unreasonably effective

you can be sloppy, as long as you are rigorous

LeetCode 867: Transpose Matrix

link

For a (3 \times 2) source matrix, we would create a (2 \times 3) transposed matrix. We copy each row of the source matrix into a column of transposed.

  • Destination row index: The length of the source row gives the row indices of the destination column.
  • Destination col index: Index of the source row is the index of the destination column.

Time: \mathcal{O}(m \cdot n), space: \mathcal{O}(\texttt{output-size}).

class Solution:
    def copy_row_to_column(self, src_row, transposed, dst_col_index) -> None:
        for dst_row_index, x in enumerate(src_row):
            transposed[dst_row_index][dst_col_index] = x

    def transpose(self, matrix: List[List[int]]) -> List[List[int]]:
        m, n = len(matrix), len(matrix[0])
        # 3 X 2 --> 2 X 3
        transposed = [
            [0] * m for _ in range(n)
        ]
        for dst_col_index, src_row in enumerate(matrix):
            self.copy_row_to_column( src_row, transposed, dst_col_index )

        return transposed

tanvirdotzaman

,

Leave a comment