Password validation

Question

I tried to solve this.. I got all the 9 conditions correct but the last 4 were wrong. Somebody help me!!

p=input("") #string input
count=0
special=['!', '@', '#', '

Answer 1

Just consider this code :

https://code.sololearn.com/cV57TqZAIuhp/?ref=app

Answer 2

This will help you if you like this to analyze

https://code.sololearn.com/cOjvkbJvYPg2/?ref=app

Answer 3

# Python code:

a, b, c = input(), 0, 0
for x in a:
if x.isnumeric(): b += 1
if x in "!@#$%&*": c += 1
if b > 1 and c > 1 and len(a) > 6:
print("Strong")
else: print("Weak")

# Hope this helps

Answer 4

What is need of condition count>=11 in if block..? May i know about the logic?

edit: nivedha srinivasan
you are counting alphabets into variable 'a' but you need length of input string which also includes special chars and digits.. so take
a = len(p) next only need special chars and digits separately..
in c : a = strlen(l);
hope it helps you..

Answer 5

In first, I understand more python than c so my comment will be for the python code.
Firstly you don't need to check if in password given are characters cause this is not a criteria of the challenge.
Secondly there are three cases you should check :
1-The length of password.
2-At least two numbers.
3-At least two special characters.

So in python you may create three different functions to check those points or you just can run the check in a single for loop. For 2 and 3 you may need a variable to store the count of them. For the 1 just use len() function.
For example:

#this is without functions
pass = str(input())
nums = list(range(10))
special_c = [#special characters needed]
n = 0
s = 0
for char in pass:
if char in nums:
n++
elif char in special_c:
s++

#the final check
if len(pass)>=7 and n>=2 and s>=2:
print('Strong')
else:
print('Weak')

You got it ?😀

Answer 6

The same for the c code

Answer 7

Thank you ,guys.completed 🥳🥳

Answer 8

No problem 👍. Ask anytime

Answer 9

// C code:

#include <stdio.h>

int main() {
char a[25];
char b[7] = {'!','@','#','#x27;,'&','*','%'};
fgets(a, 25, stdin);
int x=0, y=0, i;
for (i=0;a[i]!='\0';i++) {
for (int k=0;k<10;k++) {
if (k == a[i] - 48) x++;
if (a[i] == b[k * (k < 7)]) y++;
}
}
if (x > 1 && y > 1 && i > 6) {
puts("Strong");
}
else puts("Weak");
}

// Hope this helps

Answer 10

password = list(input())
str = " ".join(password) #a string with space b/w the items

import re
symb = re.findall("[!@#\$%&\*]+", str)
num = re.findall(r"[0-9]+", str)
filt = filter((lambda x: x in symb,str), (password))

print ("Strong" if len(symb) >= 2 and len(num) >= 2 and len(password) >=7 else "Weak")