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Password Validator
You're interviewing to join a security team. They want to see you build a password evaluator for your technical interview to validate the input. Task: Write a program that takes in a string as input and evaluates it as a valid password. The password is valid if it has at a minimum 2 numbers, 2 of the following special characters ('!', '@', '#', '
#x27;, '%', '&', '*'), and a length of at least 7 characters. If the password passes the check, output 'Strong', else output 'Weak'. Input Format: A string representing the password to evaluate. Output Format: A string that says 'Strong' if the input meets the requirements, or 'Weak', if not. Sample Input: Hello@$World19 Sample Output: Strong a = ['!','@','#','#x27;,'%','&','*'] b = ['0','1','2','3','4','5','6','7','8','9'] c = 0 d = 0 ab = input() for i in ab: if i in a: c += 1 if i in b: d += 1 if (c>1 and d>1) and (len(ab)>6): print("Strong") else: print("Weak") I could have solved it this way. is there a way with the regex method?2 ответов
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You have to use elif and if function, not only else and if. Or you can also nested it.
#I don't know for the other cases but I think this will work
a = ['!','@','#','#x27;,'%','&','*']
b = ['0','1','2','3','4','5','6','7','8','9']
c = 0
d = 0
ab = input()
for i in ab:
if i in a:
c += 1
elif i in b:
d += 1
if (c>1 and d>1) and (len(ab)>6):
print("Strong")
else:
print("Weak")
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Endalk thanks for code. My code is approved. I am eager to know the regex method to get it resolved.