sorting - Count the number of max consecutive "a"'s from a string. Python 3

Question

Say that the user inputs:

"daslakndlaaaaajnjndibniaaafijdnfijdnsijfnsdinifaaaaaaaaaaafnnasm"

How would you go about finding the highest number of consecutive "a" and how would you remove the "a"'s and leave only 2 of them instead of the large number of them before.

I was thinking of appending each letter into a new empty list but i'm not sure if that's correct or what to do after.

I really don't know where to begin with this one but this is what i'm thinking:

1. Ask the user for input.
2. Create an empty list
3. Append each letter from the input into the list

What's next I have no idea.

second edit (something along these lines):

sentence = input("Enter your text: ")
new_sentance = " ".join(sentence.split())
length = len(new_sentance)
alist = []
while (length>0):
    alist
print ()

Answer 1

Starting with the input string:

input = "daslakndlaaaaajnjndibniaaafijdnfijdnsijfnsdinifaaaaaaaaaaafnnasm"

• To get the max consecutive number of occurrences, you would use:

  max(len(s) for s in re.findall(r'a+', input))
  

• To replace only the longest unbroken sequence of "a"s with 2 "a"s, you would use:

  maxMatch = max(re.finditer(r'a+', input), key= lambda m: len(m.group()))
  output = input[:maxMatch.start()] + "aa" + input[maxMatch.end():]
  

  First, I obtain an iterable of MatchObjects by testing the input string against the regex a+, then use max to obtain the MatchObject with the greatest length. Then, I splice the portion of the original string up to the start of the match, the string "aa", and the portion of the original string after the end of the match to give you your final output.

• To replace all occurrences of more than 2 "a"s with 2 "a"s, you would use:

  output = re.sub(r'a{3,}', "aa", input)