Finding unique elements of a list

Richard A. O'Keefe ok@REDACTED
Fri Dec 3 01:52:38 CET 2004

If you want to remove duplicate elements from a list but otherwise
preserve the order of the elements, so that, e.g., 

    remove_dups([3,1,4,1,5,9,2,6,5,3,5,8,9]) -> [3,1,4,5,9,2,6,8]

then there is an O(N.lgN) method which goes like this:
    (1) pair each element with its position number in the list
    (2) sort the list of pairs (using whole pairs as keys)
    (3) remove adjacent duplicates
    (4) switch the {Elem,Posn} pairs to {Posn,Elem} pairs
	O(U) where 1 <= U <= N, U is the number of unique elements
    (5) sort the list of {Posn,Elem} pairs
    (6) strip off the positions, {Posn,Elem} -> Elem

In Haskell notation, to keep this message short,

    remove_dups :: Ord a => [a] -> [a]
    remove_dups xs = [x | (_,x) <- sort (remdups (sort (zip xs [1..])))]
      where remdups []            = []
            remdups ((x,n) : xns) = (n,x) : remdup2 xns x

            remdup2 ((y,_) : xns) x | y == x = remdup2 xns x
            remdup2 xns           _          = remdups xns

This is a little over twice as slow as simply sorting the elements and
removing adjacent duplicates, so it's only something you would use if
the original order of the elements mattered.

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