AP CSA 4.15 Debugging: The Swap That Duplicates

Unit 4: Data Collections · Lesson 4.15 · Debugging

The Swap That Duplicates

Sorting Algorithms. This is not a blank editor: it is someone else's attempt. Find what is wrong, fix it, and submit to be graded against hidden test cases.

Back to the Lesson 4.15 page All Unit 4 lessons

Why debugging is its own skill

Selection sort is correct in outline and wrong in two details, and the output tells you exactly which: values start disappearing and other values appear twice. A swap that forgets to save a value overwrites it, and the array ends up with copies where the original data used to be.

What is wrong with it

The swap assigns in the wrong order and never saves the displaced value, so one element is overwritten and another is duplicated. Use a temporary variable to swap properly.
The search for the smallest remaining element starts at index 0 every pass instead of at the current position, so it keeps reselecting elements that are already sorted. Start the inner scan at i.
The printing loop is correct.

The program reads

A count, then that many integers.

The program should print

The values in increasing order, one per line.

Worked examples

These show what the FIXED program should print. There are more cases you cannot see, and they use different values, so patching around just these numbers will fail.

Example 1 input

5
5 2 9 1 7

Example 1 output

Example 2 input

3
3 2 1

Example 2 output

1
2
3

Your answer

Main.java

import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    int n = input.nextInt();
    int[] data = new int[n];
    for (int i = 0; i < n; i++) {
      data[i] = input.nextInt();
    }

    for (int i = 0; i < data.length - 1; i++) {
      int minIndex = i;
      for (int j = 0; j < data.length; j++) {
        if (data[j] < data[minIndex]) {
          minIndex = j;
        }
      }
      data[i] = data[minIndex];
      data[minIndex] = data[i];
    }

    for (int v : data) {
      System.out.println(v);
    }
  }
}

Input for the Run button

Run sends whatever is in the code box below, bugs and all, so you can see the crash or the wrong answer for yourself before you fix anything.

Stuck?

Hint 1

Trace the two swap lines with data[i] = 5 and data[minIndex] = 2. After the first line both slots hold 2. The second line copies 2 back onto itself. The 5 is gone from the array entirely, which is why duplicates appear.

Hint 2

This is the same swap bug as 2.1, now inside a sort where it is much harder to see. Any swap needs three statements and a temporary, every time.

Hint 3

Starting the inner scan at 0 means the sorted prefix is searched again on every pass. It does not corrupt the data on its own, but it is doing work that cannot help and it hides which region is still unsorted.

Where to go next

Rework the Lesson 4.15 page Try the from-scratch version of this lesson

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