AP CSA 4.1 FRQ Practice: Aggregate, Do Not Identify

Unit 4: Data Collections · Lesson 4.1 · FRQ Practice

Aggregate, Do Not Identify

A free response question in the shape the exam uses: a stated contract, four rubric parts, and no main method handed to you. Worth 4 points.

Back to the Lesson 4.1 pageAll Unit 4 lessons

Why this question is worth four points

The ethical rule in this topic has a concrete programming shape: report the aggregate, not the individual. A count that can only be satisfied by one person is not anonymous, and this question asks you to detect exactly that, which is what a real suppression rule does.

What you are given

The driver builds an int array of counts, one per group, and passes it to your methods. Write a class named Report. Do not write a main method.

The question

  1. (a) public static int total(int[] counts) returns the sum of every count.
  2. (b) public static int groups(int[] counts) returns how many groups there are.
  3. (c) public static boolean isIdentifying(int[] counts, int minimum) returns whether ANY group has a count below minimum but above 0.
  4. (d) public static int suppressed(int[] counts, int minimum) returns the total of only the groups that are at or above minimum.

What the reader is looking for

  1. Write class Report with the four static methods described. No main method.
  2. A group with a count of 0 is not identifying, because nobody is in it. Only a small NON EMPTY group is.
  3. Part (d) leaves out the small groups entirely rather than replacing them with zero after the fact.

Worked examples

These show what a correct answer prints. There are more cases you cannot see, and they use different values, so an answer built around just these numbers will fail.

Example 1 input
4
10 20 2 30
5
Example 1 output
62
4
true
60
Example 2 input
3
5 40 50
5
Example 2 output
95
3
false
95

Your answer

Main.java

Input for the Run button

How this is scored: your answer runs against every test case, and the fraction it passes becomes your score out of 4. That is not how a human AP reader marks a rubric, so treat the score as a check on whether your code works, and the rubric above as the thing you are actually practising.


  

  
Stuck?

    
Hint 1
An enhanced for loop reads every element and is the right tool whenever you do not need the index. Part (a), (c) and (d) never need one.

    
Hint 2
Part (c) has two conditions joined by and: above zero AND below the minimum. Dropping the first makes every empty group look identifying.

    
Hint 3
Part (d) is a filtered sum. The groups below the minimum contribute nothing rather than contributing zero after being counted.



  
Before you submit

  
3 mistake(s) that lose points on this question
    
Each of these is a real error the grader catches. Check your answer against them before you submit, not instead of trying.

    
    
      
  • part (c) treats an empty group as identifying
    • 
      
  • part (d) includes the groups it was supposed to suppress
    • 
      
  • part (d) excludes a group sitting exactly on the minimum
    • 
    

  


  
Where to go next

  

    Rework the Lesson 4.1 page
    Try the from-scratch exercise for this lesson
    Try the debugging exercise for this lesson
  


  
  






  

















  
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