Code coach - Password validation

Question

I think the object of this quiz is count to find how many specific characters are in the input.
I solved it but i think there are better ways because if there are thousands of characters to find, my way will take a day or more.
If there are better ways to count something in string or to shorten my code, please let me know.
Here is my code
p=input()
count_num=p.count('0')+p.count('1')+p.count('2')+p.count('3')+p.count('4')+p.count('5')+p.count('6')+p.count('7')+p.count('8')+p.count('9')
count_chr=p.count('!')+p.count('@')+p.count('#')+p.count('%')+p.count('&')+p.count('*')+p.count('

Answer 1

Try my solution:

symbols = ['!', '@', '#', '#x27;, '%', '&', '*']
numbers = ['0','1','2','3','4','5','6','7','8','9']
i = str(input())
s = sum(1 for j in i if j in symbols)
n = sum(1 for j in i if j in numbers)
if len(i) >= 7 and s >= 2 and n >= 2:
print('Strong')
else:
print('Weak')

Answer 2

There are many ways to resolve this challenge.

You could create separate filters for nums & specials for example, then compare the input against the filters - increasing the count of the item as it appears.

Add in the length of the password and you have everything you need to determine the output

Answer 3

Tomáš Konečný you imported numpy but never used it!

Answer 4

Thanks a lot guys.

Answer 5

Thanks for noticing 😆

Answer 6

There is another approach

digits = "0123456789"
symbols = "@#₹_&-+/"

lettercount = 0
symbolcount = 0
password = input()

for lt in password:
if lt in digits:
lettercount += 1
elif lt in symbols:
symbolcount += 1

if symbolcount > 2 and lettercount > 6:
print("Strong")
else:
print("Weak")

DHANANJAY