"""
My solution:
Time Complexity:
The time complexity is O(n),
where n is the number of elements in a row.
The function iterates through each row and performs constant-time
operations for each element.
Space Complexity:
The space complexity is O(1).
The function uses a constant amount of space regardless
of the input size.
There are only a few variables (total_sum, diagonal_top_to_bottom_sum, diagonal_bottom_to_top_sum, matrix_length) used.
"""
class Solution:
    def diagonalSum(self, mat):
        total_sum = 0  # Initialize the total sum
        diagonal_top_to_bottom_sum = 0
        diagonal_bottom_to_top_sum = 0
        matrix_length = len(mat[0])  # provide length of elements in row
        print(f"matrix_length: {matrix_length}")
        for row_idx, row in enumerate(mat):
            if ((row_idx, row_idx) == (matrix_length - row_idx - 1, row_idx)):
                diagonal_top_to_bottom_sum += mat[row_idx][row_idx]
            else:
                diagonal_top_to_bottom_sum += mat[row_idx][row_idx]
                diagonal_bottom_to_top_sum += mat[matrix_length - row_idx - 1][row_idx]

        total_sum = (diagonal_bottom_to_top_sum + diagonal_top_to_bottom_sum)
        return total_sum  # Return the calculated total sum