# Solving Leetcode Interviews in Seconds with AI: Transpose Matrix


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "867" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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 <= 105 -109 <= matrix[i][j] <= 109  

	# Explanation
	*   **Determine Dimensions:** Get the number of rows (m) and columns (n) of the input matrix.
*   **Create Transpose Matrix:** Initialize a new matrix with dimensions n x m to store the transposed elements.
*   **Populate Transpose:** Iterate through the original matrix, and for each element at `matrix[i][j]`, place it at `transpose[j][i]` in the new matrix.

*   **Runtime Complexity:** O(m\*n), where m is the number of rows and n is the number of columns in the input matrix.
*   **Storage Complexity:** O(m\*n) because we create a new matrix to store the transpose.

	
	# Code
	```python
	def transpose(matrix: list[list[int]]) -> list[list[int]]:
    """
    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 <= 105
    -109 <= matrix[i][j] <= 109
    """
    rows = len(matrix)
    cols = len(matrix[0])

    transposed_matrix = [[0] * rows for _ in range(cols)]

    for i in range(rows):
        for j in range(cols):
            transposed_matrix[j][i] = matrix[i][j]

    return transposed_matrix
	```