[FIXED] Nested streams to a Map Java 11

Issue

Let's say you have an interface, a couple of implementation classes and an Enum like this: Interface:

public interface Employee {
    // No two employees will have same favorite hobbies
    List<Hobby> favoriteHobbies; 
}

Implementations:

public class Employee1 implements Employee {
    @Override
    public List<Hobby> favoriteHobbies {
        return List.of(Hobbies.swimming, Hobbies.dancing);
    }
}

public class Employee2 implements Employee {
    @Override
    public List<Hobby> favoriteHobbies {
        return List.of(Hobbies.running);
    }
}

Enum:

public enum Hobby {
    swimming,
    running,
    dancing
}

I have List<Employee> employees = List.of(Employee1, Employee2);

And using Streams, I want to Stream through employees list and again Stream through each Employee object's favoriteHobbies() and create a Map<Hobby, Employee> like below:

"swimming" -> "Employee1"
"dancing" -> "Employee1"
"running" -> "Employee2"

Solution

I would create a Map.Entry in a Stream and then collect it using its keys and values

Since java-16 this is possible using Stream#mapMulti

Map<Hobby, Employee> map = employees.stream()
        .mapMulti((Employee employee, Consumer<AbstractMap.SimpleEntry<Hobby, Employee>> consumer) -> {
            employee.favoriteHobbies().forEach(hobby -> {
                consumer.accept(new AbstractMap.SimpleEntry<>(hobby, employee));
            });
        })
        .collect(Collectors.toMap(
                AbstractMap.SimpleEntry::getKey,
                AbstractMap.SimpleEntry::getValue,
                (entryOne, entryTwo) -> entryOne // or whatever, maybe even throw new XXXException()
        ));

Before java-16 this is possible with a Stream#flatMap and Collectors#toMap

Map<Hobby, Employee> map = employees.stream()
        .flatMap(employee -> employee.favoriteHobbies()
                .stream()
                .map(hobby -> new AbstractMap.SimpleEntry<>(hobby, employee)))
        .collect(Collectors.toMap(
                AbstractMap.SimpleEntry::getKey,
                AbstractMap.SimpleEntry::getValue,
                (entryOne, entryTwo) -> entryOne // or whatever, maybe even throw new XXXException()
        ));

Which produces

Answered By - Yassin Hajaj
Answer Checked By - Katrina (JavaFixing Volunteer)