// there's a way around creating the separate sortedKeys array, and if we're allocating
          // one array of size n, we might as well allocate two -- to say nothing of the allocation
          // done in Arrays.sort.
          for (int i = 0; i < size; i++) {
            // We're careful to put only K instances in.
            if (i > 0 && comparator.compare((K) sortedKeys[i - 1], (K) sortedKeys[i]) == 0) {

    if (!iterator.hasNext()) {
      return Optional.absent();
    }

    /*
     * TODO(kevinb): consider whether this "optimization" is worthwhile. Users with SortedSets tend
     * to know they are SortedSets and probably would not call this method.
     */
    if (iterable instanceof SortedSet) {
      SortedSet<E> sortedSet = (SortedSet<E>) iterable;
      return Optional.of(sortedSet.last());
    }