Challenge 1: Comparing getClass() Across Type Arguments — Possible Solution
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ErasureDemo.java:

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

public class ErasureDemo {
    public static void main(String[] args) {
        List<String> strings = new ArrayList<>();
        List<Double> doubles = new ArrayList<>();

        // Both getClass() calls return java.util.ArrayList -- the SAME
        // Class object -- even though the declared type arguments
        // (String vs Double) are completely different. This is type
        // erasure in action: by the time the class is loaded and
        // running, there is no such thing as "ArrayList<String>" or
        // "ArrayList<Double>" anymore -- only ArrayList. The type
        // argument only ever existed for the compiler's own
        // compile-time checking and was discarded afterward.
        System.out.println(strings.getClass() == doubles.getClass());
    }
}

Output:

    true

WHY THIS WORKS AS AN ANSWER
------------------------------
This demonstrates the chapter's own opening example directly and the
comment explains the result using the chapter's own core claim: one
shared runtime class regardless of the type argument used at the
declaration site.