log n) sort done n times?

Question

Is it O(n^2 log n)? Can you show how it is derived? Is O(n^2 log n) the same as O((n^2) * (log n))?

Is this a homrwork? Basic math and wikipedia are enough to answer in 3 minutes. — Ondrej Tucny, Sep 22 '15 at 07:11
Guys, power takes preceence over multiplication... parentheses are redundant. — Ondrej Tucny, Sep 22 '15 at 07:12

score 3 · Answer 1 · answered Sep 22 '15 at 08:37

Rigorous derivation:

By definition

T(n) = O(n Log(n)) <=> for some N and C,  n > N => T(n) < C.n.log(n).

Then obviously

for these N and C, n > N => n.T(n) < C.n².log(n)

which implies

n.T(n) = O(n²log(n)).

score 0 · Accepted Answer · answered Sep 22 '15 at 07:15

Doing something n times takes O(n) time. And n log n is just another way of writing n * log n So we get:

  O(n) * O(n * log n)
= O((n) * (n * log n)) 
= O(n * n * log n)
= O(n^2 * log n)

Yes, the two ways of writing it that you have shown are the same. This is, however, something completely different: O(n^(2 log n))

2 Answers2