Unique Matrix Traversal in Kotlin

Introduction

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!

Task Statement

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 fun diagonalTraverseAndSquares(matrix: Array<IntArray>): List<Int> should implement this traversal and return a list containing the 0-indexed positions of perfect square numbers in the traversed sequence. Take a 3x4 matrix, for instance: Kotlin val matrix = arrayOf(intArrayOf(1, 2, 3, 4), intArrayOf(5, 6, 7, 8), intArrayOf(9, 10, 11, 12)) 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 7th positions in the list. Thus, our function returns: {0, 5, 8} .

Solution Building: Step 1

First, let's put on our Kotlin hats and scrutinize the dimensions of the matrix. To map the landscape of our matrix, we'll use matrix.size to determine the number of rows and matrix[0].size to determine the number of columns. Next, we initialize two MutableList<Int> : 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 0-indexed positions of perfect square numbers that can be found in the traversal list. Kotlin import kotlin.math.sqrt fun diagonalTraverseAndSquares(matrix: Array<IntArray>): List<Int> {val rows = matrix.size val cols = matrix[0].size val traversal = mutableListOf<Int>() val results = mutableListOf<Int>()

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Solution Building: Step 2

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.

Kotlin

    var row = 0
    var col = 0
    var dir = 1
    repeat(rows * cols) { // Loop runs for the total number of cells in the matrix.
        traversal.add(matrix[row][col]) // Append the current cell value to traversal.

        // Logic to control direction based on edges:
        if (dir == 1) {  // Moving down-left
            when {
                row == rows - 1 -> {
                    col += 1
                    dir = -1
                }
                col == 0 -> {
                    row += 1
                    dir = -1
                }
                else -> {
                    row += 1
                    col -= 1
                }
            }
        } else {  // Moving up-right
            when {
                col == cols - 1 -> {
                    row += 1
                    dir = 1
                }
                row == 0 -> {
                    col += 1
                    dir = 1
                }
                else -> {
                    row -= 1
                    col += 1
                }
            }
        }
    }

    var row = 0
    var col = 0
    var dir = 1
    repeat(rows * cols) { // Loop runs for the total number of cells in the matrix.
        traversal.add(matrix[row][col]) // Append the current cell value to traversal.

        // Logic to control direction based on edges:
        if (dir == 1) {  // Moving down-left
            when {
                row == rows - 1 -> {
                    col += 1
                    dir = -1
                }
                col == 0 -> {
                    row += 1
                    dir = -1
                }
                else -> {
                    row += 1
                    col -= 1
                }
            }
        } else {  // Moving up-right
            when {
                col == cols - 1 -> {
                    row += 1
                    dir = 1
                }
                row == 0 -> {
                    col += 1
                    dir = 1
                }
                else -> {
                    row -= 1
                    col += 1
                }
            }
        }
    }

Introduction

Task Statement

Solution Building: Step 1

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Solution Building: Step 2

Solution Building: Step 3

Lesson Summary