* {@link Ordering#leastOf(Iterable, int)}, which provides a simpler API for that use case.
   */
  public void offerAll(Iterable<? extends T> elements) {
    offerAll(elements.iterator());
  }

  /**
   * Adds each member of {@code elements} as a candidate for the top {@code k} elements. This

                    uq.setUrl(d.getUrl());
                    uq.setMetaData(d.getMetaData());
                    return uq;
                }).collect(Collectors.toList());
        urlQueueService.offerAll(crawlerContext.sessionId, childList);
    }

    protected void storeChildUrl(final String childUrl, final String parentUrl, final String metaData, final int depth) {

            }
            throw e;
        }
    }

    @Override
    public void delete(final String sessionId) {
        deleteBySessionId(sessionId);
    }

    @Override
    public void offerAll(final String sessionId, final List<OpenSearchUrlQueue> urlQueueList) {
        if (logger.isDebugEnabled()) {
            logger.debug("Offering URL: Session ID: {}, UrlQueue: {}", sessionId, urlQueueList);
        }

        list.subList(k, list.size()).clear();
      }
      list.trimToSize();
      return unmodifiableList(list);
    } else {
      TopKSelector<E> selector = TopKSelector.least(k, this);
      selector.offerAll(iterator);
      return selector.topK();
    }
  }

  /**
   * Returns the {@code k} greatest elements of the given iterable according to this ordering, in

        list.subList(k, list.size()).clear();
      }
      list.trimToSize();
      return unmodifiableList(list);
    } else {
      TopKSelector<E> selector = TopKSelector.least(k, this);
      selector.offerAll(iterator);
      return selector.topK();
    }
  }

  /**
   * Returns the {@code k} greatest elements of the given iterable according to this ordering, in