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Technically this is O(n^2). On average the inner loop will iterate over half of n each time it is run. So it could be written as O(n*(n/2)), and that can be simplified to O(n^2)

Robert Parker
– Robert Parker

2009-10-14 11:31:22 +00:00
Commented Oct 14, 2009 at 11:31
Hm, looks like you're right. This is not a simple problem, just to find all the 1 takes O(n), not much room for any further search/comparision with O(logn) complexity.

DaClown
– DaClown

2009-10-14 11:55:08 +00:00
Commented Oct 14, 2009 at 11:55

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