AP CSA 4.13 Debugging: The Diagonal That Is Not One

Unit 4: Data Collections · Lesson 4.13 · Debugging

The Diagonal That Is Not One

Implementing 2D Array 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 4.13 page All Unit 4 lessons

Why debugging is its own skill

A diagonal walk needs one index that moves both ways at once, not two independent loops. Written as a nested pair it visits the whole grid, and the total that comes out is a grand total wearing a diagonal label.

What is wrong with it

The diagonal sum uses a nested loop over every row and column, so it adds every element in the grid. A diagonal visits one element per row, at the position where the row and column indexes match. Rewrite it with a single loop.
The maximum starts at 0, so a grid of entirely negative numbers reports a maximum that is not in the grid. Initialize it from the first element.
The column count is already read correctly. The grid is guaranteed square for the diagonal.

The program reads

A size n, then n times n integers in row-major order, forming a square grid.

The program should print

Three lines: the sum of the main diagonal, the largest value anywhere in the grid, and the total of every element.

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

3
1 2 3 4 5 6 7 8 9

Example 1 output

15
9
45

Example 2 input

2
-5 -2 -9 -1

Example 2 output

-6
-1
-17

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[][] grid = new int[n][n];
    for (int r = 0; r < n; r++) {
      for (int c = 0; c < n; c++) {
        grid[r][c] = input.nextInt();
      }
    }

    int diagonal = 0;
    for (int r = 0; r < n; r++) {
      for (int c = 0; c < n; c++) {
        diagonal = diagonal + grid[r][c];
      }
    }
    System.out.println(diagonal);

    int max = 0;
    for (int r = 0; r < n; r++) {
      for (int c = 0; c < n; c++) {
        if (grid[r][c] > max) {
          max = grid[r][c];
        }
      }
    }
    System.out.println(max);

    int total = 0;
    for (int r = 0; r < n; r++) {
      for (int c = 0; c < n; c++) {
        total = total + grid[r][c];
      }
    }
    System.out.println(total);
  }
}

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

The main diagonal is every position where the row index equals the column index: [0][0], [1][1], [2][2]. That is n elements, so it needs n steps, so it needs one loop.

Hint 2

The starter's diagonal and total print the same number on every input. Two different quantities agreeing exactly is a strong signal that they are running the same code.

Hint 3

A maximum seeded with 0 assumes 0 belongs to the data. Seed from grid[0][0], which is guaranteed to be a real element.

Where to go next

Rework the Lesson 4.13 page Try the from-scratch version of this lesson

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