GROUNDWORK
Chapter 9 - Maps and Sets

Name: ________________________________     Date: ______________

Fill in every blank. Where a term is asked for, write the word the course
uses. Where code is asked for, write exact Java. Where output is asked
for, write what the program prints, one line per line.


VOCABULARY
----------

 1. A __________________ stores a connection between a key and a value
    directly, so a lookup doesn't depend on two arrays staying in the
    same order.

 2. In scores.put("Ada", 88), "Ada" is the __________________, which you
    look things up by, and 88 is the __________________ it leads to.

 3. A __________________ is a collection with no duplicates, best suited
    to answering one question: is this thing in here?

 4. A __________________ is a small method with no name, such as
    e -> e.getValue(), often passed to Comparator.comparing to sort a
    list.


READING CODE
------------

 5. What does this print?

        Map<String, Integer> scores = new LinkedHashMap<>();
        scores.put("Ada", 88);
        scores.put("Grace", 92);
        scores.put("Alan", 45);
        scores.put("Ada", 91);
        IO.println(scores);

    ____________________________________

 6. What does this print?

        Map<String, Integer> scores = new LinkedHashMap<>();
        scores.put("Ada", 88);
        IO.println(scores.get("Grace"));
        IO.println(scores.getOrDefault("Grace", 0));

    ____________________________________

 7. What does this print?

        String text = "a b a c b a";
        Map<String, Integer> counts = new LinkedHashMap<>();
        for (String word : text.split(" ")) {
            counts.put(word, counts.getOrDefault(word, 0) + 1);
        }
        IO.println(counts);

    ____________________________________

 8. What does this print?

        Map<String, Integer> scores = new LinkedHashMap<>();
        scores.put("Ada", 88);
        scores.put("Grace", 92);
        scores.put("Alan", 45);
        String top = null;
        for (String name : scores.keySet()) {
            if (top == null || scores.get(name) > scores.get(top)) {
                top = name;
            }
        }
        IO.println(top);

    ____________________________________

 9. What does this print?

        Set<String> band = Set.of("Ada", "Grace", "Alan");
        Set<String> chorus = Set.of("Grace", "Alan", "Katherine");
        Set<String> both = new HashSet<>(band);
        both.retainAll(chorus);
        IO.println(both.size());
        Set<String> bandOnly = new HashSet<>(band);
        bandOnly.removeAll(chorus);
        IO.println(bandOnly.contains("Ada"));

    ____________________________________

10. What does this print?

        Map<String, String> ada = new LinkedHashMap<>();
        ada.put("score", "88");
        ada.put("grade", "B");
        Map<String, String> grace = new LinkedHashMap<>();
        grace.put("score", "92");
        grace.put("grade", "A");
        Map<String, Map<String, String>> students =
                new LinkedHashMap<>();
        students.put("Ada", ada);
        students.put("Grace", grace);
        for (String name : students.keySet()) {
            Map<String, String> info = students.get(name);
            String score = info.get("score");
            String grade = info.get("grade");
            IO.println(name + ": " + score + " (" + grade + ")");
        }

    ____________________________________


WRITING CODE
------------

11. Complete the line so that it prints 0 instead of null for a key the
    map does not have:
    Map<String, Integer> scores = new LinkedHashMap<>();
    scores.put("Ada", 88);
    IO.println(scores.__________________("Grace", 0)); // 0

12. Complete the line so that it takes the "Alan" entry out of the map
    entirely:
    // scores is {Ada=88, Grace=92, Alan=45}
    scores.__________________("Alan");
    IO.println(scores); // {Ada=88, Grace=92}

13. Complete the line so that it removes the duplicates and keeps the
    numbers sorted:
    ArrayList<Integer> numbers = new ArrayList<>(List.of(9, 30, 9, 20));
    Set<Integer> unique = new __________________<>(numbers);
    IO.println(unique); // [9, 20, 30]


WHY IT WORKS
------------

14. counts.getOrDefault(word, 0) gives back the current count, or
    __________________ the very first time a word is seen, so
    counts.put(word, counts.getOrDefault(word, 0) + 1) works whether or
    not the key already __________________.

15. A loop over scores.keySet() hands you the __________________ only;
    scores.__________________() hands you each key and its value
    together, as an entry.