<jupyter_start><jupyter_text>The challenge: Alphabet Rangolisize 3----c------c-b-c--c-b-a-b-c--c-b-c------c----size 5--------e--------------e-d-e----------e-d-c-d-e------e-d-c-b-c-d-e--e-d-c-b-a-b-c-d-e--e-d-c-b-c-d-e------e-d-c-d-e----------e-d-e--------------e--------<jupyter_code>"""
What if we generate the middle row first and set the length to the number of cols
"""
N = 5
alph = 'abcdefghijklmnopqrstuvwxyz'
rows = N


def get_cols(N):
    # writing the middle row
    row = []
    for i in reversed(range(N)):
        row.append(alph[i])
    for i in range(1,N):
        row.append(alph[i])
    print(row)
    middle_row = "-".join(row)
    print(middle_row)
    return (middle_row, len(middle_row))

cols = get_cols(N)[1]


#different starting point each time
for start in range(N)
    
    
rangoli(N)<jupyter_output>['e', 'd', 'c', 'b', 'a', 'b', 'c', 'd', 'e']
e-d-c-b-a-b-c-d-e
['e', 'd', 'c', 'b', 'a', 'b', 'c', 'd', 'e']
e-d-c-b-a-b-c-d-e
{'fraction_non_alphanumeric': 0.30157068062827225, 'fraction_numerical': 0.005235602094240838, 'mean_word_length': 1.9875, 'pattern_counts': {'":': 0, '<': 4, '<?xml version=': 0, '>': 4, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 0, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
{'chain_length': '1', 'dir': 'jupyter-structured-clean-dedup', 'id': '294258', 'license': 'MIT', 'n_tokens_mistral': 443, 'n_tokens_neox': 431, 'n_words': 84, 'path': 'Hackkerank/Python Challenges/AlphabetRangoli.ipynb', 'repo_name': 'peternewman22/Python_Coding-Challenges', 'size': '955'}