OCP Java SE6 Programmer Practice Exams

[John Paterson]

Hi Folks,

Does anyone have any idea where I can get the solutions for the 'Coding Exercises' found in the book

OCP Java SE6 Programmer Practice Exams

by Bert Bates and Kathy Sierra?

Thanks.

regards
John

[saloni jhanwar]

John Paterson wrote:Hi Folks,

Does anyone have any idea where I can get the solutions for the 'Coding Exercises' found in the book

OCP Java SE6 Programmer Practice Exams

by Bert Bates and Kathy Sierra?

Thanks.

regards
John

You can post here particular exercise, if you find any doubt.Even in K&B study guide you wont find solution of exercise because that is an exercise not a mock question.

[John Paterson]

Hi saloni jhanwar

Noted. Thanks!

regards
John

[John Paterson]

Following is the question from a certain section and my attempts at one of it's sub-sections. I will post my solutions here. I hope someone can have a look and say if I am indeed answering the question in the correct way, not just compiler and run time error, but also the approach. Thanks everyone.

Section6: Collections/Generics

6.a Create an Items class that has a String name, and an int value. Note that you will use this class in several of this section's exercises, and you may need to enhance this class as you go along. For all of 6.a, use classes in java.util whenever possible

i) Write a class that makes a collection of Items that guarantees no duplicates.

My Solution:

public class Items {

    public String name;
    public int value;
   
    public static void main(String[] args) {
        
         Set s  = new HashSet();
         boolean [] ba = new boolean[5];
         
         ba[0] = s.add("a");
         ba[1] = s.add("b");
         ba[2] = s.add("c");
         
         
    }
}

ii) Write a class that makes a collection of Items that can be accessed in the order in which they were added to the collection.

public class Items {

    int value;
    String name;

    public static void main(String[] args) {

        Set s = new LinkedHashSet();
        boolean[] ba = new boolean[5];

	        
        ba[0] = s.add("a");
        ba[1] = s.add("b");
        ba[2] = s.add("c");
        ba[3] = s.add("d");       
        
        System.out.println(s);

	}
}

[Jeanne Boyarsky]

John Paterson wrote: Write a class that makes a collection of Items that guarantees no duplicates.

Two problems.

You have a collection of strings not Items

If you did create a set of Items it could contain duplicates. How does Java tell if two items in a set are duplicates? What does your Item class lack in that space?

[John Paterson]

Hi Jeanne Boyarsky,

Thanks for the reply. I took note of what you mentioned. Working on it now, will post once done. Thanks.

regards
John

[John Paterson]

Hi all,

My corrected code for Question 6 a)

i) Write a class that makes a collection of Items that guarantees no duplicates.

import java.util.*;

public class Items {

    int value;
    String name;

    Items(int value) {
        this.value = value;
    }

    
    public String toString()
    {
        this.name = Integer.toString(value);
        return name;
    }
     
    
    //Overriding the equals method
    public boolean equals(Object o) {
        Items it = (Items) o;
        if (it.value == this.value) {
            return true;
        } else {
            return false;
        }
    }
    
    //Overriding the hashCode method
    public int hashCode()
    {
        
        return value * 5;
    }
    

    public static void main(String[] args) {
        boolean[] ba = new boolean[5];


        //Write a class that makes a collection of Items that guarantees no duplicates.   
        Set s = new HashSet();
        ba[0] = s.add(new Items(1));
        ba[1] = s.add(new Items(2));
        ba[2] = s.add(new Items(3));
        ba[3] = s.add(new Items(4));

        System.out.println(s);

Though not part of the question, I overrode the toString method to produce some redable output to verify the code. To confirm that the code is properly handling duplicates, I changed ...

s.add(new Items(3)); 

to

s.add(new Items(4)); 

and the output was [4,1,2]. If not the output will be [4, 1, 2, 3]. I feel that I may have answered the question correctly. Any comments from anyone will be much appreciated. Thanks.

regards
John

[John Paterson]

Hi all,

My corrected code for Question 6 a)

ii) The only change from question i) is the following block:

Set s = new LinkedHashSet();
        
        ba[0] = s.add(new Items(1));
        ba[1] = s.add(new Items(2));
        ba[2] = s.add(new Items(3));
        ba[3] = s.add(new Items(4));       
        
        System.out.println(s);

The output follows the same order as the above code block, so I guess that fulfills the requirement of 'can be accessed in the order in which they were added to the collection' part.

[John Paterson]

Hi all,

My corrected code for Question 6 a)

ii) The only change from question i) is the following block:

Set s = new LinkedHashSet();
        
        ba[0] = s.add(new Items(1));
        ba[1] = s.add(new Items(2));
        ba[2] = s.add(new Items(3));
        ba[3] = s.add(new Items(4));       
        
        System.out.println(s);

The output follows the same order as the above code block, so I guess that fulfills the requirement of 'can be accessed in the order in which they were added to the collection' part.