AP CSA 4.15 FRQ Practice: Selection and Insertion

Unit 4: Data Collections · Lesson 4.15 · FRQ Practice

Selection and Insertion

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.

Why this question is worth four points

A swap needs a temporary. Writing a = b; b = a; loses one of the values before the second line runs, and the array ends up with a duplicate rather than a swap. That is the same bug as the one in 2.1, now inside a sort where it is much harder to see.

What you are given

The driver builds an int array and passes it to your methods. Write a class named Sorter. Do not write a main method.

The question

  1. (a) public static void swap(int[] data, int i, int j) exchanges the two elements.
  2. (b) public static int indexOfMinFrom(int[] data, int start) returns the index of the smallest element at or after start.
  3. (c) public static void selectionSort(int[] data) sorts the array ascending, using parts (a) and (b).
  4. (d) public static String show(int[] data) returns the elements separated by single spaces.

What the reader is looking for

  1. Write class Sorter with the four static methods described. No main method.
  2. Part (a) needs a temporary variable. Without one the two elements both end up holding the same value.
  3. The array is sorted IN PLACE, so the caller sees the change.

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
5
5 3 9 1 7
0 4
Example 1 output
3
7 3 9 1 5
1 3 5 7 9
Example 2 input
4
8 2 6 4
2 2
Example 2 output
1
8 2 6 4
2 4 6 8

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

The driver swaps the same pair TWICE, which should put the array back exactly as it was. A swap without a temporary fails that immediately.

Hint 2

Selection sort only needs to run to the second to last element, because once everything before it is placed the last one has nowhere else to go.

Hint 3

indexOfMinFrom seeds at start rather than at 0, which is what makes each pass ignore the part already sorted.

Before you submit

4 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 (a) swaps without a temporary, so both elements end up equal
  • part (b) seeds at index 0 rather than at start
  • part (c) sorts descending by taking the maximum instead
  • part (c) stops one pass too early

Where to go next

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