programming challenge – Advent of Code 2020 – Day 2: validating passwords (C++ version)

Original: Advent of Code 2020 – Day 2: validating passwords

I decided to translate my Rust solution to Advent of Code 2020 into C++ to increase my familiarity with C++. Here’s the task for Day 2:

Day 2: Password Philosophy


To try to debug the problem, they have created a list (your puzzle
input) of passwords (according to the corrupted database) and the
corporate policy when that password was set.

For example, suppose you have the following list:

1-3 a: abcde
1-3 b: cdefg
2-9 c: ccccccccc

Each line gives the password policy and then the password. The
password policy indicates the lowest and highest number of times a
given letter must appear for the password to be valid.
For example,
1-3 a means that the password must contain a at least 1 time and
at most 3 times.

In the above example, 2 passwords are valid. The middle password,
cdefg, is not; it contains no instances of b, but needs at least
1. The first and third passwords are valid: they contain one a or
nine c, both within the limits of their respective policies.

How many passwords are valid according to their policies?


Part Two

While it appears you validated the passwords correctly, they don’t
seem to be what the Official Toboggan Corporate Authentication System
is expecting.

The shopkeeper suddenly realizes that he just accidentally explained
the password policy rules from his old job at the sled rental place
down the street! The Official Toboggan Corporate Policy actually works
a little differently.

Each policy actually describes two positions in the password, where 1 means the first character, 2 means the second character, and so
(Be careful; Toboggan Corporate Policies have no concept of
“index zero”!) Exactly one of these positions must contain the given
Other occurrences of the letter are irrelevant for the
purposes of policy enforcement.

Given the same example list from above:

  • 1-3 a: abcde is valid: position 1 contains a and position 3 does not.
  • 1-3 b: cdefg is invalid: neither position 1 nor position 3 contains b.
  • 2-9 c: ccccccccc is invalid: both position 2 and position 9 contain c.

How many passwords are valid according to the new interpretation of the policies?

The full story can be found on the website.



#include <algorithm>
#include <cstddef>
#include <iostream>
#include <string>
#include <utility>

// must be NTBS
constexpr const char* PATH = "./data/day_2/input";

enum class Policy { Old, New };

struct Entry {
    std::pair<std::size_t, std::size_t> numbers;
    std::string password;
    char key;

    bool is_valid(Policy policy) const {
        switch (policy) {
        case Policy::Old: {
            auto frequency = static_cast<std::size_t>(
                std::count(password.begin(), password.end(), key)
            return frequency >= numbers.first && frequency <= numbers.second;
        case Policy::New: {
            auto pos_a = numbers.first - 1;
            auto pos_b = numbers.second - 1;

            if (pos_a >= password.size() || pos_b >= password.size()) {
                return false;

            // logical xor
            return (password(pos_a) == key) != (password(pos_b) == key);

// error checking omitted
inline std::istream& operator>>(std::istream& is, Entry& entry) {
    is >> entry.numbers.first;
    is.get(); // consume '-'
    is >> entry.numbers.second >> entry.key;
    is.get(); // consume ':'
    is >> entry.password;

    return is;



#include "lib.hpp"
#include <fstream>

int main() {
    std::ifstream file{PATH};

    std::size_t valid_count_old = 0;
    std::size_t valid_count_new = 0;

    for (Entry entry; file >> entry;) {
        if (entry.is_valid(Policy::Old)) {
        if (entry.is_valid(Policy::New)) {

    std::cout << "Part One: " << valid_count_old << 'n'
              << "Part Two: " << valid_count_new << 'n';