Question

Generating secure password that you can remember, lab2pr0.py It’s not easy
to come up with a secure password that one can actually remember. One of the
methods proposed is to take a line of text and take first letter of each word to form
a password. An extra step is substitute ”o” with 0, ”a” with ”@”, and ”l” with
1. For instance, using ”Cookies and Code, let’s meet on Tuesdays” would yield
”C@C,1m0T”. Write a program that asks the user to enter a phrase and outputs
the password that is generated using these rules.
You can assume that the words in the phrase will be separated by a single space or
a common punctuation sign such as ”.,?!;:” followed by a space.

use python

Answer 1

"""
Step-1) define method password_generator with input text from user as parameter to it.
Step-2)Split the words in the text based on white space using split(' ') method.
Step-3) Add first letter of each word to the password , if the first letter is 'o'->'0',
if it is 'a'->'@', if it is 'l'->'1'.
Since words in the text are seperated by a single space or
a common punctuation sign such as ”.,?!;:” followed by a space.
And we have already splitted the text based on 'white space' ,  therefore we just need to check the last letter of each word for punctuation sign and include it in the password.

'"""



def password_generator(text):
    password = ''
    temp_list = list(text.split())
    print(temp_list)
    for word in temp_list:
        if word[0] == 'o':
            password += '0'
        elif word[0] == 'l':
            password += '1'
        elif word[0] == 'a':
            password += '@'
        else:
            password += word[0]
        if word[-1] == ',' or word[-1] == ':' or word[-1] == '?' or word[-1] == '.'\
                or word[-1] == ';' or word[-1] == '!':
            password += word[-1]
    return password


if __name__ == '__main__':
    text = input('Enter text')

    print(password_generator(text))

If you have any doubts or want some modification in the code , just put the comment on the answer, i would be happy to help.