Hello, budding programmer! Are you ready to embark on a journey into the land of matrices? In this unit, we're in for a thrilling ride into the world of unique matrix traversal. We'll be navigating through the realm of 2D arrays following a distinctive order that oozes intrigue. Stay seated, and let's dive right in!

Suppose we have a matrix where each cell represents a distinct symbol or integer. Our task is to decode this matrix by reading the cells in a particular order.

The decoding begins from the top-left cell of the matrix. We move in a bottom-left downward diagonal direction until we hit the left boundary. Upon hitting the left boundary, we move one cell down (unless we're at the bottom-left corner already, in which case we move one cell to the right) and start moving in an upward diagonal direction towards the upper-right.

While moving diagonally up-right, if we hit the top boundary, we move one cell to the right and start moving in a downward diagonal direction towards the bottom-left. However, if we hit the right boundary while moving diagonally upwards, we move one cell down and start moving in a bottom-left direction. In other words, we keep zigzagging diagonally across the matrix until every cell in the matrix is visited.

Upon completing this zigzag traversal, we will have a list of traversed cell values. Next, we process this list to uncover the indices of the perfect square numbers. The function diagonalTraverseAndSquares(int[][] matrix) should implement this traversal and return a list containing the 1-indexed positions of perfect square numbers in the traversed sequence.

Take a 3x4 matrix for instance:

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1{
2    {1, 2, 3, 4},
3    {5, 6, 7, 8},
4    {9, 10, 11, 12}
5}

Upon completing the diagonal traversal, we'll get the list: {1, 5, 2, 3, 6, 9, 10, 7, 4, 8, 11, 12}. From this list, we see that 1, 9, and 4 are perfect squares and are located at the 1st, 6th, and 9th positions (1-indexed) in the list. Thus, our function returns: {1, 6, 9}.

First, let's put on our Java hats and scrutinize the dimensions of the matrix. To map the landscape of our matrix, we'll use matrix.length to determine the number of rows and matrix[0].length to determine the number of columns. Next, we initialize two ArrayList<Integer>: traversal and results. The traversal list will be responsible for keeping the cell values that we will obtain from the matrix based on our unique diagonal zigzag traversal. The results list will be populated later with the 1-indexed positions of perfect square numbers that can be found in the traversal list.

Java
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1import java.util.ArrayList;
2import java.util.List;
3
4public class DiagonalMatrixTraversal {
5
6    public static List<Integer> diagonalTraverseAndSquares(int[][] matrix) {
7        int rows = matrix.length, cols = matrix[0].length;
8        List<Integer> traversal = new ArrayList<>();
9        List<Integer> results = new ArrayList<>();

The next step involves the actual traversal of the matrix. This process is done diagonally in a zigzag pattern. We begin from the top-left corner (cell [0][0]) and make our journey through the matrix using two variables, row and col, to track the cell indices. We also initialize dir with 1, which dictates that the starting direction is the down-left direction.

However, it's not just a simple left-right or up-down movement; as per the rules, we need to ensure we change our direction whenever we hit an edge. Let dir = -1 dictate the up-right direction. To ensure we continue the correct diagonal movement and don't exceed the matrix boundaries, we use conditional checks within the loop.

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1        int row = 0, col = 0, dir = 1;
2        for (int i = 0; i < rows * cols; ++i) {  // Loop runs for the total number of cells in the matrix.
3            traversal.add(matrix[row][col]);  // Append the current cell value to traversal.
4            
5            // Logic to control direction based on edges:
6            if (dir == 1) {  // Moving down-left
7                if (row == rows - 1) {
8                    col += 1;
9                    dir = -1;
10                } else if (col == 0) {
11                    row += 1;
12                    dir = -1;
13                } else {
14                    row += 1;
15                    col -= 1;
16                }
17            } else {  // Moving up-right
18                if (col == cols - 1) {
19                    row += 1;
20                    dir = 1;
21                } else if (row == 0) {
22                    col += 1;
23                    dir = 1;
24                } else {
25                    row -= 1;
26                    col += 1;
27                }
28            }
29        }

With a completed traversal, we have obtained a list of integers. Next, we evaluate this list to find perfect squares, i.e., numbers that are squares of other integers. For every perfect square we encounter, we add its 1-indexed position in the traversal list to the results list. In Java, we can use the Math.sqrt method. If the integer square root, when squared, equals the initial number, we know the initial number is a perfect square, so we add its position to our results.

Java
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1        for (int idx = 0; idx < traversal.size(); ++idx) {
2            int val = traversal.get(idx);
3            int root = (int) Math.sqrt(val);
4            if (root * root == val) {  // Check if the value is a perfect square number.
5                results.add(idx + 1);
6            }
7        }
8
9        return results;
10    }
11}

Bravo! You've successfully navigated a challenging task involving a unique matrix traversal pattern. You've demonstrated solid skills in Java programming, specifically list manipulation, and brilliantly tackled the challenges of moving around two-dimensional arrays.

Now, it's time to embark on your journey and put your newly learned skills to the test! Try out more complex matrices and different values to truly master the concept. Keep experimenting, and you'll soon become a wizard at matrix traversals. Happy coding!