Asterisk pattern solution

# *Asterisk Pattern* solution by @CodingComputing
size = 7  # given parameter, expected to be ODD
center_idx = size // 2  # get the index of the central row/column
for row_idx in range(size):  # to construct each row and print
    if row_idx == center_idx:  # if the row is the central row...
        row_chars = ["*"] * size  # print all *s for the central row
    else:  # for a non-centre row...
        row_chars = [" "] * size  # initialization, we'll fill * where needed
        for col_idx in range(size):  # for columns within the current row...
            # check if we are on any of the diagnoals
            is_on_descending_diagonal = col_idx==row_idx
            is_on_ascending_diagonal = col_idx==(size-1-row_idx)
            # -1 is needed because row_idx and col_idx are zero indexed
            is_on_centre_col = col_idx==center_idx  # check if on the central column
            if (
                is_on_ascending_diagonal or
                is_on_descending_diagonal or
                is_on_centre_col
            ):  # Any of these conditions
                row_chars[col_idx] = '*'  # change the initialized space to *
    print(" ".join(row_chars))  # join on a space to get separation

# Output:
#
# *     *     *
#   *   *   *
#     * * *
# * * * * * * *
#     * * *
#   *   *   *
# *     *     *