DogWalker FRQ Student Handout

1. Understand the Instance Variables

private int maxDogs;
private DogWalkCompany company;
      

These variables mean:

• maxDogs = the maximum number of dogs this walker can walk in one hour
• company = the company that tracks how many dogs still need to be walked

The company gives us two helpful methods:

• company.numAvailableDogs(hour) → tells how many dogs are available that hour
• company.updateDogs(hour, dogsWalked) → updates the company after dogs are taken

2. Part A - walkDogs(int hour)

What is this method supposed to do?

This method should figure out how many dogs the walker can walk during one specific hour.

Steps to solve Part A

1. Get the number of dogs available at that hour.
2. Take the smaller of available dogs and maxDogs.
3. Tell the company those dogs have been taken.
4. Return the number of dogs walked.

Example thinking

Code for Part A

public int walkDogs(int hour)
{
    int available = company.numAvailableDogs(hour);
    int dogsWalked = Math.min(available, maxDogs);
    company.updateDogs(hour, dogsWalked);
    return dogsWalked;
}
      

Why this works

• available stores the number of dogs ready to be walked.
• Math.min(available, maxDogs) chooses the smaller number.
• updateDogs makes sure those dogs are no longer counted as available.
• The method returns the number actually walked.

3. Common Mistakes in Part A

• Forgetting to call company.updateDogs(hour, dogsWalked)
• Returning available instead of the smaller value
• Ignoring the maxDogs limit

4. Part B - dogWalkShift(int startHour, int endHour)

What is this method supposed to do?

This method calculates how much money the walker earns during an entire shift.

For each hour in the shift:

• Call walkDogs(hour)
• Earn $5 per dog
• Earn a $3 bonus if:
  • the walker walked maxDogs dogs, or
  • the hour is between 9 and 17 inclusive

Steps to solve Part B

1. Create a variable to hold the total money earned.
2. Loop from startHour to endHour, inclusive.
3. For each hour, call walkDogs(hour).
4. Add dogsWalked * 5 to the total.
5. Check if the bonus should be added.
6. Return the total money earned.

Code for Part B

public int dogWalkShift(int startHour, int endHour)
{
    int total = 0;

    for (int hour = startHour; hour <= endHour; hour++)
    {
        int dogsWalked = walkDogs(hour);
        total += dogsWalked * 5;

        if (dogsWalked == maxDogs || (hour >= 9 && hour <= 17))
        {
            total += 3;
        }
    }

    return total;
}
      

Why this works

• The loop visits every hour in the shift.
• walkDogs(hour) handles the dog-counting logic from Part A.
• Each dog earns $5.
• The bonus is added once when at least one bonus condition is true.

5. Common Mistakes in Part B

• Using < instead of <= in the loop
• Forgetting to call walkDogs(hour)
• Adding the $3 bonus twice in one hour
• Using the wrong peak hour condition

Wrong bonus example

if (dogsWalked == maxDogs)
    total += 3;
if (hour >= 9 && hour <= 17)
    total += 3;
      

This is wrong because the prompt only gives one $3 bonus per hour, not two.

6. Full Correct Solution

public int walkDogs(int hour)
{
    int available = company.numAvailableDogs(hour);
    int dogsWalked = Math.min(available, maxDogs);
    company.updateDogs(hour, dogsWalked);
    return dogsWalked;
}

public int dogWalkShift(int startHour, int endHour)
{
    int total = 0;

    for (int hour = startHour; hour <= endHour; hour++)
    {
        int dogsWalked = walkDogs(hour);
        total += dogsWalked * 5;

        if (dogsWalked == maxDogs || (hour >= 9 && hour <= 17))
        {
            total += 3;
        }
    }

    return total;
}
      

7. Easy Memory Trick