Find and delete different characters between strings

Question

lst1 = ["carro"]
lst2 = ["carroa", "carrooao"]

How do I check if the word in lst1 exists in any substring of lst2. If it exists delete the excess characters of the word that has the substring, making the string lst2, equal to lst1. Do not use size comparison logic or just play the right value on the other variables. I need the character or substring to be identified to be deleted.

Expected result:

lst2 = ["carro", "carro"]

Answer 1

Here is my solution (based on reply of José Henrique, but with some amendments):

def remove_caracteres(modelo, palavra):
    i = 0 # Índice na palavra a ser testada
    n = 0 # Número de caracteres do modelo encontrados na palavra testada
    for letra in modelo:
        while i < len(palavra):
            if letra == palavra[i]:
                i += 1
                n += 1
                break
            i += 1
    if n == len(modelo):
        return modelo
    return palavra

The difference is that this function is able to treat some more cases.

Cases where extra characters are removed:

• remove_caracteres("carro", "carroa")

  "carro"

• remove_caracteres("carro", "carrooao")

  "carro"

• remove_caracteres("carro", "carraao")

  "carro"

Cases where the extra characters nay are removed:

• remove_caracteres("carro", "carraaa")

  "carraaa"

• remove_caracteres("carro", "acrro")

  "acrro"

• remove_caracteres("carro", "caro")

  "caro"

• remove_caracteres("carro", "carr")

  "carr"

Extreme cases:

• remove_caracteres("", "carroa")

  ""

• remove_caracteres("carro", "")

  "carro"

• remove_caracteres("", "")

  ""

Answer 2

I didn’t understand why to do it, but anyway, it can be done like this:

str1 = "carro"
str2 = "carroa"
str3 = "carrooao"

if ((len(str1) < len(str2)) and (str2[:len(str1)] == str1)):
    str2 = str1;

if ((len(str1) < len(str3)) and (str3[:len(str1)] == str1)):
    str3 = str1;

print str1
print str2
print str3

Exit:

str1 = "car"
str2 = "car"
str3 = "car"

Answer 3

Since you cannot simply set one variable equal to another, I created a function to manually remove the excesses, follow an example:

def remove_letras(modelo, palavra):
  #passa letra por letra verificando se 
  #existe alguma correspondende e vai removendo 
  #as que não correspondem
  atual = 0
  for letra in modelo:
    palavra = palavra[:atual] + palavra[palavra.find(letra):]
    atual += 1
  #remove os excessos do final
  return  palavra[:atual] 


#TESTES
str1 = "carro"
str2 = "carroa"
str3 = "carrooao"
str4 = "carraao"

str2 = remove_letras(str1, str2)
str3 = remove_letras(str1, str3)
str4 = remove_letras(str1, str4)

print str1
print str2
print str3
print str4

Follow the test link: Repl.it

Answer 4

I solved my problem using two functions I developed. The first one would take a list as a comparison parameter that would be modelos and the word that would be compared as palavra. It would return what was found as anomaly, in case "Carroa", would return "a".

def clnw(modelos, palavra):

    for a in modelos:

       palavra = palavra.replace(a, "")

    lx = palavra

   if lx != "":
       return lx 
   else: return ""  

The function below would take the "a" as rst and the word that would be corrected as plvr.

def fixw(rst, plvr):
       st = plvr.rfind(rst)
       fn = len(plvr) -1

       plvrl = list(plvr)
       rstl = list(rst)

    while fn >= st:

        del plvrl[fn]

        fn = fn-1
    fix = ''.join(plvrl)

    return fix

Returning "Carro", with these logics can fix the strings as long as the anomaly is left, as was my case.