867. Transpose Matrix
Problem Description
The problem requires us to transpose a given 2D integer array, matrix
. Transposing a matrix involves flipping the matrix over its main diagonal. This process converts rows to columns and vice versa, which leads to the interchange of the matrix's row and column indexes.
For example, let's consider a matrix as follows:
1 2 3 4 5 6
The main diagonal of this matrix is the set of elements that extend from the top left to the bottom right (elements 1
and 5
in this case).
When we transpose the matrix, the rows become columns, and the columns become rows. The transposed matrix will look like this:
1 4 2 5 3 6
The element that was originally at the second row, first column (4
), is now at the first row, second column. The solution requires us to perform this operation on any given matrix and return the new transposed matrix.
Intuition
For the given solution, Python's built-in functions simplify the process of transposing a matrix. Here is the intuition behind the used approach:
-
The
*
operator, when used in the context of function argument unpacking, will unpack the argument list. For a 2D matrix, this effectively unpacks the rows of the matrix, making them available as individual arguments. -
The
zip
function takes iterables (can be zero or more), aggregates them in a tuple, and returns it. When used with a 2D matrix unpacked into rows,zip
essentially combines the elements of the rows that have the same index, thus forming the columns of the transposed matrix. -
Finally, the
list
function converts the resulting tuples back into lists, as required for the solution. In Python, thezip
function returns an iterator of tuples. To match the expected format of the solution, we convert each tuple into a list.
Putting these elements together, the single-line python code return list(zip(*matrix))
takes the original matrix, sends each row as a separate argument to zip
, which pairs up the elements with the same index from each row, forming the new rows of the transposed matrix. These tuples are then converted into lists to get the final transposed matrix.
Solution Approach
Implementing the solution for transposing a matrix in Python is quite straightforward thanks to Python's powerful syntax and built-in functions. The provided reference solution uses almost no explicit algorithms because high-level function calls handle the necessary operations. Nevertheless, it's beneficial to break down the solution to understand the underlying patterns and behavior.
Here's the provided solution for reference:
class Solution:
def transpose(self, matrix: List[List[int]]) -> List[List[int]]:
return list(zip(*matrix))
Let's walk through how it works, step by step:
-
Function Argument Unpacking (
*
Operator):- The first step involves an advanced Python pattern called argument unpacking. In the expression
zip(*matrix)
, the*
operator is used to unpack the 2Dmatrix
list. - Essentially, if the
matrix
is a list of lists like[ [a1, a2], [b1, b2], [c1, c2] ]
, callingzip(*matrix)
is the same as callingzip([a1, a2], [b1, b2], [c1, c2])
. Every individual list inmatrix
is passed as a separate argument tozip
.
- The first step involves an advanced Python pattern called argument unpacking. In the expression
-
zip
Function:- The
zip
function takes any number of iterables and returns an iterator of tuples, where each tuple contains the i-th element from each of the input iterables. - When applied to rows of a matrix,
zip
effectively groups together the elements of the matrix by their column indices. For instance,a1
will be paired withb1
andc1
, forming the first tuple of the new row in the transposed matrix.
- The
-
list
Conversion:- The output from
zip
is an iterator of tuples. Thelist()
function is used to convert these tuples into lists, as the problem expects a list of lists structure for the transposed matrix.
- The output from
Since there is no nested loop or manual iteration, the entire operation is quite efficient. The actual transposition — the core algorithmic task — is completely delegated to the zip
function, which is a built-in, highly optimized component of Python. The clever use of argument unpacking with *
allows us to avoid manual index handling or iterating through rows and columns of the matrix, which would be needed in a more traditional, lower-level language solution.
Ultimately, this solution showcases the power of Python in terms of writing concise and readable code that leverages high-level functions and language features to perform complex operations with minimal code.
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Start EvaluatorExample Walkthrough
Let's take a small example to illustrate the solution approach. Consider the following 2D integer array, matrix
:
matrix = [ [1, 2, 3], [4, 5, 6] ]
We want to transpose this matrix, which will result in flipping the matrix over its main diagonal. To do this, we'll apply the steps in the solution approach.
-
Function Argument Unpacking (
*
Operator):- We use the
*
operator to unpack the matrix's rows as arguments to thezip
function. We can visualize this step as taking the two rows[1, 2, 3]
and[4, 5, 6]
and unpacking them such that they're passed tozip
likezip([1, 2, 3], [4, 5, 6])
.
- We use the
-
zip
Function:- The
zip
function then takes these two lists and pairs elements at the same positions together, resulting in tuples. The output ofzip
given our two rows would be an iterator that generates the following tuples one by one:(1, 4)
,(2, 5)
,(3, 6)
. - These tuples represent the rows of the new, transposed matrix. The first tuple
(1, 4)
will be the first row, the second tuple(2, 5)
will be the second row, and so on.
- The
-
list
Conversion:- Next, we convert each of these tuples back into lists using the
list
function since the expected output format is a list of lists. - Applying the
list
function to the iterator of tuples fromzip
, we obtain the transposed matrix as a list of lists:[[1, 4], [2, 5], [3, 6]]
.
- Next, we convert each of these tuples back into lists using the
After going through the steps with our example matrix
, the final transposed matrix is:
[ [1, 4], [2, 5], [3, 6] ]
This walkthrough demonstrates how the provided Python solution transposes a matrix efficiently by utilizing function argument unpacking, the zip
function, and list conversion to transform the matrix's rows into the transposed matrix's columns. The elegance of the solution lies in its simplicity and effective use of Python's built-in functionality to accomplish the task with a single line of code.
Solution Implementation
1# Import typing module to use type hints
2from typing import List
3
4class Solution:
5 def transpose(self, matrix: List[List[int]]) -> List[List[int]]:
6 # Transpose the input matrix.
7 # This is done by unpacking the rows of the matrix as arguments to the zip function.
8 # zip(*matrix) couples elements with the same index from each row together, effectively transposing the elements.
9 # Then, the zip object is converted into a list of lists, which is the transposed matrix.
10 transposed_matrix = [list(row) for row in zip(*matrix)]
11
12 # Return the transposed matrix.
13 return transposed_matrix
14
1class Solution {
2
3 // Function to transpose a given matrix
4 public int[][] transpose(int[][] matrix) {
5 // 'rows' is the number of rows in the input matrix
6 int rows = matrix.length;
7 // 'cols' is the number of columns in the input matrix which is derived from the first row
8 int cols = matrix[0].length;
9
10 // 'transposedMatrix' is the transposed matrix where the rows and columns are swapped
11 int[][] transposedMatrix = new int[cols][rows];
12
13 // Iterate over each column of the transposed matrix
14 for (int i = 0; i < cols; i++) {
15 // Iterate over each row of the transposed matrix
16 for (int j = 0; j < rows; j++) {
17 // Assign the value from the original matrix to the correct position in the transposed matrix
18 transposedMatrix[i][j] = matrix[j][i];
19 }
20 }
21 // Return the transposed matrix
22 return transposedMatrix;
23 }
24}
25
1#include <vector> // Include vector from Standard Template Library (STL)
2
3// Solution class
4class Solution {
5public:
6 // Function to transpose a given matrix
7 // @param originalMatrix: the original matrix to be transposed
8 // @return: a new matrix which is the transpose of the original matrix
9 vector<vector<int>> transpose(vector<vector<int>>& originalMatrix) {
10 int rowCount = originalMatrix.size(); // Number of rows in the matrix
11 int columnCount = originalMatrix[0].size(); // Number of columns in the matrix
12
13 // Create a new matrix with dimensions swapped (columns x rows)
14 vector<vector<int>> transposedMatrix(columnCount, vector<int>(rowCount));
15
16 // Iterate over each element in the new matrix
17 for (int i = 0; i < columnCount; ++i) {
18 for (int j = 0; j < rowCount; ++j) {
19 // Assign the value from the original matrix to the new position
20 // in the transposed matrix by swapping indices
21 transposedMatrix[i][j] = originalMatrix[j][i];
22 }
23 }
24 return transposedMatrix; // Return the transposed matrix
25 }
26};
27
1/**
2 * Transposes a given matrix (converts rows to columns and vice versa).
3 * @param {number[][]} matrix The matrix to be transposed.
4 * @return {number[][]} The transposed matrix.
5 */
6function transpose(matrix: number[][]): number[][] {
7 // Get the number of rows in the matrix.
8 const rowCount: number = matrix.length;
9 // Get the number of columns in the matrix.
10 const columnCount: number = matrix[0].length;
11
12 // Initialize a new matrix with dimensions swapped (rows become columns and vice versa).
13 const transposedMatrix: number[][] = new Array(columnCount)
14 .fill(0)
15 .map(() => new Array(rowCount).fill(0));
16
17 // Iterate over each column of the new transposed matrix.
18 for (let i: number = 0; i < columnCount; ++i) {
19 // Iterate over each row of the new transposed matrix.
20 for (let j: number = 0; j < rowCount; ++j) {
21 // Assign the transposed value from the original matrix to the new matrix.
22 transposedMatrix[i][j] = matrix[j][i];
23 }
24 }
25
26 // Return the newly formed transposed matrix.
27 return transposedMatrix;
28}
29
Time and Space Complexity
The provided code receives a matrix and transposes it using Python's built-in zip
function combined with argument unpacking (*
).
Time Complexity:
The time complexity for transposing a matrix involves iterating over each element exactly once. In this code, zip
takes m
sequences (rows), where m
is the number of rows of the matrix, and combines them into n
tuples, where n
is the number of columns. Since each element is touched once during the operation, the time complexity is O(m*n)
where m
is the number of rows and n
is the number of columns in the original matrix.
Space Complexity:
The zip
function creates n
tuples (where n
is the number of columns of the input matrix), each containing m
elements (where m
is the number of rows of the input matrix), and list()
then converts these tuples into lists. This operation creates a new list of lists with the same number of elements as the original matrix. Therefore, the space complexity is also O(m*n)
, as it requires additional space proportional to the size of the input matrix.
Learn more about how to find time and space complexity quickly using problem constraints.
What does the following code do?
1def f(arr1, arr2):
2 i, j = 0, 0
3 new_arr = []
4 while i < len(arr1) and j < len(arr2):
5 if arr1[i] < arr2[j]:
6 new_arr.append(arr1[i])
7 i += 1
8 else:
9 new_arr.append(arr2[j])
10 j += 1
11 new_arr.extend(arr1[i:])
12 new_arr.extend(arr2[j:])
13 return new_arr
14
1public static List<Integer> f(int[] arr1, int[] arr2) {
2 int i = 0, j = 0;
3 List<Integer> newArr = new ArrayList<>();
4
5 while (i < arr1.length && j < arr2.length) {
6 if (arr1[i] < arr2[j]) {
7 newArr.add(arr1[i]);
8 i++;
9 } else {
10 newArr.add(arr2[j]);
11 j++;
12 }
13 }
14
15 while (i < arr1.length) {
16 newArr.add(arr1[i]);
17 i++;
18 }
19
20 while (j < arr2.length) {
21 newArr.add(arr2[j]);
22 j++;
23 }
24
25 return newArr;
26}
27
1function f(arr1, arr2) {
2 let i = 0, j = 0;
3 let newArr = [];
4
5 while (i < arr1.length && j < arr2.length) {
6 if (arr1[i] < arr2[j]) {
7 newArr.push(arr1[i]);
8 i++;
9 } else {
10 newArr.push(arr2[j]);
11 j++;
12 }
13 }
14
15 while (i < arr1.length) {
16 newArr.push(arr1[i]);
17 i++;
18 }
19
20 while (j < arr2.length) {
21 newArr.push(arr2[j]);
22 j++;
23 }
24
25 return newArr;
26}
27
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C# Solution: Matrix Transpose
js solution:
Interesting, never really used the zip function with list comprehension before. Neat!