Sorting a number of lists based on two lists in Java

Sorting of multiple lists of the same size is possible following these steps:

1. Create an array/list of indexes according to the sorting rules (first by names, then by birthdays in appropriate YYYYMMDD format, etc.)

List<String> names  = Arrays.asList("bbb", "aaa", "bbb", "aaa");
List<String> phones = Arrays.asList("1023456", "1324560", "1227890", "1446752");
List<String> bdays  = Arrays.asList("20001122", "19980105", "20010614", "19990507");

int[] indexes = IntStream.range(0, names.size())
    .boxed()
    .sorted(Comparator.comparing(names::get).thenComparing(bdays::get))
    .mapToInt(Integer::intValue)
    .toArray(); // -> [1, 3, 0, 2]

2. Reorder each of the input lists according to the indexes. A separate method may be created for this:

private static List<String> sort(List<String> list, int[] indexes) {
    return Arrays.stream(indexes).mapToObj(list::get).collect(Collectors.toList());
}
// ...
System.out.println(sort(names,  indexes)); -> [aaa, aaa, bbb, bbb]
System.out.println(sort(phones, indexes)); -> [1324560, 1446752, 1023456, 1227890]
System.out.println(sort(bdays,  indexes)); -> [19980105, 19990507, 20001122, 20010614]

However, it would be more natural to create a container object to combine the values from the different input lists, which has become quite simple after introducing record classes in Java 16, and the sort these objects:

record Person(String name, String phone, String bday) {}

List<Person> persons = IntStream.range(0, names.size())
    .boxed()
    .map(i -> new Person(names.get(i), phones.get(i), bdays.get(i)))
    .sorted(Comparator.comparing(Person::name).thenComparing(Person::bday))
    .collect(Collectors.toList());

persons.forEach(System.out::println);

Output:

Person[name=aaa, phone=1324560, bday=19980105]
Person[name=aaa, phone=1446752, bday=19990507]
Person[name=bbb, phone=1023456, bday=20001122]
Person[name=bbb, phone=1227890, bday=20010614]