Challenge 3: instanceof Pattern Matching Without an Explicit Cast — Possible Solution
====================================================================

PatternDemo.java:

public class PatternDemo {
    static void describe(Object obj) {
        if (obj instanceof String s) {
            System.out.println("String length: " + s.length());
        } else if (obj instanceof Integer i) {
            System.out.println("Integer doubled: " + (i * 2));
        } else {
            System.out.println("Unrecognized type");
        }
    }

    public static void main(String[] args) {
        describe("hello");
        describe(21);
        describe(3.14);
    }
}

Output:

    String length: 5
    Integer doubled: 42
    Unrecognized type

Explanation:
`obj instanceof String s` checks obj's type AND binds it to a new
variable `s` of type String in the very same expression -- no separate
`String s = (String) obj;` cast line is needed, unlike the old
two-step check-then-cast approach. The same pattern applies to the
Integer branch with `i`, which is auto-unboxed for the multiplication.

WHY THIS WORKS AS AN ANSWER
------------------------------
Both branches use the pattern-matching form of instanceof the chapter
introduces, binding directly to a typed variable with no explicit
cast anywhere in the method, exactly the "no separate cast needed"
behavior the chapter describes.