So I created this code that takes in a list and returns True or False depending on whether they meet these criterias or not. Here is my code

def passwordlength(password: str):
    upper = digit = special = False
    for char in password:
        if char in "ABCDEFGHIJKLMNOPQRSTUVWXYZ" :
            upper = True
        elif char in "1234567890":
            digit = True
        elif char in "$#@%!":
            special = True
    return upper and digit and special

def passwordOK(password: str):
    if passwordlength(password):
        if not (6 < len(password) < 12):
            return False
        else:
            return True
    return False

Now I want to add a new twist, which is that if any letter, number, character in the list appears 3 times in a row, the result will return False.. here is my attempt at it...

    def passwordlength(password: str):
    upper = digit = special = False
    for char in password:
        if char in "ABCDEFGHIJKLMNOPQRSTUVWXYZ" :
            upper = True
        elif char in "1234567890":
            digit = True
        elif char in "$#@%!":
            special = True
    return upper and digit and special

def passwordOK(password: str):
    if passwordlength(password):
        if not (6 < len(password) < 12):
            return False
        for char in password:
            if password.count(char) > 3:
                return False
        return True
    return False

However, this clearly doesn't work at all.. what am I doing wrong here?? what should my code be in order for my output to be correct??

CodePudding user response：

Use groupby from itertools. With this version, you set the number of characters in a row.

from itertools import groupby

def passwordinarow(password: str, n: int=3):
    """Return True if password contains 'n' characters in a row else False."""
    return any([len(list(g)) >= n for c, g in groupby(password)])

Test:

>>> passwordinarow('AAABC123$')
True

>>> passwordinarow('ABC123$')
False

CodePudding user response：

If you want to return false when a character, number appears 3 times in the string, clearly you should change your if condition like this:

[...]
for char in password:
    if password.count(char) == 3:
        return False
[...]

If you meant at least 3 times appearance:

[...]
for char in password:
    if password.count(char) >= 3:
        return False
[...]

However password.count(char) will loop over the whole password for each char, so why don't we loop once, and get the result:

[...]
from collections import defaultdict
defdict = defaultdict(lambda: 0)
for char in password:
    defdict[char]  = 1
    if defdict[char] >= 3:
        return false
[...]

Update:

I mean like 3 times in a row btw, like if its AAA, BBB, CCC then it should return false.. if it's ABANA.. then it doesn't matter

Define a function like this:

def has_3consecutive_char(password: str) -> bool:
    from collections import defaultdict
    defdict = defaultdict(lambda: 0)
    prev_char = ""
    for char in password:
        if prev_char == char:
            defdict[char]  = 1
            if defdict[char] >= 2:
                return True
        else:
            defdict[char] = 0
        prev_char = char
    return False

Then use it in your passwordOK function:

def passwordOK(password: str):
    if passwordlength(password):
        if not (6 < len(password) < 12):
            return False
        if has_3consecutive_char(password):
                return False
        return True
    return False

CodePudding user response：

There are many different ways to approach this.

Personally I might do something like:

def has_repeated_chars(password: str, n_repetitions: int=3) -> bool:
    ngrams = zip(*(password[i] for i in range(n_repetitions)))
    return min(len(set(trigram)) for trigram in ngrams) == 1

Generally the logic is to create an iterable of ngrams of length 3, create a set for each ngram and if the set is of length 1 then we know the same character has been repeated 3 times.