0867 - Transpose Matrix (Easy)

Problem Link

https://leetcode.com/problems/transpose-matrix/

Problem Statement

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.

Example 1:

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

Example 2:

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

Constraints:

• $m == matrix.length$
• $n == matrix[i].length$
• $1 <= m, n <= 1000$
• $1 <= m * n <= 10 ^ 5$
• $-10 ^ 9 <= matrix[i][j] <= 10 ^ 9$

Approach 1: Iterating over the columns and rows

The solution used was iterating over the columns and rows of the original matrix switching their indexes and creating a new row for each column from the original matrix in order to obtain its transposed matrix.

For example, consider the given matrix $[[1, 2, 4], [5, 7, 8]]$ as the input. Starting from the first element of the first column we assign it to the first element of the first row of the transposed matrix $transposed[0][0] = matrix[0][0]$. Then we go to the second element of the first column and assign it to the second element of the first row of the transposed matrix $transposed[0][1] = matrix[1][0]$. And we repeat this process for every column of the original matrix until we get the final result of $transposed$ which will be $[[1, 5], [2, 7], [4, 8]]$.

Python

Written by @jessicaribeiroalves

class Solution:
    def transpose(self, matrix: List[List[int]]) -> List[List[int]]:
        transposed = []
        for row in range(len(matrix[0])):
            transposed.append([])
            for column in range(len(matrix)):
                transposed[row].append(matrix[column][row])
        return transposed

Written by @wingkwong

class Solution:
    def transpose(self, matrix: List[List[int]]) -> List[List[int]]:
        return [[matrix[row][col] for row in range(0, len(matrix))] for col in range(0, len(matrix[0]))]

JavaScript

Written by @jessicaribeiroalves

const transpose = (matrix) => {
    const transposed = [];
    for (const row in matrix[0]) {
        transposed.push([]);
        for (const column of matrix) {
            transposed[row].push(column[row]);
        }
    }
    return transposed;
};

Time Complexity: $O(m*n)$

The time complexity for this solution is $O(m*n)$ considering $m$ as the number of rows and $n$ as the number of columns in the original matrix.

Space Complexity: $O(m*n)$

The space complexity for this solution is also $O(m*n)$ considering $m$ as the number of rows and $n$ as the number of columns in the original matrix.