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Transpose Matrix

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Given a 2D integer array matrix, return the transpose of matrix.

The transpose of a matrix is the matrix flipped over its main diagonal, switching the matrix's row and column indices.

For example:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[1,4,7],[2,5,8],[3,6,9]]

In this example, the original matrix is a 3x3 matrix. The transpose is created by swapping rows and columns. So, the element at matrix[0][1] which is '2' becomes the element at transposed_matrix[1][0].

Here's another example with a non-square matrix:

Input: matrix = [[1,2,3],[4,5,6]]
Output: [[1,4],[2,5],[3,6]]

In this case, the input is a 2x3 matrix, and the output is a 3x2 matrix. Again, the rows and columns have been swapped.

Your task is to write a function that takes a 2D integer array (a list of lists) as input and returns its transpose.

Consider these constraints:

  • 1 <= m, n <= 1000 where 'm' is the number of rows and 'n' is the number of columns.
  • 1 <= m * n <= 10^5
  • -10^9 <= matrix[i][j] <= 10^9

Can you implement an efficient solution for this problem, considering both time and space complexity?

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What is the expected data type of the matrix elements (e.g., integers, floats)?
  2. Is the input matrix guaranteed to be rectangular (i.e., all rows have the same number of columns)?
  3. What should I return if the input matrix is empty or null?
  4. Can I assume the matrix is a valid 2D array, or should I handle potential exceptions like jagged arrays?
  5. Should the transposition be done in-place, or should I return a new matrix?

Brute Force Solution

Approach

The goal is to swap rows and columns in a table. The brute force way to do this is to create a new empty table, then systematically copy elements from the original table into their new, transposed positions in the new table.

Here's how the algorithm would work step-by-step:

  1. First, create a completely new, empty table. This new table will hold the transposed result.
  2. Go through the original table, one element at a time, starting from the top left.
  3. For each element you find in the original table, determine its new position in the transposed table.
  4. Take the element and put it into its calculated position in the new, empty table you created earlier.
  5. Repeat this process for every single element in the original table. This ensures that you've considered them all.
  6. Once you've gone through every element and copied it into the new table in its transposed position, the new table represents the transposed matrix, and you're done.

Code Implementation

def transpose_matrix_brute_force(matrix):
    number_of_rows = len(matrix)
    number_of_columns = len(matrix[0])

    # Initialize the transposed matrix with swapped dimensions
    transposed_matrix = [[0 for i in range(number_of_rows)] for j in range(number_of_columns)]

    # Iterate through the original matrix
    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):

            # Swap row and column indices to transpose
            transposed_matrix[column_index][row_index] = matrix[row_index][column_index]

    return transposed_matrix

Big(O) Analysis

Time Complexity
O(m*n)The algorithm iterates through each element of the original matrix to create the transposed matrix. The dimensions of the original matrix are assumed to be m rows and n columns. Therefore, the algorithm accesses and copies each element once to the new transposed matrix, resulting in a total of m*n operations. Thus, the time complexity is O(m*n). If m=n, then it will simplify to O(n^2).
Space Complexity
O(N*M)The algorithm creates a new table (matrix) to store the transposed result. The size of this new table will be the number of columns of the original matrix multiplied by the number of rows of the original matrix. If the original matrix has dimensions N x M (N rows and M columns), the transposed matrix will have dimensions M x N, thus requiring space proportional to N*M. Therefore, the auxiliary space used is proportional to the size of the transposed matrix itself. This results in a space complexity of O(N*M).

Optimal Solution

Approach

The goal is to swap the rows and columns of a table. We can do this efficiently by creating a new table where the original rows become columns and vice versa. A key consideration is handling tables that aren't square, meaning the number of rows and columns are different.

Here's how the algorithm would work step-by-step:

  1. First, figure out the dimensions of the new table: the number of rows will be equal to the original number of columns, and the number of columns will be the original number of rows.
  2. Create a brand new, empty table with these new dimensions.
  3. Now, systematically copy each value from the original table to the new table, but swap the row and column positions. For example, the value at row 1, column 2 in the original table goes to row 2, column 1 in the new table.
  4. Continue copying all values until the new table is completely filled. The new table is the transposed version of the original.

Code Implementation

def transpose_matrix(matrix):
    number_of_rows = len(matrix)
    number_of_columns = len(matrix[0])

    # Dimensions are swapped for the transposed matrix.
    transposed_number_of_rows = number_of_columns
    transposed_number_of_columns = number_of_rows

    # Initialize the transposed matrix with swapped dimensions.
    transposed_matrix = [([0] * transposed_number_of_columns) for _ in range(transposed_number_of_rows)]

    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):
            # Copy each element, swapping row and column indices.
            transposed_matrix[column_index][row_index] = matrix[row_index][column_index]

    return transposed_matrix

Big(O) Analysis

Time Complexity
O(m*n)The algorithm iterates through each element of the original matrix to copy it to the transposed matrix. Let 'm' be the number of rows and 'n' be the number of columns in the original matrix. The dominant operation is accessing each element of the original matrix once, and placing it into the new matrix. Therefore, the time complexity is proportional to the number of elements in the original matrix, which is m * n, resulting in a time complexity of O(m*n).
Space Complexity
O(M*N)The algorithm creates a new table to store the transposed matrix. The dimensions of this new table are determined by the original matrix's dimensions: if the original matrix has M rows and N columns, the new table will have N rows and M columns. Therefore, the auxiliary space required is proportional to the product of M and N, representing the total number of elements in the transposed matrix. Thus, the space complexity is O(M*N), where M is the number of rows and N is the number of columns of the original matrix.

Edge Cases

CaseHow to Handle
Null or undefined matrix inputReturn an empty matrix or throw an IllegalArgumentException to signal invalid input.
Empty matrix (0 rows or 0 columns)Return an empty matrix or the original matrix if an in-place operation is expected and the input is already empty.
Non-rectangular matrix (rows with different lengths)Throw an IllegalArgumentException as a non-rectangular matrix cannot be transposed or pad the shorter rows with a default value.
1x1 matrix (single element)Return the original matrix as the transpose will be the same.
Very large matrix exceeding memory limitationsConsider an out-of-memory error and explore block-wise transposition or external memory algorithms if possible.
Matrix with integer overflow potential during index calculationsEnsure indices are within acceptable ranges and consider using larger integer types if necessary to prevent overflow during calculations.
Matrix with negative numbers or extreme valuesThe transposition algorithm itself is generally not affected by negative numbers or extreme values, but consider potential overflow if these values are used in further calculations after transposition.
Matrix containing floating point numbers with precision issuesWhile the transposition itself doesn't cause new precision problems, subsequent operations involving these transposed floats may require consideration for potential precision errors.