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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.

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

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

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 "$@"

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

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)
}

Kudos Beluga • Jan 23 '21 • Edited

I started learning Rust and immediately felt satisfied coding in it, and Advent of Code seemed like the perfect place to get better at rust. Probably not the best or cleanest way to do it (didn't use XOR), but here's day 2 in Rust:

use std::fs;
fn part1(stringtocompute: Vec<&str>, chartofind: &str, min: u16, max: u16) -> u32 {
    let mut occurences = 0;
    for i in stringtocompute {
        if &i == &chartofind {
            occurences += 1;
        }
    }
    if occurences >= min && occurences <= max {
        return 1;
    } else {
        return 0;
    }
}
fn part2(stringtocompute: Vec<&str>, chartofind: &str, min: u16, max: u16) -> u32 {
    let mut positionsfound = 0;
    for i in 0..stringtocompute.len() {
        if stringtocompute[i] == chartofind && i == min.into() {
            positionsfound += 1;
        } else if stringtocompute[i] == chartofind && i == max.into() {
            positionsfound += 1;
        }
    }
    if positionsfound == 1 {
        return 1;
    } else {
        return 0;
    }
}
fn main() {
    let contents = fs::read_to_string("./day2.txt").expect("Bad file read");
    let mut values = contents.split("\n").collect::<Vec<_>>();
    values.pop();
    let mut valids: u32 = 0;
    let mut valids2: u32 = 0;
    for i in &values {
        let nums = &i.split("-").collect::<Vec<_>>();
        let min = nums[0].parse::<u16>().unwrap();
        let max = nums[1].split(" ").collect::<Vec<_>>()[0]
            .parse::<u16>()
            .unwrap();
        let chartofind = nums[1].split(":").collect::<Vec<_>>()[0]
            .split(" ")
            .collect::<Vec<_>>()[1];
        let stringtocompute = i.split(": ").collect::<Vec<_>>()[1]
            .split("")
            .collect::<Vec<_>>();
        valids += part1(stringtocompute.clone(), chartofind, min, max);
        valids2 += part2(stringtocompute, chartofind, min, max);
    }
    println!("Part 1: {:?}\nPart 2: {:?}", valids, valids2);
}

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'))))

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

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);

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);
    }
}

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()
}

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)

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")