Discussion on Alphabet Rangoli Challenge

9 years ago+ 1 comment

For what it's worth, here is my solution in Python 2. I did NOT focus on getting it as short as possible, since that tends to also make it confusing -- instead, I focussed on making it as simple/understandable as possible.

from string import ascii_lowercase as letters
# ascii_lowercase, i.e. letters, is the following string: 'abcdefghijklmnopqrstuvwxyz'

def build_line(i, char_list):
    
    # Eg. If i = 2 and size = 5 ...
    
    # ... reduced_list = [e, d]
    reduced_list = char_list[ : i]
   
    # ... reversed_reduced_list = [d, e]
    reversed_reduced_list = reduced_list[: : -1]
   
    # ... reduced_list = [e, d, c]
    reduced_list.append(char_list[i])
   
    # ... reduced_list = [e, d, c, d, e]
    reduced_list.extend(reversed_reduced_list)
    
    # ... curr_baseline = 'e-d-c-d-e'
    curr_baseline = DASH.join(curr_baseline_list)
    
    # Eg. If line_length = 17, then we are saying:   print '{:-^17}'.format(curr_baseline) 
    # which means that the resulting string has to be padded on the left and right with the character '-' such that the string curr_baseline is in the middle and the total length of the final string is 17.
    # This is the final string that is printed:   ----e-d-c-d-e----
    print '{:{}^{}}'.format(curr_baseline, DASH, line_length)
    
DASH = '-'

size = int(raw_input())

# Formula for length of each line (including dashes) is: (2*x) - 1, where x = (2*(size-1)) + 1
x = (2 * (size - 1)) + 1
line_length = (2 * x) - 1

char_list = []

# Eg. If size = 5, values of letter_index are: 4, 3, 2, 1, 0
for letter_index in range(size - 1, -1, -1):    
        char_list.append(letters[letter_index])
        
# Eg. If size = 5, char_list is now [e, d, c, b, a]. Note that len(char_list) = size.

# Eg. If size = 5, values of i: 0, 1, 2, 3, 4
for i in range(size):
    build_line(i, char_list)
    
# Eg. If size = 5, values of i: 3, 2, 1, 0
for i in range(size - 2, -1, -1):
    build_line(i, char_list)

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