AP CSA 2.9 Debugging: The Maximum That Cannot Be Negative

Unit 2: Selection and Iteration · Lesson 2.9 · Debugging

The Maximum That Cannot Be Negative

Implementing Selection and Iteration 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 2.9 page All Unit 2 lessons

Why debugging is its own skill

Initializing a running maximum to zero works on every test case you are likely to type by hand, and fails the moment every value is negative. This is the classic silent bug: correct-looking code, correct output all week, wrong answer on the one input that matters.

What is wrong with it

Run the program on a list where every value is negative. It reports a maximum of 0, which is not even in the list. Initialize the running maximum from the data itself rather than from 0.
The count of values equal to the maximum is always 1. The counter is being reset inside the loop instead of accumulated. Fix it so a list with three copies of the largest value reports 3.
The minimum is computed correctly. Compare how it is initialized to how the maximum is initialized, and notice that the same fix pattern applies.

The program reads

An integer count of at least 1, then that many integers.

The program should print

Three lines: the largest value, the smallest value, and how many times the largest value appears.

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

4
-5 -2 -9 -7

Example 1 output

-2
-9
1

Example 2 input

5
3 9 4 9 1

Example 2 output

9
1
2

Your answer

Main.java

import java.util.Scanner;

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

    int max = 0;
    for (int i = 0; i < data.length; i++) {
      if (data[i] > max) {
        max = data[i];
      }
    }
    System.out.println(max);

    int min = data[0];
    for (int i = 1; i < data.length; i++) {
      if (data[i] < min) {
        min = data[i];
      }
    }
    System.out.println(min);

    int times = 0;
    for (int i = 0; i < data.length; i++) {
      if (data[i] == max) {
        times = 1;
      }
    }
    System.out.println(times);
  }
}

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

A running maximum initialized to 0 is really claiming that 0 is a member of the list. If every value is negative, no element ever beats 0, and the loop reports a number that was never in the data. Start from data[0] and scan from index 1.

Hint 2

times = 1 assigns; times++ accumulates. Assignment throws away everything counted so far, which is why the answer is always exactly 1 whenever the maximum appears at all.

Hint 3

The minimum loop is the shape you want: initialize from the first element, then start the scan at index 1. Make the maximum match it.

Where to go next

Rework the Lesson 2.9 page Try the from-scratch version of this lesson

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