Problem Description

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

The transpose of a matrix is obtained by flipping the matrix over its main diagonal. This means that the row and column indices are switched - what was at position (i, j) in the original matrix will be at position (j, i) in the transposed matrix.

For example, if you have a matrix with m rows and n columns, the transposed matrix will have n rows and m columns. Each element at position (i, j) in the original matrix becomes the element at position (j, i) in the transposed matrix.

The visual example shows how the first row [1, 2, 3] becomes the first column in the transposed matrix, the second row [4, 5, 6] becomes the second column, and so on. The main diagonal (elements where row index equals column index) remains in the same position after transposition.

Intuition

When we think about transposing a matrix, we're essentially converting rows into columns and columns into rows. The first row becomes the first column, the second row becomes the second column, and so on.

If we have a matrix with dimensions m x n, after transposition it becomes n x m. The key insight is that the element at position (i, j) in the original matrix needs to move to position (j, i) in the transposed matrix. This is like looking at the matrix from a different angle - we're rotating our perspective by 90 degrees along the main diagonal.

To build the transposed matrix, we can create a new matrix with swapped dimensions and systematically copy each element from its original position to its transposed position. For each row i and column j in the original matrix, we place that element at row j and column i in the new matrix.

In Python, there's an elegant way to achieve this using the zip(*matrix) pattern. The * operator unpacks the rows of the matrix as separate arguments to zip. The zip function then takes the first element from each row to form the first tuple, the second element from each row to form the second tuple, and so on. This effectively groups elements by their column index, which is exactly what we want for transposition - each column becomes a new row in the transposed matrix.

Solution Approach

Following the simulation approach, we need to create a new matrix where each element at position (i, j) in the original matrix is placed at position (j, i) in the transposed matrix.

Let m be the number of rows and n be the number of columns in the input matrix. The transposed matrix will have dimensions n x m.

The implementation uses Python's zip(*matrix) to achieve the transpose:

1. Unpacking the matrix: The *matrix syntax unpacks the rows of the matrix as individual arguments. If matrix = [[1,2,3], [4,5,6]], then *matrix expands to [1,2,3], [4,5,6].

2. Applying zip: The zip() function takes these unpacked rows and pairs up elements at the same index position from each row. It creates tuples where:

   • The first tuple contains all first elements: (1, 4)
   • The second tuple contains all second elements: (2, 5)
   • The third tuple contains all third elements: (3, 6)

3. Converting to list: Since zip() returns an iterator of tuples, we use list() to convert it to a list of lists (though technically it's a list of tuples, which works the same way for our purposes).

This elegantly handles the index swapping for us - what were columns (same index across different rows) become rows in the result. The time complexity is O(m * n) since we need to visit each element once, and the space complexity is also O(m * n) for storing the transposed matrix.

Example Walkthrough

Let's walk through transposing a small matrix step by step:

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

This is a 2×3 matrix (2 rows, 3 columns). After transposition, we expect a 3×2 matrix (3 rows, 2 columns).

Step 1: Unpack the matrix When we use *matrix, it unpacks the rows:

• First row: [1, 2, 3]
• Second row: [4, 5, 6]

Step 2: Apply zip function zip([1, 2, 3], [4, 5, 6]) pairs up elements at the same index:

• Index 0: Takes 1 from first row and 4 from second row → (1, 4)
• Index 1: Takes 2 from first row and 5 from second row → (2, 5)
• Index 2: Takes 3 from first row and 6 from second row → (3, 6)

Step 3: Convert to list The result becomes: [[1, 4], [2, 5], [3, 6]]

Verification: Let's verify the transpose is correct by checking the index mapping:

• Original matrix[0][0] = 1 → Transposed result[0][0] = 1 ✓
• Original matrix[0][1] = 2 → Transposed result[1][0] = 2 ✓
• Original matrix[0][2] = 3 → Transposed result[2][0] = 3 ✓
• Original matrix[1][0] = 4 → Transposed result[0][1] = 4 ✓
• Original matrix[1][1] = 5 → Transposed result[1][1] = 5 ✓
• Original matrix[1][2] = 6 → Transposed result[2][1] = 6 ✓

The first row [1, 2, 3] has become the first column, and the second row [4, 5, 6] has become the second column, which is exactly what transposition should achieve.

Solution Implementation

Time and Space Complexity

The time complexity is O(m × n), where m is the number of rows and n is the number of columns in the matrix. This is because transposing a matrix requires visiting every element exactly once to construct the transposed result, regardless of whether using zip(*matrix) or a traditional nested loop approach.

The space complexity is O(m × n) for the output space needed to store the transposed matrix. However, if we exclude the space required for the output (as mentioned in the reference answer), the auxiliary space complexity is O(1), since the zip(*matrix) operation uses iterators and doesn't create additional intermediate data structures proportional to the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Confusing In-Place Transposition with General Transposition

A common mistake is attempting to transpose a non-square matrix in-place. While square matrices (n×n) can be transposed in-place by swapping elements across the diagonal, rectangular matrices (m×n where m≠n) cannot be transposed in-place because the dimensions change.

Incorrect approach for non-square matrices:

# This ONLY works for square matrices!
def transpose(matrix):
    n = len(matrix)
    for i in range(n):
        for j in range(i + 1, n):
            matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
    return matrix

Solution: Always create a new matrix for the transpose unless you're certain you're dealing with square matrices:

def transpose(matrix):
    rows, cols = len(matrix), len(matrix[0])
    result = [[0] * rows for _ in range(cols)]
    for i in range(rows):
        for j in range(cols):
            result[j][i] = matrix[i][j]
    return result

2. Incorrectly Initializing the Result Matrix

When manually creating the transposed matrix, a frequent error is using the wrong dimensions or sharing references between rows.

Incorrect initialization:

# Wrong dimensions - keeps original dimensions
result = [[0] * cols for _ in range(rows)]  # Should be swapped!

# Shared reference problem
result = [[0] * rows] * cols  # All rows reference the same list!

Solution: Ensure dimensions are swapped (cols × rows becomes the new size) and each row is independently created:

result = [[0] * rows for _ in range(cols)]  # Correct dimensions
# OR use list comprehension that creates new lists
result = [[matrix[i][j] for i in range(rows)] for j in range(cols)]

3. Edge Case Handling with Empty Matrices

Failing to handle empty matrices or matrices with empty rows can cause index errors.

Problematic code:

def transpose(matrix):
    cols = len(matrix[0])  # IndexError if matrix is empty
    # ...

Solution: Check for empty matrices before accessing elements:

def transpose(matrix):
    if not matrix or not matrix[0]:
        return []
    rows, cols = len(matrix), len(matrix[0])
    # ... rest of the implementation