#!/usr/bin/python

def unique(s):

    """Return a list of the elements in s, but without duplicates.

    For example, unique([1,2,3,1,2,3]) is some permutation of [1,2,3],

    unique("abcabc") some permutation of ["a", "b", "c"], and

    unique(([1, 2], [2, 3], [1, 2])) some permutation of

    [[2, 3], [1, 2]].

    For best speed, all sequence elements should be hashable.  Then

    unique() will usually work in linear time.

    If not possible, the sequence elements should enjoy a total

    ordering, and if list(s).sort() doesn't raise TypeError it's

    assumed that they do enjoy a total ordering.  Then unique() will

    usually work in O(N*log2(N)) time.

    If that's not possible either, the sequence elements must support

    equality-testing.  Then unique() will usually work in quadratic

    time.

    """

    n = len(s)

    if n == 0:

        return []

    # Try using a dict first, as that's the fastest and will usually

    # work.  If it doesn't work, it will usually fail quickly, so it

    # usually doesn't cost much to *try* it.  It requires that all the

    # sequence elements be hashable, and support equality comparison.

    u = {}

    try:

        for x in s:

            u[x] = 1

    except TypeError:

        del u  # move on to the next method

    else:

        return u.keys()

    # We can't hash all the elements.  Second fastest is to sort,

    # which brings the equal elements together; then duplicates are

    # easy to weed out in a single pass.

    # NOTE:  Python's list.sort() was designed to be efficient in the

    # presence of many duplicate elements.  This isn't true of all

    # sort functions in all languages or libraries, so this approach

    # is more effective in Python than it may be elsewhere.

    try:

        t = list(s)

        t.sort()

    except TypeError:

        del t  # move on to the next method

    else:

        assert n > 0

        last = t[0]

        lasti = i = 1

        while i < n:

            if t[i] != last:

                t[lasti] = last = t[i]

                lasti += 1

            i += 1

        return t[:lasti]

    # Brute force is all that's left.

    u = []

    for x in s:

        if x not in u:

            u.append(x)

    return u