O(nlogn) Algorithm - Find three evenly spaced ones within binary string

Question

I had this question on an Algorithms test yesterday, and I can\'t figure out the answer. It is driving me absolutely crazy, because it was worth about 40 points. I figure

Answer 1

Assumption:

Just wrong, talking about log(n) number of upper limit of ones

EDIT:

Now I found that using Cantor numbers (if correct), density on set is (2/3)^Log_3(n) (what a weird function) and I agree, log(n)/n density is to strong.

If this is upper limit, there is algorhitm who solves this problem in at least O(n*(3/2)^(log(n)/log(3))) time complexity and O((3/2)^(log(n)/log(3))) space complexity. (check Justice's answer for algorhitm)

This is still by far better than O(n^2)

This function ((3/2)^(log(n)/log(3))) really looks like n*log(n) on first sight.

How did I get this formula?

Applaying Cantors number on string.
Supose that length of string is 3^p == n
At each step in generation of Cantor string you keep 2/3 of prevous number of ones. Apply this p times.

That mean (n * ((2/3)^p)) -> (((3^p)) * ((2/3)^p)) remaining ones and after simplification 2^p. This mean 2^p ones in 3^p string -> (3/2)^p ones . Substitute p=log(n)/log(3) and get
((3/2)^(log(n)/log(3)))