old_contents: #!/usr/bin/env python3.6
  """Jerod Gawne, 2017-10-13
  
  Alphabet Rangoli
  
  https://www.hackerrank.com/challenges/alphabet-rangoli
  
  Editorial:
  This can be solved using center.
  
  n = int(raw_input())
  for i in range(n):
      s = "-".join(chr(ord('a')+n-j-1) for j in range(i+1))
      print((s+s[::-1][1:]).center(n*4-3, '-'))
  
  for i in range(n-1):
      s = "-".join(chr(ord('a')+n-j-1) for j in range(n-i-1))
      print((s+s[::-1][1:]).center(n*4-3, '-'))
  
  Sample Input:
  5
  
  Sample Output:
  --------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--------
  """
  
  
  def print_rangoli(size):
      """
      If n is the number of unique letters we will be printing,
      then we end up printing twice this many in any given row.
      We are also printing a '-' for each letter, so now we're
      at 4n characters total. BUT we do no re-print the "middle"
      character, so it and its character are substracted from the
      total. So 4n-2. Now, we also don't want to print the trailing
      '-' so subtract 1 more. Thus, 4n-3 is the width.
  
      credit: https://www.hackerrank.com/challenges/alphabet-rangoli/forum/comments/208741
      credit: https://www.hackerrank.com/challenges/alphabet-rangoli/forum/comments/254211
  
      :param size: int
      :return: None
      """
      import string
      alpha = string.ascii_lowercase
  
      ls = []
      for i in range(size):
          s = '-'.join(alpha[i:size])
          ls.append((s[::-1] + s[1:]).center(4 * size - 3, '-'))
  
      print(*(ls[:0:-1] + ls), sep='\n')
  
  
  def main():
      """
      Main/Tests
      """
      n = int(input())
      print_rangoli(n)
  
  
  if __name__ == '__main__':
      try:
          main()
      except Exception:
          import sys
          import traceback
  
          print(traceback.print_exception(*sys.exc_info()))
  
new_contents: #!/usr/bin/env python3.6
  """Jerod Gawne, 2017-10-13
  
  Alphabet Rangoli
  
  https://www.hackerrank.com/challenges/alphabet-rangoli
  
  Editorial:
  This can be solved using center.
  
  n = int(raw_input())
  for i in range(n):
      s = "-".join(chr(ord('a')+n-j-1) for j in range(i+1))
      print((s+s[::-1][1:]).center(n*4-3, '-'))
  
  for i in range(n-1):
      s = "-".join(chr(ord('a')+n-j-1) for j in range(n-i-1))
      print((s+s[::-1][1:]).center(n*4-3, '-'))
  
  Sample Input:
  5
  
  Sample Output:
  --------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--------
  """
  
  
  def print_rangoli(size):
      """Jerod Gawne, 2017-10-13
  
      If n is the number of unique letters we will be printing,
      then we end up printing twice this many in any given row.
      We are also printing a '-' for each letter, so now we're
      at 4n characters total. BUT we do no re-print the "middle"
      character, so it and its character are substracted from the
      total. So 4n-2. Now, we also don't want to print the trailing
      '-' so subtract 1 more. Thus, 4n-3 is the width.
  
      credit: https://www.hackerrank.com/challenges/alphabet-rangoli/forum/comments/208741
      credit: https://www.hackerrank.com/challenges/alphabet-rangoli/forum/comments/254211
  
      :param size: int
      :return: None
      """
      import string
      alpha = string.ascii_lowercase
  
      ls = []
      for i in range(size):
          s = '-'.join(alpha[i:size])
          ls.append((s[::-1] + s[1:]).center(4 * size - 3, '-'))
  
      print(*(ls[:0:-1] + ls), sep='\n')
  
  
  def main():
      """
      Main/Tests
      """
      n = int(input())
      print_rangoli(n)
  
  
  if __name__ == '__main__':
      try:
          main()
      except Exception:
          import sys
          import traceback
  
          print(traceback.print_exception(*sys.exc_info()))
  
subject: improve documentation