Checking repeated character in Python list

Question

Hello!

I’m trying to develop a program that reads a string and returns a boolean if there are repeated characters.

I need enough on that score:

("") // False  
("oo") // False  
("po") // True
("EEEttOo") // False  
("PoiX2@op") // False  
("PoiX2@aP") // False
("PoiX2@p@") // False
("PoiX2@2pa") // False
("PoiX2@pa") // True

---

I thought it would be easier to work with string, passing it to a list and doing a check with for and if however, I’m not able to develop a valid logic.

s = 'Banana'

listastr = list(s)
out = True

Could someone give me a light on how to solve?

I am basing that "p" is different from "P" and the fact that " has given False, is due to the system requiring an entry.

In my system, I’m using Regex, so it searches for the characters [a-z],[A-Z],[0-9] and some special characters.

Answer 1

To search for the occurrence of a character pattern in a string use a regular expression.

The pattern (?P<char>.).*?(?P=char) means:

• (?P<char>.) setting up a catch group, identified as char, corresponding to any character.
• .*? corresponds to zero or more of any characters.
• (?P=char) matches the first character matched by the capture group char.

The algorithm is simple it uses the method re.compile() to compile the pattern (?P<char>.).*?(?P=char). Defines the function permitido(s) that accepts a string s as a parameter, this function first tests whether s is empty or composed of returning spaces Fase if the test is affirmative and then make a search with search() by the first occurrence of the pattern returns a Boolean based on the search result compared to equality with None.

Example:

import re                                              #Importa o módulo re para trabalhar com regex.

regex = re.compile(r'(?P<char>.).*?(?P=char)', re.I)   #Compila uma expressão regular para servir como padrão de busca.

def permitido(s):
    if s.strip() == "": return False                   #Se a string for vazia ou composta por espaços retorna False.
    return regex.search(s) == None                     #Busca por correspondência do padrão em s e testa igualdade com None e retorna o resultado desse teste. 
    
print(permitido('Banana'))   # False
print(permitido(""))         # False  
print(permitido("oo"))       # False  
print(permitido("po"))       # True
print(permitido("EEEttOo"))  # False  
print(permitido("PoiX2@op")) # False  
print(permitido("PoiX2@aP")) # False
print(permitido("PoiX2@p@")) # False
print(permitido("PoiX2@2pa"))# False
print(permitido("PoiX2@pa")) # True

Answer 2

An alternative is to create a set from the string. As a set does not allow duplicated elements, its size will be smaller if the string has some repeated character:

def permitido(s):
    if not s: # se a string é vazia, retorna False
        return False
    return len(s) == len(set(s))
 
print(permitido('Banana'))   # False
print(permitido(""))         # False  
print(permitido("oo"))       # False  
print(permitido("po"))       # True
print(permitido("EEEttOo"))  # False  
print(permitido("PoiX2@op")) # False  
print(permitido("PoiX2@aP")) # False
print(permitido("PoiX2@p@")) # False
print(permitido("PoiX2@2pa"))# False
print(permitido("PoiX2@pa")) # True

I only had to include a check at the beginning for the empty casyde string. If it is not empty, then create the set, and if it is the same size as the string it is because there are no repeat characters.