* It will never change, even if the maps change at a later time.
   *
   * <p>Since this method uses {@code TreeMap} instances internally, the keys of the right map must
   * all compare as distinct according to the comparator of the left map.
   *
   * <p><b>Note:</b>If you only need to know whether two sorted maps have the same mappings, call
   * {@code left.equals(right)} instead of this method.
   *

        /*
         * sortAndDedup may have made enough room for this element, but that's not necessarily good
         * enough. Consider, for example, the case where we have a buffer of size (n+1), add n
         * distinct elements, and add the last element over again many times over. We don't want a
         * situation where we re-sort the entire buffer every time the last element is re-added.
         *

    @WeakOuter
    final class EntrySetImpl extends EntrySetView {
      @Override
      public Spliterator<Entry<K, V>> spliterator() {
        return Spliterators.spliterator(this, Spliterator.ORDERED | Spliterator.DISTINCT);
      }
    }
    return new EntrySetImpl();
  }

  @Override
  Set<K> createKeySet() {
    @WeakOuter
    final class KeySetImpl extends KeySetView {
      @Override

  }

  @Override
  @GwtIncompatible // Spliterator
  Spliterator<Entry<K, V>> entrySpliterator() {
    return CollectSpliterators.map(
        delegate.entrySet().spliterator(),
        Spliterator.DISTINCT | Spliterator.ORDERED | Spliterator.IMMUTABLE | Spliterator.NONNULL,
        Maps::unmodifiableEntry);
  }

  @Override
  public void forEach(BiConsumer<? super K, ? super V> action) {
    delegate.forEach(action);

      // extra zero bits we need to add to the end of bitsPerChar to get 3 in total.
      // The logic here would be wrong for bitsPerChar > 8, but since we require distinct ASCII
      // characters that can't happen.
      int zeroesInBitsPerChar = Integer.numberOfTrailingZeros(bitsPerChar);
      this.charsPerChunk = 1 << (3 - zeroesInBitsPerChar);

  }

  @Override
  public Spliterator<E> spliterator() {
    return CollectSpliterators.indexed(
        size(),
        ImmutableList.SPLITERATOR_CHARACTERISTICS | Spliterator.SORTED | Spliterator.DISTINCT,
        delegateList()::get,
        comparator());
  }

  // redeclare to help optimizers with b/310253115
  @SuppressWarnings("RedundantOverride")
  @Override
  @J2ktIncompatible
  @GwtIncompatible

        final String permissionKey = userBean.map(user -> StreamUtil.stream(user.getPermissions())
                .get(stream -> stream.sorted().distinct().collect(Collectors.joining("\n")))).orElse(StringUtil.EMPTY);

        try {
            return facetCache.get(query + "\n" + permissionKey, () -> {

 */
@GwtCompatible
abstract class ImmutableMapEntrySet<K, V> extends ImmutableSet.CachingAsList<Entry<K, V>> {
  private static final int SPLITERATOR_CHARACTERISTICS =
      Spliterator.ORDERED | Spliterator.DISTINCT | Spliterator.NONNULL | Spliterator.IMMUTABLE;

  static final class RegularEntrySet<K, V> extends ImmutableMapEntrySet<K, V> {
    private final transient ImmutableMap<K, V> map;

   * href="http://en.wikipedia.org/wiki/Adobe_Photoshop#File_format">Wikipedia</a>; this is the
   * regular output/input of Photoshop (which can also export to various image formats; note that
   * files with extension "PSB" are in a distinct but related format).
   *
   * <p>This is a more recent replacement for the older, experimental type {@code x-photoshop}: <a
   * href="http://tools.ietf.org/html/rfc2046#section-6">RFC-2046.6</a>.
   *
   * @since 15.0

   * It will never change, even if the maps change at a later time.
   *
   * <p>Since this method uses {@code TreeMap} instances internally, the keys of the right map must
   * all compare as distinct according to the comparator of the left map.
   *
   * <p><b>Note:</b>If you only need to know whether two sorted maps have the same mappings, call
   * {@code left.equals(right)} instead of this method.
   *