Journey through 2D Arrays in Kotlin: Finding Ideal Positions for Game Pieces

Introduction

Hello, and welcome, code explorer! Today's journey takes us through the intricate paths within a 2-dimensional array, often likened to a game board. Our mission is to identify ideal spots for game piece placement. Sounds like an adventure, doesn't it? Let's embark!

Task Statement

Visualize a chessboard in the form of a 2D array, where each cell could be marked E for empty or P for a piece. Our quest involves creating a Kotlin function named findPositions() . Upon examining this 2D array, this function identifies all the spots where a new piece could be placed so that it can move to another empty cell in one move. The catch is that a piece can move only to an immediately neighboring cell directly above, below, to the left, or right, but not diagonally. Consider this 4x4 board, for instance: P E E P E P E P P E P P P E P E P E E P E P E P P E P P P E P E The function should output the following pairs representing positions: (0, 1), (0, 2), (1, 2), (2, 1), (3, 1) . This output represents the positions where a new piece can fit perfectly and then be able to move in the next turn.

Solution Building: Step 1

Let's step right into action by starting with an empty positions list to help us log the sought positions. Understanding the dimensions of our "board" is key to defining boundaries in our exploration mission. In Kotlin, we determine the size of a 2D array using the size property. Using a list is crucial here because we don't know beforehand how many positions will be found. A MutableList in Kotlin is dynamic and can resize itself to accommodate any number of elements, making it a perfect choice for storing an unknown number of positions. Kotlin fun findPositions(board: Array<CharArray>): MutableList<IntArray> {val positions = mutableListOf<IntArray>() val rows = board.size val cols = board[0].size // Further exploration follows fun findPositions(board: Array<CharArray>): MutableList<IntArray> {val positions = mutableListOf<IntArray>() val rows = board.size val cols = board[0].size // Further exploration follows

Solution Building: Step 2

With our boundary map, we begin our expedition across the board using two nested for loops to do this, traversing the entire board one cell at a time. Kotlin fun findPositions(board: Array<CharArray>): MutableList<IntArray> {val positions = mutableListOf<IntArray>() val rows = board.size val cols = board[0].size for (i in 0 until rows) {for (j in 0 until cols) {// ensuing exploration fun findPositions(board: Array<CharArray>): MutableList<IntArray> {val positions = mutableListOf<IntArray>() val rows = board.size val cols = board[0].size for (i in 0 until rows) {for (j in 0 until cols) {// ensuing exploration

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P E E P
E P E P
P E P P
P E P E

Kotlin

fun findPositions(board: Array<CharArray>): MutableList<IntArray> {
    val positions = mutableListOf<IntArray>()

    val rows = board.size
    val cols = board[0].size
    // Further exploration follows

Kotlin

fun findPositions(board: Array<CharArray>): MutableList<IntArray> {
    val positions = mutableListOf<IntArray>()

    val rows = board.size
    val cols = board[0].size

    for (i in 0 until rows) {
        for (j in 0 until cols) {
            // ensuing exploration

Kotlin

fun findPositions(board: Array<CharArray>): MutableList<IntArray> {
    val positions = mutableListOf<IntArray>()

    val rows = board.size
    val cols = board[0].size

    for (i in 0 until rows) {
        for (j in 0 until cols) {
            if (board[i][j] == 'E') {
                if ((i > 0 && board[i - 1][j] == 'E') ||
                    (i < rows - 1 && board[i + 1][j] == 'E') ||
                    (j > 0 && board[i][j - 1] == 'E') ||
                    (j < cols - 1 && board[i][j + 1] == 'E')) {
                    positions.add(intArrayOf(i, j))
                }
            }
        }
    }
    return positions
}

fun main() {
    val board = arrayOf(
        charArrayOf('P', 'E', 'E', 'P'),
        charArrayOf('E', 'P', 'E', 'P'),
        charArrayOf('P', 'E', 'P', 'P'),
        charArrayOf('P', 'E', 'P', 'E')
    )

    val positions = findPositions(board)

    for (pos in positions) {
        println("(${pos[0]}, ${pos[1]})")
    }
}

// The output will be:
// (0, 1)
// (0, 2)
// (1, 2)
// (2, 1)
// (3, 1)

Kotlin

fun findPositions(board: Array<CharArray>): MutableList<IntArray> {
    val positions = mutableListOf<IntArray>()

    val rows = board.size
    val cols = board[0].size

    for (i in 0 until rows) {
        for (j in 0 until cols) {
            if (board[i][j] == 'E') {
                if ((i > 0 && board[i - 1][j] == 'E') ||
                    (i < rows - 1 && board[i + 1][j] == 'E') ||
                    (j > 0 && board[i][j - 1] == 'E') ||
                    (j < cols - 1 && board[i][j + 1] == 'E')) {
                    positions.add(intArrayOf(i, j))
                }
            }
        }
    }
    return positions
}

fun main() {
    val board = arrayOf(
        charArrayOf('P', 'E', 'E', 'P'),
        charArrayOf('E', 'P', 'E', 'P'),
        charArrayOf('P', 'E', 'P', 'P'),
        charArrayOf('P', 'E', 'P', 'E')
    )

    val positions = findPositions(board)

    for (pos in positions) {
        println("(${pos[0]}, ${pos[1]})")
    }
}

// The output will be:
// (0, 1)
// (0, 2)
// (1, 2)
// (2, 1)
// (3, 1)

Introduction

Task Statement

Solution Building: Step 1

Solution Building: Step 2

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Solution Building: Step 3

Lesson Summary