public Builder<E> add(E... elements) {
      for (E element : elements) {
        add(element);
      }
      return this;
    }

    /**
     * Adds each element of {@code elements} to the {@code ImmutableCollection} being built.
     *
     * <p>Note that each builder class overrides this method in order to covariantly return its own
     * type.
     *
     * @param elements the elements to add

        if (newLength > elements.length) {
          elements = Arrays.copyOf(elements, newLength);
        }
      }
      elements[n++] = element;
      return this;
    }

    /**
     * Adds each element of {@code elements} to the {@code ImmutableSortedSet}, ignoring duplicate
     * elements (only the first duplicate element is added).
     *
     * @param elements the elements to add

 * according to when the first occurrence of the element was added. When the multiset contains
 * multiple instances of an element, those instances are consecutive in the iteration order. If all
 * occurrences of an element are removed, after which that element is added to the multiset, the
 * element will appear at the end of the iteration.
 *

func TestFindFunc(t *testing.T) {
	emptyElement := []string{}
	elements := []string{"a", "b", "c"}
	tests := []struct {
		name     string
		elements []string
		fn       func(string) bool
		want     *string
	}{
		{
			elements: emptyElement,
			fn: func(s string) bool {
				return s == "b"
			},
			want: nil,
		},
		{
			elements: elements,
			fn: func(s string) bool {
				return s == "bb"
			},

        }
    };
    private final List<Stoppable> elements = new ArrayList<Stoppable>();

    public CompositeStoppable() {
    }

    public static CompositeStoppable stoppable(Object... elements) {
        return new CompositeStoppable().add(elements);
    }

    public static CompositeStoppable stoppable(Iterable<?> elements) {
        return new CompositeStoppable().add(elements);
    }

    E[] elements = createSamplesArray();
    E duplicate = elements[(elements.length / 2) - 1];
    elements[(elements.length / 2) + 1] = duplicate;
    return new ArrayWithDuplicate<>(elements, duplicate);
  }

  // Helper methods to improve readability of derived classes

  protected int getNumElements() {
    return getSubjectGenerator().getCollectionSize().getNumElements();
  }

    }

    @Override
    public boolean add(@ParametricNullness E element) {
      checkArgument(predicate.apply(element));
      return unfiltered.add(element);
    }

    @Override
    public boolean addAll(Collection<? extends E> collection) {
      for (E element : collection) {
        checkArgument(predicate.apply(element));
      }
      return unfiltered.addAll(collection);
    }

   * have been removed.
   *
   * <p><b>Warning:</b> Typical uses of the resulting iterator may produce an infinite loop. You
   * should use an explicit {@code break} or be certain that you will eventually remove all the
   * elements.
   *
   * <p>To cycle over the elements {@code n} times, use the following: {@code
   * Iterables.concat(Collections.nCopies(n, Arrays.asList(elements)))}
   *

    E[] elements = createSamplesArray();
    E duplicate = elements[(elements.length / 2) - 1];
    elements[(elements.length / 2) + 1] = duplicate;
    return new ArrayWithDuplicate<>(elements, duplicate);
  }

  // Helper methods to improve readability of derived classes

  protected int getNumElements() {
    return getSubjectGenerator().getCollectionSize().getNumElements();
  }

      return unfiltered.add(element, occurrences);
    }

    @Override
    public int remove(@CheckForNull Object element, int occurrences) {
      checkNonnegative(occurrences, "occurrences");
      if (occurrences == 0) {
        return count(element);
      } else {
        return contains(element) ? unfiltered.remove(element, occurrences) : 0;
      }