Challenge 3: Sorting Strings by Length with Comparator.comparing() — Possible Solution
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LengthSortDemo.java:

import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;

public class LengthSortDemo {
    public static void main(String[] args) {
        List<String> names = new ArrayList<>(List.of("Alexandra", "Bo", "Charlotte", "Sam"));

        names.sort(Comparator.comparing(String::length));

        System.out.println(names);
    }
}

Output:

    [Bo, Sam, Alexandra, Charlotte]

Explanation:
String::length is a method reference supplying the sort key --
Comparator.comparing() builds a Comparator that orders elements by
comparing the RESULT of calling length() on each one, rather than
comparing the strings themselves alphabetically. This is exactly the
"external ordering logic" the chapter describes: String's own natural
ordering (alphabetical, via Comparable) is never touched here at all.

WHY THIS WORKS AS AN ANSWER
------------------------------
This uses Comparator.comparing() to sort by a property (length) other
than the type's natural ordering, matching the chapter's own stated
use case for Comparator over Comparable -- when the natural ordering
isn't the one actually needed.