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Advent of Code 2020 Solution Megathread - Day 2: Password Philosophy

Ryan Palo on December 02, 2020

With Day 1 in the books, hopefully you’re starting to get into the swing of things. I know that I definitely didn’t spend a few hours fiddling with...

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Anna • Dec 3 '20

COBOL (second one is on my GitHub):

   IDENTIFICATION DIVISION.
   PROGRAM-ID. AOC-2020-02-1.
   AUTHOR. ANNA KOSIERADZKA.

   ENVIRONMENT DIVISION.
   INPUT-OUTPUT SECTION.
   FILE-CONTROL.
       SELECT INPUTFILE ASSIGN TO "d2.input"
       ORGANIZATION IS LINE SEQUENTIAL.

   DATA DIVISION.
   FILE SECTION.
     FD INPUTFILE
     RECORD IS VARYING IN SIZE FROM 8 to 50
     DEPENDING ON REC-LEN.
     01 INPUTRECORD PIC X(50).
   WORKING-STORAGE SECTION.
     01 FILE-STATUS PIC 9 VALUE 0.
     01 REC-LEN  PIC 9(2) COMP.
     01 WS-MIN PIC 9(4).
     01 WS-MAX PIC 9(4).
     01 WS-CHAR PIC A.
     01 WS-STRING-EMPTY PIC X.
     01 WS-PASSWORD PIC A(50).
     01 WS-SUBSTR-1 PIC X(5). 
     01 WS-CHAR-COUNT PIC 9(2).

   LOCAL-STORAGE SECTION.
     01 CORRECT-ROWS UNSIGNED-INT VALUE 0.

   PROCEDURE DIVISION.
   001-MAIN.
        OPEN INPUT INPUTFILE.
        PERFORM 002-READ UNTIL FILE-STATUS = 1.
        CLOSE INPUTFILE.
        DISPLAY CORRECT-ROWS.
        STOP RUN.

   002-READ.
        READ INPUTFILE
            AT END MOVE 1 TO FILE-STATUS
            NOT AT END PERFORM 003-PROCESS-RECORD
        END-READ.

   003-PROCESS-RECORD.
       MOVE 0 TO WS-CHAR-COUNT.
       UNSTRING INPUTRECORD DELIMITED BY SPACE OR "-" OR ":" INTO 
           WS-MIN
           WS-MAX
           WS-CHAR
           WS-STRING-EMPTY
           WS-PASSWORD.
       INSPECT WS-PASSWORD TALLYING WS-CHAR-COUNT FOR ALL WS-CHAR.
       IF WS-CHAR-COUNT >= WS-MIN AND WS-CHAR-COUNT <= WS-MAX THEN 
          ADD 1 TO CORRECT-ROWS
       END-IF.

Patryk Woziński • Dec 3 '20 • Edited

IMO you’re a real hipster using Cobol! Great job! :)

Anna • Dec 4 '20

Thanks! Gotta stay busy in the pandemic somehow. :)

Patryk Woziński • Dec 4 '20

May I ask you - have you used Cobol before? Or it's just your pandemic-advent challenge? :D

Anna • Dec 4 '20 • Edited

I started in the spring and wrote a couple things like FizzBuzz: github.com/GaloisGirl/Coding/tree/...
Then I had a bit of a life, and now I'm having a second wave of COBOL.

Ryan Palo • Dec 2 '20

Ooh boy it's a little early in the season to already be stumped by an off-by-one bug. 1-BASED INDICES RYAN! 1-BASED! IT'S RIGHT THERE IN THE PROMPT!

Also, I started a little parsing module to help me with some of the repetitive tedia.

Day2.h:

/// Day 2: Password Philosophy 
/// 
/// Find the passwords that aren't compliant.

#include <stdlib.h>
#include <stdint.h>

/// A Policied Password is a password that is accompanied by a policy
/// consisting of two positive integers and a letter.  These components
/// can be used to validate the password.
typedef struct {
  int a;
  int b;
  char letter;
  char password[30];
} PoliciedPassword;

/// Part 1 calculates how many valid passwords there are.
/// A password is valid if the # of occurrences of 'letter' is between
/// 'a' and 'b,' inclusive.
///
/// passes: the list of policied passwords to check
/// count: the number of passwords to check
int part1(PoliciedPassword** passes, size_t count);

/// Part 2 calculates how many valid passwords there are.
/// A password is valid of either the 'a' index or 'b' index character 
/// (1-based!) is equal to 'letter,' but not both.
int part2(PoliciedPassword** passes, size_t count);

/// day2 runs both parts in sequence and outputs their results.
int day2();

Day2.c:

/// Day 2: Password Philosophy 
/// 
/// Find the passwords that aren't compliant.

#include "Day2.h"
#include "parsing.h"

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>

/// Parse the input file,
static PoliciedPassword** parse(size_t* lines) {
  FILE* fp;
  fp = fopen("data/day2.txt", "r");
  if (fp == NULL) {
    printf("Couldn't open input file.\n");
    exit(EXIT_FAILURE);
  }

  *lines = count_lines(fp);

  PoliciedPassword** passes = malloc(sizeof(PoliciedPassword*) * *lines);

  for (size_t i = 0; i < *lines; i++) {
    PoliciedPassword* p = malloc(sizeof(PoliciedPassword));
    fscanf(fp, "%d-%d %c: %s\n", &p->a, &p->b, &p->letter, p->password);
    passes[i] = p;
  }

  fclose(fp);
  return passes;
}

static void freePoliciedPasswordList(PoliciedPassword** passes, size_t count) {
  for (size_t i = 0; i < count; i++) {
    free(passes[i]);
    passes[i] = NULL;
  }
  free(passes);
}

int part1(PoliciedPassword** passes, size_t count) {
  size_t valid = 0;

  for (size_t i = 0; i < count; i++) {
    PoliciedPassword* p = passes[i];
    uint8_t matches = 0;
    for (size_t j = 0; p->password[j]; j++) {
      if (p->password[j] == p->letter) matches++;
    }
    if (p->a <= matches && matches <= p->b) valid++;
  }
  return valid;
}

int part2(PoliciedPassword** passes, size_t count) {
  size_t valid = 0;

  for (size_t i = 0; i < count; i++) {
    PoliciedPassword* p = passes[i];
    int matches = 0;
    if (p->password[p->a - 1] == p->letter) matches++;
    if (p->password[p->b - 1] == p->letter) matches++;
    if (matches == 1) valid++;
  }
  return valid;
}

int day2() {
  size_t count;
  PoliciedPassword** passes = parse(&count);
  printf("====== Day 2 ======\n");
  printf("Part 1: %d\n", part1(passes, count));
  printf("Part 2: %d\n", part2(passes, count));
  freePoliciedPasswordList(passes, count);
  return EXIT_SUCCESS;
}

parsing.h:

#ifndef AOC2020_PARSING_H
#define AOC2020_PARSING_H

#include <stdio.h>
#include <stdlib.h>

/// Counts the number of newline characters in a text file.
/// Assumes no newline at the end of the last line (so adds +1 more)
size_t count_lines(FILE* fp);

#endif

parsing.c:

#include "parsing.h"

#include <stdlib.h>
#include <stdio.h>

size_t count_lines(FILE* fp) {
  size_t lines = 0;
  while (!feof(fp)) {
    if (getc(fp) == '\n') lines++;
  }

  rewind(fp);
  return lines + 1;
}

Also a little tooling to help generate the daily files:

bin/newday:

#!/usr/bin/env bash

# newday: Creates a new day of files for the Advent of Code Challenge
#
# Date:   12/2/2020
# Author: Ryan Palo

function usage() {
  echo "usage: newday NUMBER"
  echo
  echo "    NUMBER: The number of the day to create.  Will be inserted"
  echo "            into the templates and used for filenames."
}

function help() {
  echo "newday: Creates a new day of files for the Advent of Code challenge."
  echo
  usage
  echo
}

function make_day() {
  day="$1"
  echo "Creating Day ${day}." >&2
  sed "s/{X}/$day/g" "templates/DayX.c" > "src/Day${day}.c"
  sed "s/{X}/$day/g" "templates/DayX.h" > "src/Day${day}.h"
  sed "s/{X}/$day/g" "templates/TestDayX.c" > "test/TestDay${day}.c"
  sed "s/{X}/$day/g" 'templates/main.c' > 'src/main.c'
  touch "data/day${day}.txt"
  echo "Complete." >&2
}

function main() {
  if [[ "$#" -ne 1 ]]; then
    usage
    exit 1
  fi

  if [[ "$1" == '-h' ]]; then
    help
    exit 0
  fi

  if ! [[ "$1" =~ [:digit:]+ ]]; then
    echo "Input must be a number"
    exit 1
  fi

  make_day "$1"
}

main "$@"

Patryk Woziński • Dec 2 '20 • Edited

Hi guys!
The second day was awesome. I've really enjoyed the task. :) I'm still learning Elixir, but I think my solution was not that bad. :) Of course, I did it in Elixir.

This time I've split the day into two modules - one per part. The full solution is on my GitHub repository.

The first part of the day looks like:

defmodule AdventOfCode.Day2Part1 do
  @pattern ~r/(?<min>\d*)-(?<max>\d*)\s(?<char>\w):\s*(?<password>\w*)/

  def calculate(file_path) do
    file_path
    |> File.stream!()
    |> Stream.map(&String.replace(&1, "\n", ""))
    |> Stream.map(fn line ->
      Regex.named_captures(@pattern, line)
    end)
    |> Stream.filter(fn %{"char" => char, "min" => min, "max" => max, "password" => password} ->
      occurrences =
        password
        |> String.graphemes()
        |> Enum.count(&(&1 == char))

      occurrences >= String.to_integer(min) and occurrences <= String.to_integer(max)
    end)
    |> Enum.count()
  end
end

The second part of the day looks like:

defmodule AdventOfCode.Day2Part2 do
  @pattern ~r/(?<x>\d*)-(?<y>\d*)\s(?<char>\w):\s*(?<password>\w*)/

  def calculate(file_path) do
    file_path
    |> File.stream!()
    |> Stream.map(&String.replace(&1, "\n", ""))
    |> Stream.map(fn line ->
      Regex.named_captures(@pattern, line)
    end)
    |> Stream.filter(fn %{"char" => char, "x" => x, "y" => y, "password" => password} ->
      at_position?(char, x, password) != at_position?(char, y, password)
    end)
    |> Enum.count()
  end

  defp at_position?(char, position, password) do
    String.at(password, String.to_integer(position) - 1) == char
  end
end

Christopher Kruse • Dec 2 '20

Rust solution for Day 2. Felt happy and clever at the inclusion of the XOR for part 2.

Full repo on Github.

use aoc_runner_derive::{aoc, aoc_generator};
use regex::Regex;

struct PasswordChallenge {
    password: String,
    min: usize,
    max: usize,
    search_char: char,
}

#[aoc_generator(day2)]
fn parse_input_day1(input: &str) -> Vec<PasswordChallenge> {
    input.lines().map(|v| parse_line(v)).collect()
}

fn parse_line(line: &str) -> PasswordChallenge {
    let pattern = Regex::new(r"^(\d+)-(\d+)\s(\w):\s(\w+)$").expect("couldn't create Regex");
    let captures = pattern.captures(line).unwrap();

    PasswordChallenge {
        min: str::parse(captures.get(1).unwrap().as_str()).unwrap(),
        max: str::parse(captures.get(2).unwrap().as_str()).unwrap(),
        search_char: (captures.get(3).unwrap().as_str()).chars().next().unwrap(),
        password: String::from(captures.get(4).unwrap().as_str()),
    }
}

#[aoc(day2, part1)]
fn valid_password_count(input: &Vec<PasswordChallenge>) -> usize {
    input
        .iter()
        .filter(|challenge| is_valid_password_by_char_count(challenge))
        .count()
}

fn is_valid_password_by_char_count(challenge: &PasswordChallenge) -> bool {
    let pw_chars_count = challenge
        .password
        .chars()
        .filter(|ch| ch == &challenge.search_char)
        .count();

    challenge.min <= pw_chars_count && challenge.max >= pw_chars_count
}

#[aoc(day2, part2)]
fn valid_password_positions(input: &Vec<PasswordChallenge>) -> usize {
    input
        .iter()
        .filter(|challenge| is_valid_password_by_pos(challenge))
        .count()
}

fn is_valid_password_by_pos(challenge: &PasswordChallenge) -> bool {
    challenge.password.char_indices().fold(false, |memo, (idx, ch)| {
        if idx == challenge.min-1 || idx == challenge.max-1 {
            memo ^ (ch == challenge.search_char)
        } else {
            memo
        }
    })
}

Patryk Woziński • Dec 2 '20

Rust looks intriguing! I hope that one day I will find time to learn this language. :D

Kara Carrell • Dec 3 '20

Hey Y'all!! This time, I actually brought the input.txt file in as it was, and again, made a ruby solution that im sure could be a bit more refactored, but i got to play with regex! so that was fun...
Here's my full code, and here's a snippet of what i did to solve it:

class PasswordChecker

  attr_reader :parse_passwords, :check_passwords

  def initialize(passwords)
    @passwords = passwords
    @parsed_passwords = parse_passwords
  end

  def parse_passwords
    parsed = @passwords.map do |item|
      item_split = item.match(/(^\d*)-(\d*) (\w): (\w*)/)
      {
        min: item_split[1].to_i,
        max: item_split[2].to_i,
        letter: item_split[3],
        password: item_split[4]
      }
    end
    parsed
  end

  def validate_password_frequency(log)
    # puts "rules are minimum #{log[:min]} and max #{log[:max]}"
    # puts "there are #{log[:password].count(log[:letter])} letter #{log[:letter]}'s"
    log[:password].count(log[:letter]).to_i.between?(log[:min],log[:max])
  end

  def validate_password_accuracy(log)
    return false unless log[:password].include?(log[:letter])
    return false if log[:password][log[:min] - 1] == log[:letter] && log[:password][log[:max] - 1] == log[:letter]

    # puts "rules are letter #{log[:letter]} must be at position #{log[:min]} and then at position #{log[:max]}"
    # puts "there's a #{log[:password][log[:min] - 1]} at position #{log[:min]} and a #{log[:password][log[:max] - 1]} at position #{log[:max]}"
    log[:password][log[:min] - 1] == log[:letter] || log[:password][log[:max] - 1] == log[:letter]
  end

  def check_passwords(way="frequency")
    @parsed_passwords.map do |log|
        if way == "accuracy"
          validate_password_accuracy(log)
        else
          validate_password_frequency(log)
        end
    end.count(true)
  end

end

Ali Spittel • Dec 4 '20

Mine!

def check_count_in_range(password, letter, start, stop):
    return password.count(letter) >= start and password.count(letter) <= stop


def check_indexes(password, letter, start, stop):
    return (password[start] == letter or password[stop] == letter) and (password[start] != password[stop])


with open('input.txt') as file:
    letters = [num.split(": ") for num in file]
    part1_tally = 0
    part2_tally = 0
    for policy, password in letters:
        length, letter = policy.split(" ")
        start, stop = length.split("-")
        start, stop = int(start), int(stop)

        # part 1
        if check_count_in_range(password, letter, start, stop): part1_tally += 1

        # part 2
        if check_indexes(password, letter, start - 1, stop - 1): part2_tally += 1


    print("Part 1", part1_tally)
    print("Part 2", part2_tally)

Neil Gall • Dec 3 '20

Using AoC to brush up my Rust this year.

I adapted Bodil's parser combinators into a little parser module, as I suspect there'll be much more parsing to come.

And then the solution is:


mod parser;

use std::fs::File;
use std::io::prelude::*;
use std::ops::Range;
use parser::*;


// ---- model

#[derive(Debug, Eq, PartialEq)]
struct Password {
    position1: usize,
    position2: usize,
    character: char,
    password: String
}

// ---- model parser

fn password(input: &str) -> ParseResult<Password> {
    let pos1p = first(integer, string("-"));
    let pos2p = first(integer, whitespace);
    let charp = first(letter, string(": "));
    let passp = one_or_more(letter);
    let parser = map(seq(pos1p, seq(pos2p, seq(charp, passp))), |(n1, (n2, (c, p)))| Password {
        position1: n1 as usize,
        position2: n2 as usize,
        character: c,
        password: p.into_iter().collect()
    });
    parser.parse(input)
}

// --- input file

fn read_file(filename: &str) -> std::io::Result<String> {
    let mut file = File::open(filename)?;
    let mut contents = String::new();
    file.read_to_string(&mut contents)?;
    Ok(contents)
}

fn parse_input(input: &str) -> ParseResult<Vec<Password>> {
    let p = one_or_more(first(password, whitespace));
    p.parse(input)
}


// --- problem

impl Password {
    fn part1_is_valid(&self) -> bool {
        let n = self.password.chars().filter(|c| c == &self.character).count();
        self.position1 <= n && n <= self.position2
    }

    fn part2_is_valid(&self) -> bool {
        let c1 = self.password.chars().nth(self.position1 - 1) == Some(self.character);
        let c2 = self.password.chars().nth(self.position2 - 1) == Some(self.character);
        (c1 || c2) && !(c1 && c2)
    }
}

fn part1(passwords: &Vec<Password>) -> usize {
    passwords.iter().filter(|p| p.part1_is_valid()).count()
}

fn part2(passwords: &Vec<Password>) -> usize {
    passwords.iter().filter(|p| p.part2_is_valid()).count()
}

fn main() {
    let input = read_file("../input.txt").unwrap();
    let (_, passwords) = parse_input(&input).unwrap();
    println!("part1 {}", part1(&passwords));
    println!("part2 {}", part2(&passwords));
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_parse_passwords() {
        let (rest, passwords) = parse_input("1-3 a: abcde\n1-3 b: cdefg\n2-9 c: ccccccccc").unwrap();
        assert_eq!(rest, "");
        assert_eq!(passwords,
            vec![
                Password { position1: 1, position2: 3, character: 'a', password: String::from("abcde") },
                Password { position1: 1, position2: 3, character: 'b', password: String::from("cdefg") },
                Password { position1: 2, position2: 9, character: 'c', password: String::from("ccccccccc") }
            ]
        );
    }

    #[test]
    fn test_part1_is_valid_1() {
        let p = Password { position1: 1, position2: 3, character: 'a', password: String::from("abcde") };
        assert_eq!(p.part1_is_valid(), true);
    }

    #[test]
    fn test_part1_is_valid_2() {
        let p = Password { position1: 1, position2: 3, character: 'b', password: String::from("cdefg") };
        assert_eq!(p.part1_is_valid(), false);
    }

    #[test]
    fn test_part1_is_valid_3() {
        let p = Password { position1: 2, position2: 9, character: 'c', password: String::from("ccccccccc") };
        assert_eq!(p.part1_is_valid(), true);
    }

    #[test]
    fn test_part2_is_valid_1() {
        let p = Password { position1: 1, position2: 3, character: 'a', password: String::from("abcde") };
        assert_eq!(p.part2_is_valid(), true);
    }

    #[test]
    fn test_part2_is_valid_2() {
        let p = Password { position1: 1, position2: 3, character: 'b', password: String::from("cdefg") };
        assert_eq!(p.part2_is_valid(), false);
    }

    #[test]
    fn test_part2_is_valid_3() {
        let p = Password { position1: 2, position2: 9, character: 'c', password: String::from("ccccccccc") };
        assert_eq!(p.part2_is_valid(), false);
    }
}

David Clothier • Dec 15 '20 • Edited

SQL (0 lines of code huehue)

Get this table naming 'day2'

2.1. Solution

SELECT COUNT(*) FROM 
(
SELECT
   letra
 , pass
 , nmin
 , nmax
 ,(LENGTH(pass) - LENGTH(REPLACE(pass, letra, ''))) AS times
FROM
   day2
HAVING
   times BETWEEN nmin AND nmax
) puzzle_answer

2.2. Solution

SELECT COUNT(*) 
FROM   day2 
WHERE  LOCATE(letra, pass, nmin) = nmin
       XOR LOCATE(letra, pass, nmax) = nmax

Benjamin Trent • Dec 3 '20 • Edited

My Rust solution for Day 2. Also using an XOR :). Great minds think alike.

#[derive(Debug, Eq, PartialEq)]
pub struct PasswordPolicy {
    character: char,
    min: usize,
    max: usize,
} 

impl From<&String> for PasswordPolicy {
    fn from(s: &String) -> Self {
        let spaces:Vec<&str> = s.split(" ").collect();
        let character:char = spaces[1].chars().next().unwrap();
        let vec:Vec<usize> = spaces[0].split("-").map(|i| i.parse().unwrap()).collect();
        PasswordPolicy {
            character,
            min: vec[0],
            max: vec[1]
        }
    }
}


impl PasswordPolicy {

    pub fn satisfied_1(&self, password: &str) -> bool {
        let ct = password.chars().filter(|c| *c == self.character).count();
        (self.min - 1) < ct && ct < (self.max + 1)
    }

    pub fn satisfied_2(&self, password: &str) -> bool {
        let mut ans = false;
        for (i, c) in password.chars().enumerate() {
            if (i + 1) == self.min  || (i + 1) == self.max {
                ans ^= self.character == c;
            }
        }
        ans
    }
}

#[aoc_generator(day2)]
fn input_to_vec(input: &str) -> Vec<(PasswordPolicy, String)> {
    input.lines().map(|i| {
        let splt = i.split(": ").map(|s| String::from(s)).collect::<Vec<String>>();
        (PasswordPolicy::from(&splt[0]), splt[1].to_string())
    }).collect()
}

#[aoc(day2, part1)]
fn valid_password_count(input: &Vec<(PasswordPolicy, String)>) -> usize {
    input.iter().filter(|(policy, password)| policy.satisfied_1(password.as_str())).count()
}

#[aoc(day2, part2)]
fn valid_password_count2(input: &Vec<(PasswordPolicy, String)>) -> usize {
    input.iter().filter(|(policy, password)| policy.satisfied_2(password.as_str())).count()
}

pihentagy • Dec 10 '20 • Edited

Short and hopefully non-cryptic python. Ideas to improve?

def explode(line):
    policy, password = line.split(': ',1)
    nums, letter = policy.split()
    a, b = map(int, nums.split('-'))
    return a,b , letter, password

def compliant(line):
    min, max, letter, password = explode(line)
    return min <= password.count(letter) <= max

def compliant2(line):
    p1, p2, letter, password = explode(line)
    return (password[p1-1] == letter) != (password[p2-1] == letter)

print(sum(1 for _ in filter(compliant, open('input2'))))
print(sum(1 for _ in filter(compliant2, open('input2'))))

Derk-Jan Karrenbeld • Dec 4 '20 • Edited

What I came up with in Ruby:

require 'benchmark'

class PasswordPolicy
  def self.from_line(line)
    length, char, password = line.split(' ')
    first, second = length.split('-').map(&:to_i)
    char = char.delete(':')

    PasswordPolicy.new(positions: [first - 1, second - 1], char: char, password: password)
  end

  def initialize(positions:, char:, password:)
    self.positions = positions
    self.char = char
    self.password = password
  end

  def valid?
    positions.count { |i| password[i] == char } == 1
  end

  private

  attr_accessor :positions, :char, :password
end

lines = File.readlines('input.txt')

valids = 0

Benchmark.bmbm do |x|
  x.report do
    valids = lines.count do |line|
      PasswordPolicy.from_line(line).valid?
    end
  end
end

puts valids

And for comparison, here is the inlined version:

entries = File.readlines('input.txt').map do |line|
  positions, char, password = line.split
  left, right = positions.split(?-).map(&:to_i)
  [left, right, char.first, password.chomp]
}

puts entries.count do |left, right, char, password|
  (password[left - 1] == char) != (password[right - 1] == char)
}

Ignacior • Dec 8 '20

Javascript solution:

let fs = require("fs");

var entryList = fs.readFileSync("input.txt", "utf8").split("\n");

function checkCharacter(str, chr, min, max) {
  let count = 0;
  for (let c of str) {
    if (c == chr) {
      count++;
    }
  }
  return !(count < min || count > max);
}

let validPasswords = 0;

for (let entry of entryList) {
  let patterns = entry.split(": ");
  let validatorPattern = patterns[0].split(" ");
  let minMax = validatorPattern[0].split("-");

  if (
    checkCharacter(
      patterns[1],
      validatorPattern[1],
      parseInt(minMax[0]),
      parseInt(minMax[1])
    )
  ) {
    validPasswords++;
  }
}
console.log("Part one");
console.log("Valid passwords: ", validPasswords);

function isValidPassword(str, chr, validPos, invalidPos) {
  return (
    (str.charAt(validPos) === chr || str.charAt(invalidPos) === chr) &&
    str.charAt(validPos) != str.charAt(invalidPos)
  );
}

validPasswords = 0;

for (let entry of entryList) {
  let patterns = entry.split(": ");
  let validatorPattern = patterns[0].split(" ");
  let validInvalid = validatorPattern[0].split("-");

  if (
    isValidPassword(
      patterns[1],
      validatorPattern[1],
      parseInt(validInvalid[0]) - 1,
      parseInt(validInvalid[1]) - 1
    )
  ) {
    validPasswords++;
  }
}
console.log("Part two");
console.log("Valid passwords: ", validPasswords);

Harry Gibson • Dec 8 '20

I can't remember the last time I had a genuine use for XOR in my day job!

def parse_line(line):
    nums, char, pw = line.split(' ')
    num_1, num_2 = [int(i) for i in nums.split('-')]
    char = char.strip(':')
    return num_1, num_2, char, pw

def is_valid_part1(line):
    min_occ, max_occ, char, pw = parse_line(line)
    n = pw.count(char)
    return min_occ <= n <= max_occ

def is_valid_part2(line):
    loc_1, loc_2, char, pw = parse_line(line)
    if max(loc_1, loc_2) > len(pw) + 1: return False
    return (pw[loc_1-1] == char) ^ (pw[loc_2-1] == char)


with open('input.txt') as f:
    input_lines = [l.strip() for l in f]

print (f"Part 1: {sum(map(is_valid_part1, input_lines))} passwords were valid")
print (f"Part 2: {sum(map(is_valid_part2, input_lines))} passwords were valid")

Harshavardhan • Dec 8 '20

Here's my solution in Python

def partOne(low, high, letter, password):
    count = 0
    for char in password:
        if char == letter:
            count += 1
    return high >= count >= low


def partTwo(position1, position2, letter, password):
    return (password[position1] == letter) ^ (password[position2] == letter)


out1 = 0
out2 = 0
with open("Problem-2/Day-2-Password-Philosophy.txt") as file:
    for line in file:
        freq, letter, password = line.split(" ")
        letter = letter[:len(letter) - 1]
        low, high = map(int, freq.split('-'))
        if partOne(low, high, letter, password):
            out1 += 1
        if partTwo(low - 1, high - 1, letter, password):
            out2 += 1

print(out1, out2)