Count number of valid password based on some given restrictions - c++

Question

I have to count how many valid passwords are possible based on some given restrictions. The restrictions are following:
Minimum 10 characters and Maximum 14 characters.
Characters can include small English letters ('a'-'z'), Capital letters ('A'-'Z'), digits ('0'-'9') and special characters ('!','#','#','$','%','^','&','*','(',')')
Valid passwords must contain at least one of small letter, capital letter, digit and special character each.
Passwords cannot contain own student ID. Students ID are of 7 digits. First two digits denote year (00, 01, ... , 99), next two digits denote department code (00, 01, ... 99), and last three digits denote roll inside a department (000 - 180). So a student ID can be like: 1210142, where 12 denotes he is from batch 12, 10 is department code, and 142 is roll number. A student with ID 1210142 cannot have a password like Ti#s1210142mE but can have password like Ti#s121014m2E.
A student can use another student's ID in his/her own password.
Given the restrictions how many valid passwords can be generated?
I wrote a simple C++ program to simulate it. But as it is just a naive implementation, it would require a whole lot of time (maybe greater than my lifetime) to spit out an answer. Is there any clever way to figuring out the answer using code, possibly using regex or something like that?
My effort so far:
#include <iostream>
#include <algorithm>
inline bool is_valid_password(const std::string &generated_password, const std::string &student_id){
bool small = false, captial = false, digit = false, special = false;
for(const char &c : generated_password){
if(c >= 'a' && c <= 'z') small = true;
else if(c >= 'A' && c <= 'Z') captial = true;
else if(c >= '0' && c <= '9') digit = true;
else if(c == '!' || c == '#' || c == '#' || c == '$' || c == '%' || c == '^' || c == '&' || c == '*' || c == '(' || c == ')') special = true;
}
if(small && captial && digit && special){
return generated_password.find(student_id) == std::string::npos;
}
return false;
}
char valid_character_set [] = {
'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z',
'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z',
'0','1','2','3','4','5','6','7','8','9',
'!','#','#','$','%','^','&','*','(',')'
};
long long comb(int N, int K, const std::string &student_id)
{
std::string bitmask(K, 1); // K leading 1's
bitmask.resize(N, 0); // N-K trailing 0's
long long counter = 0;
// print integers and permute bitmask
do {
std::string generated_password = "";
for (int i = 0; i < N; ++i) // [0..N-1] integers
{
if (bitmask[i]){
generated_password += valid_character_set[i];
}
}
if(is_valid_password(generated_password, student_id)) {
//std::cout << "valid password found: " << generated_password << '\n';
counter++;
}
} while (std::prev_permutation(bitmask.begin(), bitmask.end()));
return counter;
}
int main(int argc, char const *argv[])
{
std::cout << "Enter your student id: ";
std::string student_id; std::cin >> student_id;
std::cout << "Your student id is: " << student_id << "\n";
// all possible 10 character passwords
std::cout << comb(72, 10, student_id) << '\n';
return 0;
}

Answer 1

I don't know if you are familiar with writing regex, but if not, you can learn and try out a lot on this website. In my opinion, regex is a very powerful concept and it is worth it to invest some time to learn it.
Anyway, when you are familiar with regex, there is a very useful standard library for regex in C++. Just include
#include<regex>
And then you can instantiated your own regex object:
std::regex r = regex("yourRegexPattern");
With that object, you can invoke some useful functions, like regex_match() (which could be the right one for your use case), regex_search() or regex_replace().
For more information, I have linked the official reference above, or you can just google "c++ regex"
Edit: fixed links