Envío 2019

Código

word = input()

# solvable with len(set(word)) == len(word) which relies on hashes so, here's a more explicit and simple version.

# 512 is enough for strings on the a-z to not collide when using hash() and modulo
HT_LEN = 512 
hash_tbl = [False] * HT_LEN

repeated = 'yes'
for char in word:
  idx = hash(char) % HT_LEN
  if hash_tbl[idx]:
    break
  else:
    hash_tbl[idx] = True
else:
  repeated = 'no'

print(repeated)