Transpose matrix problem in C#

General
1

Transposing a matrix involves flipping the matrix over its diagonal, effectively converting its rows into columns and vice versa. This is a common operation in linear algebra and programming tasks. Below, I’ll explain how to transpose a matrix in C# and provide a sample implementation.

Problem Statement

Given a 2D array (matrix) with dimensions m x n, the goal is to create a new matrix that is the transpose of the original matrix.

Steps to Transpose the Matrix

  1. Create a New Matrix: The dimensions of the transposed matrix will be n x m, where m is the number of rows and n is the number of columns in the original matrix.

  2. Fill the Transposed Matrix: Iterate through the original matrix, and for each element at position (i, j), place it in the transposed matrix at position (j, i).

C# Implementation

Here’s how you can implement the transposition of a matrix in C#:

public class MatrixTranspose
{
    public int[][] Transpose(int[][] matrix)
    {
        // Get dimensions of the original matrix
        int m = matrix.Length;          // Number of rows
        int n = matrix[0].Length;      // Number of columns

        // Create a new matrix for the transpose
        int[][] transposedMatrix = new int[n][];
        for (int i = 0; i < n; i++)
        {
            transposedMatrix[i] = new int[m]; // Initialize each row
        }

        // Fill the transposed matrix
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                transposedMatrix[j][i] = matrix[i][j];
            }
        }

        return transposedMatrix;
    }
}

Explanation of the Code

  1. Matrix Declaration: The method Transpose takes a jagged array (int[][] matrix) as an input, representing the original matrix.

  2. Get Dimensions: We determine the number of rows (m) and columns (n) of the original matrix.

  3. Create a New Matrix: We create a new jagged array called transposedMatrix with dimensions n x m. Each row is initialized in a loop.

  4. Fill the Transposed Matrix:

    • We use nested loops to iterate through each element of the original matrix.
    • For each element at (i, j), we place it in the transposed matrix at (j, i).

  5. Return the Result: Finally, we return the newly created transposed matrix.

Example

Given the input matrix:

1  2  3
4  5  6

The transposed matrix will be:

1  4
2  5
3  6

Complexity

  • Time Complexity: O(m * n), where m is the number of rows and n is the number of columns.
  • Space Complexity: O(n * m) for the new transposed matrix.

This implementation effectively transposes the given matrix while maintaining clarity and simplicity.