12 - Data Structures: Maps



  • learn the purpose of Maps and two most commonly used implementations: HashMap and TreeMap
  • learn how HashMap is implemented internally
  • Practice!


Class Properties

Class can have properties:

class Main {
    private String name = "Rachel";

Properties can be accessed from each method of the class:

class Main {
    private static String name = "Rachel";

    static void foo() {

    static void bar() {
        name = "Monica";


Common Map Operations

// creating new map
Map<String, String> map = new HashMap<>();

// adding new element to the map
map.put("key1", "value1");
map.put("key2", "value2");

// getting element from map
String value = map.get("key2");

// getting all the keys
Set<String> keys = map.keySet();

// getting all the values
Collection<String> values = map.values();

// map size

// removing element from map

// iterating over map
for (Map.Entry<String, String> entry : map.entrySet()) {
    System.out.println(entry.getKey() + ":" + entry.getValue());


Exercise 1

Make a Map that associates the following employee IDs with names.

Keys and values of Maps can be any Object type, so in real life you would probably have the key be a String and the associated value be a Person or Employee object. To make things simpler on this exercise, you can use String for both the ID and the name, rather than bothering to create a Person or Employee class.

The point here is to associate keys with values, then retrieve values later based on keys.

Id Name
a1234 Steve Jobs
a1235 Bill Gates
a1236 Jeff Bezos
a1237 Larry Page
a1238 Sergey Brin

😎 Toggle Solution
public class MapsExercise1 {

    public static void main(String[] args) {
        Map<String, String> employees = new HashMap<>();
        employees.put("a1234", "Steve Jobs");
        employees.put("a1235", "Bill Gates");
        employees.put("a1236", "Jeff Bezos");

        // lets get Steve
        String name = employees.get("a1234");

        // iterate over all keys and values
        for (Map.Entry<String, String> entry : employees.entrySet()) {
            System.out.println(entry.getValue() + " has an id = " + entry.getKey());

Exercise 2

Go back to the previous problem and make your lookup method work with keys regardless of whether they are lower or uppercase. For example, both “a1234” and “A1234” should match Steve Jobs. Hint: very similar to the previous exercise, so if your solution is complex, you are overlooking the obvious.

😎 Toggle Solution
public class MapsExercise2 {
    static Map<String, String> employees = new HashMap<>();

    public static void main(String[] args) {
        addEmployee("a1234", "Steve Jobs");
        addEmployee("a1235", "Bill Gates");
        addEmployee("a1236", "Jeff Bezos");

        String employee = findEmployee("A1234");

    private static void addEmployee(String id, String name) {
        employees.put(id.toLowerCase(), name);

    private static String findEmployee(String id) {
        return employees.get(id.toLowerCase());

Exercise 3

Write a program that creates a map of students and the country they are from. Add 10 students from our class to this map.

  • use HashMap
  • what are the data types for key and value?
  1. Print to the console where the person next to you comes from (using apropriate Map method)
  2. Print each entry in a format name: country
  3. Print all unique country names.

😎 Toggle Solution
public class MapsExercise3 {

    public static void main(String[] args) {
        Map<String, String> students = new HashMap<>();
        students.put("Josh", "USA");
        students.put("Oliver", "Germany");
        students.put("Elena", "Russia");
        students.put("Stephan", "France");
        students.put("Sergei", "Russia");


        for (Map.Entry<String, String> entry : students.entrySet()) {
            System.out.println(entry.getKey() + ":" + entry.getValue());

        Set<String> uniqueCountries = new HashSet<>(students.values());

Exercise 4

Write a method that acts as a english-german dictionary. It takes one parameter - english word - and returns german translation. If word is not found it returns "Sorry, I don't know such word".

😎 Toggle Solution
public class MapsExercise4 {

    public static void main(String[] args) {

    private static String translate(String word) {
        Map<String, String> dictionary = new HashMap<>();
        dictionary.put("table", "der tisch");
        dictionary.put("flower", "die blume");
        dictionary.put("lamp", "die lampe");

        if (dictionary.containsKey(word)) {
            return dictionary.get(word);
        } else {
            return "Sorry, I don't know such word";

Exercise 5

Write a method program that contains a pizza menu - we are interested only in pizza name and it's price. For simplification we can assume that all pizzas are sold in the same size and price is an Integer. Write a method that takes how much money there is in your wallet and it returns Map of pizzas and their prices, that you can afford.

😎 Toggle Solution
public class MapsExercise5 {

    public static void main(String[] args) {

    private static Map<String, Integer> pizzas(Integer money) {
        Map<String, Integer> pizzas = new HashMap<>();
        pizzas.put("margherita", 10);
        pizzas.put("funghi", 12);
        pizzas.put("capriciosa", 14);
        pizzas.put("tonno", 12);

        Map<String, Integer> result = new HashMap<>();

        for (Map.Entry<String, Integer> pizza : pizzas.entrySet()) {
            if (pizza.getValue() <= money) {
                result.put(pizza.getKey(), pizza.getValue());

        return result;

Exercise 6

Write a program that calculates average price of a second hand car based on the list of prices found on EBay. Example list:

  • Toyota: 10000, 15000, 18000
  • BMW: 20000, 23000, 50000
  • Audi: 35000, 43000, 18000, 50000

The method should return a map where key is a car name and value is a average price.

😎 Toggle Solution
public class MapsExercise6 {

    public static void main(String[] args) {
        Map<String, List<Integer>> prices = new HashMap<>();
        prices.put("Toyota", Arrays.asList(10000, 15000, 18000));
        prices.put("BMW", Arrays.asList(20000, 23000, 50000));
        prices.put("Audi", Arrays.asList(35000, 43000, 18000, 50000));

    private static Map<String, Integer> averagePrices(Map<String, List<Integer>> carPrices) {
        Map<String, Integer> result = new HashMap<>();

        for (Map.Entry<String, List<Integer>> car : carPrices.entrySet()) {
            result.put(car.getKey(), average(car.getValue()));

        return result;

    private static int average(List<Integer> list) {
        int result = 0;
        for (Integer i : list) {
            result += i;
        return result / list.size();

Exercise 7

Peppa Pig has following friends: Suzy Sheep, Emily Elephant, Rebecca Rabbit, Danny Dog, Pedro Pony. Zoe Zebra has following friends: Freddy Fox, Rebecca Rabbit, Gabriella Goat, Kylie Kangaroo, Danny Dog

Write a method that prints their common friends.


Write it in a flexible way, so that we can easily add new character with their friends without changing too much code.

😎 Toggle Solution
public class MapsExercise7 {
    private static Map<String, Set<String>> friends = new HashMap<>();

    public static void main(String[] args) {
        friends.put("Peppa Pig", new HashSet<>(Arrays.asList("Suzy Sheep", "Emily Elephant", "Rebecca Rabbit", "Danny Dog", "Pedro Pony")));
        friends.put("Zoe Zebra", new HashSet<>(Arrays.asList("Freddy Fox", "Rebecca Rabbit", "Gabriella Goat", "Kylie Kangaroo", "Danny Dog")));

        System.out.println(commonFriends("Peppa Pig", "Zoe Zebra"));
        System.out.println(commonFriends("Peppa Pig", "Freddy Fox"));

    private static Set<String> commonFriends(String hero1, String hero2) {
        if (!friends.containsKey(hero1) || !friends.containsKey(hero2)) {
            return Collections.emptySet();

        Set<String> common = new HashSet<>(friends.get(hero1));

        return common;

Additional Resources

Last Updated: 10/2/2019, 2:31:25 PM