Phone book programming - TreeSet or HashSet

Question

I have a nice problem - create a phone book - containing list of contacts. As a phonebook goes,

1. Contacts are to be always sorted.(by name)

2. Can star mark certain contacts, so they have to be above all the rest.(the * contacts are ordered by the time of contact creation)

   class PhoneBook{
   //require an always sorted d.s
   TreeSet<Contact> contacts = new TreeSet<Contact>();
   
   @Override
   public String toString() {
       return "PhoneBook [contacts=" + contacts + "]";
   }
   
   public boolean addContact(Contact contact){
       //validate before adding the contact.
       return contacts.add(contact);
   }
   

   }

   class Contact implements Comparable<Contact>{
   String name;
   int phoneNo;
   Date timeAdded;
   boolean starContact;
   
   
   
   public Contact(String name, int phoneNo, Date timeAdded, boolean starContact) {
       super();
       this.name = name;
       this.phoneNo = phoneNo;
       this.timeAdded = timeAdded;
       this.starContact = starContact;
   }
   
   
   
   @Override
   public int compareTo(Contact otherContact) {
       if(this.starContact && otherContact.starContact){
           return this.timeAdded.before(otherContact.timeAdded)?-1:1; //impossible to add 2 contacts at the same time
       }else if(this.starContact){
           return -1;
       }else if(otherContact.starContact){
           return 1;
       }else{
           //simple Contacts
           return this.name.compareTo(otherContact.name);
       }
   }
   
   
   
   
   
   
   @Override
   public String toString() {
       return "\nContact [name=" + name + ", timeAdded=" + timeAdded
               + ", starContact=" + starContact + "]";
   }
   
   
   }
   

Test Code

    public class MobilePhoneBookDemo {

/**
 * @param args
 */
public static void main(String[] args) {

    PhoneBook phoneBook = new PhoneBook();

    Contact frnd1 = new Contact("Z",56,new Date(),false);
    phoneBook.addContact(frnd1);
    Contact frnd2 = new Contact("A",3,new Date(),false);
    phoneBook.addContact(frnd2);
    Contact frnd3 = new Contact("C",30,new Date(),false);
    phoneBook.addContact(frnd3);
    try {
        TimeUnit.SECONDS.sleep(1);
    } catch (InterruptedException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    Contact ta = new Contact("Ta", 5, new Date(), true);
    phoneBook.addContact(ta);
    Contact ma = new Contact("Ma", 31, new Date(), true);
    phoneBook.addContact(ma);
    Contact baba = new Contact("Baba", 300, new Date(), true);
    phoneBook.addContact(baba);

    //change the priority later for one of my friends.
    System.out.println(phoneBook);
    frnd1.starContact = true;
    System.out.println(phoneBook.contacts.contains(frnd1));

    if(phoneBook.contacts.remove(frnd1)){
        System.out.println("removed");
        phoneBook.contacts.add(frnd1);
    }

    System.out.println(phoneBook);
}

}

Problems faced:

1. The contains doesn't find the entry anymore, what's amiss? I did try and put an equals and a hashcode on Contact, apparently, if there is a Comparator/Comparable present, the compare* is only invoked.
2. Is it fair to use a TreeSet here, or should any other datastructure be used? For eg. HashSet and then convert to a TreeSet?
3. The contains() doesn't even compare for all entries in the map, it just compared against C,Ma and Ta entries. Why was that?

Questions priority according to order. I appreciate all the answers, but this is indeed a complete test case, so please try and run PhoneBook just once before providing an answer. Thanks a lot.

Answer 1

This line:

return this.timeAdded.before(otherContact.timeAdded)?-1:1;

will never return 0 if you compare a contact with itself. Therefore the set will not be able to find the object with contains()

Answer 2

You had me at

contacts are to be always sorted.(by name)

If order is important use TreeSet. Elements are stored in a variety of binary search tree, this means that they are sorted by default.

HashSet on the other hand, does not guarantee any order, sorting or even insert order.

---

Edit - Try to write your conditions in this way, more readable and more robust as well. The order of conditions is the order of priority of comparison.

if(a != b){
    return a<b;
}

if(c != d) {
    return c<d;
}

//and so on.

return 0;

Answer 3

The solution is simple, first remove, then change the value and add it back again. I comment the modification first,I have also put in the modified compareTo() since there were lots of confusion around it.

//frnd1.starContact = true;
    System.out.println("Entry present>"+phoneBook.contacts.contains(frnd1));

    if(phoneBook.contacts.remove(frnd1)){
        System.out.println("removed");
        frnd1.starContact = true;
        phoneBook.contacts.add(frnd1);
    }


@Override
public int compareTo(Contact otherContact) {
    if(otherContact.phoneNo == this.phoneNo){
        return 0;
    }
    if(this.starContact && otherContact.starContact){
        return this.timeAdded.before(otherContact.timeAdded)?-1:1; //impossible to add 2 contacts at the same time
    }else if(this.starContact){
        return -1;
    }else if(otherContact.starContact){
        return 1;
    }else{
        //simple Contacts
        return this.name.compareTo(otherContact.name);
    }
}