"analyzer": "empty_analyzer"
          }
        }
      },
      {
        "lang_pt": {
          "match": "*_pt",
          "mapping": {
            "type": "text",
            "analyzer": "portuguese_analyzer"
          }
        }
      },
      {
        "lang_pt-br": {
          "match": "*_pt-br",
          "mapping": {
            "type": "text",

            "analyzer": "empty_analyzer"
          }
        }
      },
      {
        "lang_pt": {
          "match": "*_pt",
          "mapping": {
            "type": "text",
            "analyzer": "portuguese_analyzer"
          }
        }
      },
      {
        "lang_pt-br": {
          "match": "*_pt-br",
          "mapping": {
            "type": "text",

            "analyzer": "empty_analyzer"
          }
        }
      },
      {
        "lang_pt": {
          "match": "*_pt",
          "mapping": {
            "type": "text",
            "analyzer": "portuguese_analyzer"
          }
        }
      },
      {
        "lang_pt-br": {
          "match": "*_pt-br",
          "mapping": {
            "type": "text",

            normalized[i] = parts[i].replaceFirst("^0+(?!$)", "");
            if (normalized[i].isEmpty()) {
                normalized[i] = "0";
            }
        }

        // Find longest sequence of consecutive zeros
        int longestStart = -1;
        int longestLength = 0;
        int currentStart = -1;
        int currentLength = 0;

        for (int i = 0; i < 8; i++) {

   * either it has exceeded the keep alive limit or the idle connections limit.
   *
   * Returns the duration in nanoseconds to sleep until the next scheduled call to this method.
   * Returns -1 if no further cleanups are required.
   */
  fun closeConnections(now: Long): Long {
    // Find the longest-idle connections in 2 categories:
    //

    for (n = len; n < size - n; n <<= 1) {
      System.arraycopy(array, 0, array, n, n);
    }
    System.arraycopy(array, 0, array, n, size - n);
    return new String(array);
  }

  /**
   * Returns the longest string {@code prefix} such that {@code a.toString().startsWith(prefix) &&
   * b.toString().startsWith(prefix)}, taking care not to split surrogate pairs. If {@code a} and
   * {@code b} have no common prefix, returns the empty string.

    for (n = len; n < size - n; n <<= 1) {
      System.arraycopy(array, 0, array, n, n);
    }
    System.arraycopy(array, 0, array, n, size - n);
    return new String(array);
  }

  /**
   * Returns the longest string {@code prefix} such that {@code a.toString().startsWith(prefix) &&
   * b.toString().startsWith(prefix)}, taking care not to split surrogate pairs. If {@code a} and
   * {@code b} have no common prefix, returns the empty string.

        "it.unimi.dsi.fastutil.objects.ObjectOpenHashSet",
        "it.unimi.dsi.fastutil.objects.Reference2ObjectOpenHashMap",
        // For build operation listener and logging
        "it.unimi.dsi.fastutil.longs.LongSet",
        "it.unimi.dsi.fastutil.longs.LongOpenHashSet",
        "it.unimi.dsi.fastutil.longs.LongSets",
        // For dependency management
        "it.unimi.dsi.fastutil.longs.Long2ObjectMap",

            // Leading zeros removed and consecutive zeros compressed
            assertEquals("2001:db8::1", method.invoke(null, "2001:0db8:0:0:0:0:0:1"));

            // Longest sequence of zeros compressed (3 zeros at end vs 1 zero at start)
            assertEquals("2001:db8:0:1::1", method.invoke(null, "2001:0db8:0:1:0:0:0:1"));

/**
 * A bounded {@linkplain BlockingQueue blocking queue} backed by an array. This queue orders
 * elements FIFO (first-in-first-out). The head of the queue is that element that has been
 * on the queue the longest time. The tail of the queue is that element that has been on
 * the queue the shortest time. New elements are inserted at the tail of the queue, and the queue
 * retrieval operations obtain elements at the head of the queue.