Free Response - Horse Barn B

The following is a variation of part b of a free response question from 2012. It was question 3 on the exam. You can see all the free response questions from past exams at https://apstudents.collegeboard.org/courses/ap-computer-science-a/free-response-questions-by-year.

The original question had an interface called Horse, but the problem below has been changed to a class Horse instead of the interface. Interfaces are no longer covered on the AP CS A exam. However, you can still solve problems that have interfaces in them by changing them to a class, since an interface just describes the methods that a class must have.

Question 3. Consider a software system that models a horse barn.

public class Horse
{
   /** @return the horse's name */
   public String getName()
   { implementation not shown }

   /** @return the horse's weight */
   public int getWeight()
   { implementation not shown }

   // There may be other methods that are not shown

}

A horse barn consists of N numbered spaces. Each space can hold at most one horse. The spaces are indexed starting from 0; the index of the last space is N - 1. No two horses in the barn have the same name. The declaration of the HorseBarn class is shown below.

public class HorseBarn
{
   /** The spaces in the barn. Each array element holds a reference to the horse
    * that is currently occupying the space. A null value indicates an empty
    * space.
    */
   private Horse[] spaces;

   /** Consolidates the barn by moving horses so that the horses are in
    *  adjacent spaces, starting at index 0, with no empty space between
    *  any two horses.
    * Postcondition: The order of the horses is the same as before the
    *  consolidation.
    */
   public void consolidate()
   { /* to be implemented in part (b) */ }
}

Part b. Write the HorseBarn method consolidate. This method consolidates the barn by moving horses so that the horses are in adjacent spaces, starting at index 0, with no empty spaces between any two horses. After the barn is consolidated, the horses are in the same order as they were before the consolidation.

Figure 1: Example calls and results

How to solve this problem

One way to solve this problem is to create a temporary array the same size as spaces and then loop through the current spaces array and if the current element isn’t null copy it to the temporary array. What kind of loop should you use? A for loop or an enhanced for loop would work for this problem. You will need an index for at least the temporary array.

While we are looping through the spaces array, we need to check for non-null positions.

7-7-1: How do we check if the space at the current index isn’t null?

• if (spaces.get(index) != null)
• This is the syntax for checking an element within an ArrayList.
• if (!spaces[index].null())
• Is null() a standard Java method? Comparing an object with a null value is simpler.
• if (spaces[index] != null)
• "!=" is the best way to compare an element with a null value.

Try to write the code for the method consolidate in the HorseBarn class. When you are ready click “Run” to test your solution.

class Horse { private String name; private int weight; public Horse(String theName, int theWeight) { this.name = theName; this.weight = theWeight; } public String getName() { return this.name;} public int getWeight() { return this.weight; } public String toString() { return "name: " + this.name + " weight: " + this.weight; } } public class HorseBarn { private Horse[] spaces; /** Constructor that takes the number of stalls * @param numStalls - the number of stalls in the barn */ public HorseBarn(int numStalls) { spaces = new Horse[numStalls]; } /** Consolidates the barn by moving horses so that the horses are * in adjacent spaces, starting at index 0, with no empty space * between any two horses. * Postcondition: The order of the horses is the same as before * the consolidation. */ public void consolidate() { } public String toString() { String result = ""; Horse h = null; for (int i = 0; i < spaces.length; i++) { h = spaces[i]; result = result + "space " + i + " has "; if (h == null) result = result + " null \n"; else result = result + h.toString() + "\n"; } return result; } public static void main (String[] args) { HorseBarn barn = new HorseBarn(7); barn.spaces[0] = new Horse("Trigger", 1340); barn.spaces[2] = new Horse("Silver",1210); barn.spaces[5] = new Horse("Patches", 1350); barn.spaces[6] = new Horse("Duke", 1410); System.out.println("before consolidate"); System.out.println(barn); barn.consolidate(); System.out.println("after consolidate"); System.out.println(barn); } }

Video - One way to code the solution

The following video is also on YouTube at https://youtu.be/3HytvgdLCNI. It walks through coding a solution.

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