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    <p>If your elements are somehow Comparable (the fact that the order has any real meaning is indifferent -- it just needs to be consistent with your definition of equality), the fastest duplicate removal solution is going to sort the list ( 0(n log(n)) ) then to do a single pass and look for <em>repeated</em> elements (that is, equal elements that follow each other) (this is O(n)).</p> <p>The overall complexity is going to be O(n log(n)), which is roughly the same as what you would get with a Set (n times long(n)), but with a much smaller constant. This is because the constant in sort/dedup results from the cost of comparing elements, whereas the cost from the set is most likely to result from a hash computation, plus one (possibly several) hash comparisons. If you are using a hash-based Set implementation, that is, because a Tree based is going to give you a O( n log²(n) ), which is even worse.</p> <p>As I understand it, however, you do not need to <em>remove</em> duplicates, but merely test for their existence. So you should hand-code a merge or heap sort algorithm on your array, that simply exits returning true (i.e. "there is a dup") if your comparator returns 0, and otherwise completes the sort, and traverse the sorted array testing for repeats. In a merge or heap sort, indeed, when the sort is completed, you will have compared every duplicate pair unless both elements were already in their final positions (which is unlikely). Thus, a tweaked sort algorithm should yield a huge performance improvement (I would have to prove that, but I guess the tweaked algorithm should be in the O(log(n)) on uniformly random data)</p>
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    1. POJava: Detect duplicates in ArrayList?
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    1. COIn this case, n is 6 so I wouldn't waste a lot of time on implementation details, but I'll keep your idea of the special heap sort if I ever need to do something like that.
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      1. USPaul Tomblin
    2. COI don't understand the third paragraph. Mergesort and heapsort are both O(nlog(n)), not O(log(n)) as you write; even if you exit once you identify a duplicate, that still doesn't change your time complexity...
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