Allocate, Copy and Transpose 2D Matrix

Avoid shallow copy bugs and stick to list comprehensions when copying and allocating 2D arrays.

Allocating a 2D array

Allocate target Matrix of num_rows* num_cols intialized to 0

Allocate a 2D matrix

target = [ [0]*num_cols for _ in range(num_rows) ]

Copying a 2D Matrix

Create copy of 2D array

class Solution:
  def get_copy(self,matrix: List[List[int]] ) -> List[List[int]]  : 
      num_rows, num_cols = len(matrix), len(matrix[0])
      # Using Nested List Comprehension.
      copy = [[matrix[r][c] for c in range(num_cols)] for r in range(num_rows)]
      return copy

Transposing a 2D matrix

In a transposed matrix, the rows become columns and columns become rows. Which means :

src[r][c] == target[c][r]

OR

target[r][c] == src[c][r]

Transpose of a matrix

"""
Solution for : https://leetcode.com/problems/transpose-matrix/
You could alternately preallocate an n*m target matrix and copy over each element from src to target matrix using a nested loop over the original matrix.
"""
class Solution:
    def transpose(self, matrix: List[List[int]]) -> List[List[int]]:
        num_rows = len(matrix)
        num_cols = len(matrix[0])
        # Using Nested List Comprehension. Number of Columns equal number of rows.
        # Remember : target[r][c] = src[c][r]
        transpose = [[matrix[c][r] for c in range(num_rows)] for r in range(num_cols)]
        return transpose