Problem statement

Given a 2D integer array matrix, return its transpose — the matrix flipped over its main diagonal, swapping 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

• 1 <= rows, cols <= 8
• -100 <= matrix[i][j] <= 100

Hints for Transpose Matrix

1. Hint 1

   The transposed matrix has dimensions cols × rows.

2. Hint 2

   out[j][i] = matrix[i][j].

How to solve Transpose Matrix

Transposing swaps the roles of the two indices: the value at row i, column j becomes the value at row j, column i. The only wrinkle is that the result has the opposite dimensions — a rows × cols matrix transposes to cols × rows — so it cannot be done in place unless the matrix is square.

Approach

1. Read the dimensions rows and cols from the input.
2. Allocate the output with cols rows and rows columns.
3. For every i and j, write out[j][i] = matrix[i][j].
4. Return the new matrix.

Why it works

The transpose is defined entrywise, so a single pass over every cell is both necessary and sufficient — there is no structure to exploit and no way to avoid touching each value once. Building a fresh matrix sidesteps the aliasing problem: an in-place swap only works when the shape is preserved, which is to say only for square matrices.

Complexity

• Time — O(rows · cols)
• Space — O(rows · cols) for the output

Pitfalls

• The output dimensions are swapped — allocating rows × cols overflows or truncates for non-square input.
• Write out[j][i], not out[i][j]; the latter just copies the matrix.
• In-place transposition is only valid for square matrices, and even then must swap only one triangle or it undoes itself.

Reference solution

Python

from typing import List

def transpose(matrix: List[List[int]]) -> List[List[int]]:
    return [list(row) for row in zip(*matrix)]

Java

public static int[][] transpose(int[][] matrix) {
    int rows = matrix.length, cols = matrix[0].length;
    int[][] out = new int[cols][rows];
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) out[j][i] = matrix[i][j];
    }
    return out;
}

C++

vector<vector<int>> transpose(vector<vector<int>>& matrix) {
    int rows = (int) matrix.size(), cols = (int) matrix[0].size();
    vector<vector<int>> out(cols, vector<int>(rows, 0));
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) out[j][i] = matrix[i][j];
    }
    return out;
}

JavaScript

var transpose = function(matrix) {
    const rows = matrix.length, cols = matrix[0].length;
    const out = [];
    for (let j = 0; j < cols; j++) {
        const row = [];
        for (let i = 0; i < rows; i++) row.push(matrix[i][j]);
        out.push(row);
    }
    return out;
};

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