Password validator

# Created by Cépagrave [busyCarpenting]

print((lambda p:4<len(p)<11*(1-p.isalnum())*(len(p.split())<2)*bool({*p}&{*'0123456789'}))(input()))

"""
Explanation of this code:

the challenge rules are:

1 - min length of 5
2 - max length of 10
3 - at least 1 digit
4 - at least 1 special character
5 - no space

// First, the code global structure is:

print((lambda p: -- long expression -- )(input()))

the result here is: input is taken, passed as "p" argument to the lambda, this argument is used in the long expression, which will return a boolean (True or False)

// Now, the long expression:
4<len(p)<11*(1-p.isalnum())*(len(p.split())<2)*bool({*p}&{*'0123456789'})

the center of it is:

4<len(p)<11
it will return True if rules (1) and (2) are respected

on the right, 3 elements are multiplying 11:
11*(--1--)*(--2--)*bool(--3--)
if one of the 3 elements is evaluated to False (or 0), then 11*..*..*.. will take the value 0, and len(p)<0 will return "False"

// So now, the 3 elements:

(1-p.isalnum()) will give 0 if the password is only made of alphanumeric characters (a,b,c,d,...0,1,2,3,...) => special characters needed, or else 1-p.isalphanum() takes value 0 => 11*0=0, len(p)<0 returns False

(len(p.split())<2) will give 0 if there is one or more space in the password => 11*0, etc...

bool({*p}&{*'0123456789'}) this is evaluating a sets intersection:
{*p} is the set of characters in p
{*'0123456789'} is the set of string digits
if they share one element, it will be in the intersection, and bool({"5"}) for ex will evaluate to True, while bool(--empty set--) will evaluate to False, thus 0, => 11*0, etc...
"""

# old version:
# print((lambda p:bool(4<len(p)<11 and set(p)&{*'0123456789'}and not(p.isalnum()+len(p.split())>1))*'Ok'or'No')(input()))