Lesson 3
Transposing Matrices with Kotlin
Introduction

Hello there! Are you ready to enhance your programming skills with an exciting exercise using Kotlin? In this unit, we are delving into the world of matrices. Specifically, we'll be transposing a given matrix. Let's dive into this matrix manipulation game without delay!

Task Statement

Let's dive into the task at hand. Your objective is to write a Kotlin function named transformMatrix(). This function will accept a 2D Int array (which represents a matrix) containing integers as input. Your responsibility is to return another 2D Int array, which is the transposed version of the given matrix.

Remember, transposing a matrix involves switching its rows and columns. In other words, all the rows of the original matrix should convert into columns in the transposed matrix, and vice versa.

For instance, if the original matrix (input 2D array) is:

Kotlin
1val matrix = arrayOf( 2 intArrayOf(1, 2, 3), 3 intArrayOf(4, 5, 6) 4)

Then the transposed matrix (output 2D array) will be:

Kotlin
1val transposed = arrayOf( 2 intArrayOf(1, 4), 3 intArrayOf(2, 5), 4 intArrayOf(3, 6) 5)

It is crucial for your result to maintain the integrity of the data type in the original matrix. The values in the input matrix are integers, and they should remain integers in the output matrix as well.

Solution Building: Step 1

The initial step in building our solution involves determining the dimensions of the matrix. We need to know the number of rows and columns present in it. In Kotlin, the size properties can provide this information. The number of rows corresponds to the size of the outer array, and the number of columns matches the size of any inner array.

Kotlin
1fun transformMatrix(matrix: Array<IntArray>): Array<IntArray> { 2 val rows = matrix.size 3 val cols = if (rows > 0) matrix[0].size else 0 4 // If rows > 0, the above line sets cols to the size of the first row, else sets to 0 5}
Solution Building: Step 2

The next step is to create a "placeholder" for the transposed matrix in alignment with its required dimensions. This will be a new 2D array, but with the number of rows and columns swapped. Initially, we can populate this matrix with all zeros.

Kotlin
1fun transformMatrix(matrix: Array<IntArray>): Array<IntArray> { 2 val rows = matrix.size 3 val cols = if (rows > 0) matrix[0].size else 0 4 val result = Array(cols) { IntArray(rows) } 5}
Solution Building: Step 3

It's time to get to the heart of the matter — transposing the matrix. For every element in the original matrix, we want to move it from the ith row and jth column to the jth row and ith column of the transposed matrix. A straightforward nested for-loop can efficiently execute this swap for all elements of the matrix.

Kotlin
1fun transformMatrix(matrix: Array<IntArray>): Array<IntArray> { 2 val rows = matrix.size 3 val cols = if (rows > 0) matrix[0].size else 0 4 val result = Array(cols) { IntArray(rows) } 5 6 for (i in 0 until rows) { 7 for (j in 0 until cols) { 8 result[j][i] = matrix[i][j] 9 } 10 } 11 return result 12} 13 14fun main() { 15 // Initial matrix (2D array) 16 val matrix = arrayOf( 17 intArrayOf(1, 2, 3), 18 intArrayOf(4, 5, 6) 19 ) 20 21 // Call our function on the matrix and output the result 22 val transposed = transformMatrix(matrix) 23 for (row in transposed) { 24 for (elem in row) { 25 print("$elem ") 26 } 27 println() 28 } 29 // Expected output: 30 // 1 4 31 // 2 5 32 // 3 6 33}

This step concludes our solution!

Lesson Summary

With this, we've brought this lesson to a close! Congratulations on successfully implementing a function that can transpose matrices! This task isn't simple, but by accomplishing it, you've demonstrated your understanding of and proficiency in Kotlin multidimensional arrays, loops, and the concept of matrix transposition.

Your hard work continues, though. Now that you have acquired this valuable skill, it's time to reinforce it with more practice. During the next session, you will encounter practice problems that build on this concept. So, get ready and start coding!

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