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Advent of Code 2020 Solution Megathread - Day 1: Report Repair

Ryan Palo on December 01, 2020

It's back! It's finally back! One of my favorite things on the internet is the yearly Advent of Code challenge. It's a programming challenge tha...

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Ali Spittel • Dec 1 '20 • Edited

Hey!! So excited that it's this time again here is my code. I would have just done loops without sets but the servers were down when I was trying to get my full input so I just went ahead and did a tiny optimization :).

def part_one(nums, set_nums):
    for num in nums:
        if 2020 - num in set_nums:
            return num * (2020 - num)


def part_two(nums, set_nums):
    for num in nums:
        for num2 in nums:
            if (2020 - num - num2) in set_nums:
                return num * (2020 - num - num2) * num2


with open("input.txt") as _file:
    nums = [int(line) for line in _file]
    set_nums = set(nums)

    print("Part 1", part_one(nums, set_nums))
    print("Part 2", part_two(nums, set_nums))

Ryan Palo • Dec 1 '20

Oooh! That's a really slick way to drop out a level of nesting. I like it!

Patryk Woziński • Dec 2 '20 • Edited

Hi! It was also my first time in Advent of Code.

There is my solution for the first day in Elixir: click, GitHub

Except input setup the code looks like:

  def part1() do
    numbers = Enum.map(@numbers, &String.to_integer(&1))

    [res | _] = for x <- numbers,
      y <- numbers,
      x + y == 2020,
      do: x * y

    res
  end

  def part2() do
    numbers = Enum.map(@numbers, &String.to_integer(&1))

    [res | _] = for x <- numbers,
      y <- numbers,
      z <- numbers,
      x + y + z == 2020,
      do: x * y * z

    res
  end

David Clothier • Dec 15 '20

SQL works for me. Easy and fast ;)

Get this table naming 'day1':

1.2. Solution

SELECT
       a1.num*a2.num*a3.num
FROM
       day1 a1
     , day1 a2
     , day1 a3
WHERE
       a1.num + a2.num + a3.num = 2020
LIMIT 1

Benjamin Trent • Dec 1 '20

Here is my solution in rust (minus all the code required to grab the input and parse it)

fn sums_to<'a>(
    sum: &'_ usize,
    curr: &'a usize,
    rest: &'a [usize],
) -> Option<(&'a usize, &'a usize)> {
    if rest.len() == 0 {
        return Option::None;
    }
    for re in rest {
        if re + curr == *sum {
            return Option::Some((curr, re));
        }
    }
    sums_to(sum, &rest[0], &rest[1..])
}

fn tri_sums_to<'a>(
    sum: &'_ usize,
    x: &'a usize,
    rest: &'a [usize],
) -> Option<(&'a usize, &'a usize, &'a usize)> {
    if rest.len() == 0 {
        return Option::None;
    }
    if x < sum {
        let sub_sum = *sum - x;
        for i in 0..(rest.len() - 1) {
            if let Some((a, b)) = sums_to(&sub_sum, &rest[i], &rest[i + 1..]) {
                return Option::Some((x, a, b));
            }
        }
    }
    tri_sums_to(sum, &rest[0], &rest[1..])
}

async fn day1_1() -> Result<usize> {
    let input = input_to_vec(1).await?;
    let (v1, v2) = sums_to(&2020, &input[0], &input[1..]).unwrap();
    Ok(v1 * v2)
}

async fn day1_2() -> Result<usize> {
    let input = input_to_vec(1).await?;
    let (v1, v2, v3) = tri_sums_to(&2020, &input[0], &input[1..]).unwrap();
    Ok(v1 * v2 * v3)
}

Christopher Kruse • Dec 2 '20

Hey - I'm working on the problems in Rust this year, too! I'm putting all mine in Github, but will share here, too.

(I found a cool cargo plugin that provides helpful macros and makes doing AoC stuff easier - cargo-aoc)

use aoc_runner_derive::{aoc, aoc_generator};

#[aoc_generator(day1)]
fn parse_input_day1(input: &str) -> Vec<usize> {
    input
        .lines()
        .map(|v| v.parse().expect("Failed to parse input!"))
        .collect()
}

#[aoc(day1, part1)]
fn find_2020_entries(input: &Vec<usize>) -> usize {
    let pair = input.iter().enumerate().find_map(|(idx, &item)| {
        match input
            .iter()
            .enumerate()
            .find(|(idx2, &second)| idx2 != &idx && (item + second == 2020))
        {
            Some((_, second)) => Some(item * second),
            None => None,
        }
    });
    pair.unwrap_or(1)
}

#[aoc(day1, part2)]
fn find_2020_entries_with_three(input: &Vec<usize>) -> usize {
    let result = input.iter().enumerate().find_map(|(idx, &item)| {
        input.iter().enumerate().find_map(|(idx2, &second)| {
            match input.iter().enumerate().find(|(idx3, &third)| {
                idx2 != idx && &idx != idx3 && &idx2 != idx3 && (item + second + third == 2020)
            }) {
                Some((_, third)) => Some(item * second * third),
                None => None,
            }
        })
    });
    result.unwrap_or(1)
}

Florian Rohrer • Dec 1 '20

I am also gonna be doing aoc this year. Here is my Python solution:

with open("input.txt") as f:
    l = [int(x.strip()) for x in f]

for i, n1 in enumerate(l):
    for j, n2 in enumerate(l[i+1:]):
        if n1 + n2 == 2020:
            part1 = n1 * n2
        for n3 in l[i+j+1:]:
            if n1 + n2 + n3 == 2020:
                part2 = n1 * n2 * n3

print(part1)
print(part2)

willsmart • Dec 1 '20 • Edited

Yep, this'll be another year fighting the urge to pre-optimise xmas code 🤦

Anyway, here's a nice efficient implementation in JS... (linear for part one, which is nice. Square for part two which could be improved but, like, why?)

const input = JSON.parse(require('fs').readFileSync('1a.json', 'utf-8'));

// sort ascending
input.sort((a, b) => a - b);


function go(c, tgt) {
  // tgt is the target sum we're aiming for
  // a is an index which moves through the array, 
  // b is ahead of a but retreats to maintain a sum <= to target.

  for (let a = c, b = input.length - 1; a < input.length; a++) {
    let { [a]: av, [b]: bv } = input;
    while (av + bv > tgt && b >= a) bv = input[--b];
    if (av + bv == tgt) return { a, b };
  }
}

let res;
// Part 1
(res = go(0, 2020)) &&
  console.log({
    part: 1,
    sum: input[res.a] + input[res.b],
    prod: input[res.a] * input[res.b],
  });

// Part 2
for (const c in input)
  (res = go(c, 2020 - input[c])) &&
    console.log({
      part: 2,
      sum: input[res.a] + input[res.b] + input[c],
      prod: input[res.a] * input[res.b] * input[c],
    });

willsmart • Dec 1 '20

I'm liking Ryan's idea of using this to practice new languages and get out of the comfort zone (though JS is just so comfy now 😌 )
Plan is to make myself a twister-style spinner with 5 or so languages, and force myself to use whichever one the arrow lands on on the day.

Ryan Palo • Dec 1 '20

I'm going to try to do my solutions in C. C is something I'm learning, so if anybody has any pointers (LOL), please don't hesitate to critique my code.

I didn't do anything crazy today. I just threw loops at it to see if it was fast enough. It solved before I could blink, so I'm not going to work any harder at improving speed.

/// Day 1: Report Repair
/// 
/// My expense report for my trip is messed up and entries need
/// adjusting.

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

/// Parse the input file, which contains one positive integer per line.
int* parse(const char* filename, size_t* count) {
  FILE *fp;

  fp = fopen(filename, "r");
  if (fp == NULL) {
    printf("Could not open input file.\n");
    exit(EXIT_FAILURE);
  }

  // Count how many we need
  size_t lines = 0;
  while (!feof(fp)) {
    if (fgetc(fp) == '\n') lines++;
  }
  lines++;  // The last line has no newline, but we should count it.
  rewind(fp);

  int* entries = malloc(sizeof(int)*lines);

  char buff[5];  // No numbers greater than 4 digits
  for (size_t i = 0; i < lines; i++) {
    fgets(buff, 6, fp);
    entries[i] = atoi(buff);
  }
  fclose(fp);

  *count = lines;
  return entries;
}

/// Part 1 finds the two entries that add up to 2020 exactly and returns
/// their product, or -1 if none are found.
int part1(int* entries, size_t count) {
  for (size_t i = 0; i < count; i++) {
    for (size_t j = i + 1; j < count; j++) {
      if (entries[i] + entries[j] == 2020) {
        return entries[i] * entries[j];
      }
    }
  }
  return -1;
}

/// Part 2 is the same as Part 1, but we're looking for a trio of numbers
/// that sum up to 2020.  Returns their product or -1 if no trio is found.
int part2(int* entries, size_t count) {
  for (size_t i = 0; i < count; i++) {
    for (size_t j = i + 1; j < count; j++) {
      for (size_t k = j + 1; k < count; k++) {
        if (entries[i] + entries[j] + entries[k] == 2020) {
          return entries[i] * entries[j] * entries[k];
        }
      }
    }
  }
  return -1;
}

int main() {
  size_t count;
  int* entries = parse("day1/day1.txt", &count);
  printf("The 2020 pair product is: %d\n", part1(entries, count));
  printf("The 2020 triplet product is: %d\n", part2(entries, count));
  return EXIT_SUCCESS;
}

Johnny • Dec 2 '20 • Edited

I'm also doing it in C this year! I'm also fairly new to the language and pretty much the only thing I've done differently is use fscanf instead of fgets in conjunction with atoi

(+ I've optimised a little bit to do both parts in the same loop)

pihentagy • Dec 10 '20

Python, why make it complicate if it is so siple?

for (a,b) in itertools.combinations([int(n) for n in open('input')], 2): 
    if a+b == 2020: 
        print(a*b)

ttyago • Dec 1 '21

My solution is the same but yours is also more slim!
Good job!

Jimmy Klein • Dec 4 '20 • Edited

Hi !
Here is my solution in PHP.

I try to make a generic function that can handle 2, 3 or more number in the expense report.

Full size here : Advent of Code - Day 1

Anna • Dec 3 '20

COBOL:

   IDENTIFICATION DIVISION.
   PROGRAM-ID. Advent-of-Code-2020-day-1-part-2.
   AUTHOR. ANNA KOSIERADZKA.

   ENVIRONMENT DIVISION.
   INPUT-OUTPUT SECTION.
   FILE-CONTROL.
       SELECT INPUTFILE ASSIGN TO "d1.input"
       ORGANIZATION IS LINE SEQUENTIAL.

   DATA DIVISION.
   FILE SECTION.
     FD INPUTFILE.
     01 INPUTRECORD PIC 9(4).
   WORKING-STORAGE SECTION.
     01 FILE-STATUS PIC 9 VALUE 0.
     01 WS-ARRAY-LEN PIC 9(3) VALUE 200.
     01 WS-ARRAY PIC 9(4) OCCURS 0 to 200 DEPENDING ON WS-ARRAY-LEN.
     01 WS-SUM PIC 9(4).
     01 WS-PRODUCT-1 PIC 9(8).
     01 WS-PRODUCT-2 PIC 9(12).

   LOCAL-STORAGE SECTION.
     01 I USAGE UNSIGNED-INT VALUE 1.
     01 J USAGE UNSIGNED-INT VALUE 1.
     01 K USAGE UNSIGNED-INT VALUE 1.

   PROCEDURE DIVISION.
   001-MAIN.
        OPEN INPUT INPUTFILE.
        PERFORM 002-READ UNTIL FILE-STATUS = 1.
        CLOSE INPUTFILE.
        PERFORM 004-LOOP.
        STOP RUN.

   002-READ.
        READ INPUTFILE
            AT END MOVE 1 TO FILE-STATUS
            NOT AT END PERFORM 003-WRITE-TO-TABLE
        END-READ.

   003-WRITE-TO-TABLE.
       COMPUTE WS-ARRAY(I) = INPUTRECORD
       ADD 1 to I.

   004-LOOP.
       PERFORM VARYING I FROM 1 BY 1 UNTIL I > WS-ARRAY-LEN
       AFTER J FROM 1 BY 1 UNTIL J > WS-ARRAY-LEN
       AFTER K FROM 1 BY 1 UNTIL K > WS-ARRAY-LEN
         ADD WS-ARRAY(I) WS-ARRAY(J) WS-ARRAY(K) GIVING WS-SUM
         IF WS-SUM = 2020 THEN
           MULTIPLY WS-ARRAY(I) BY WS-ARRAY(J) GIVING WS-PRODUCT-1 
           MULTIPLY WS-PRODUCT-1 BY WS-ARRAY(K) GIVING WS-PRODUCT-2
           DISPLAY WS-PRODUCT-2
           EXIT PERFORM
         END-IF
       END-PERFORM.

Ryan Palo • Dec 3 '20

This is awesome! I'm so happy I get to add COBOL to the list 😁

Kara Carrell • Dec 2 '20

Hey Y'all!!! I thought I'd actually try making this happen this year and I completed the first challenges today!! I think in later challenges I'll actually try importing the text file as it is, but I wanted to make it fun for myself. I did the solution this time in ruby, and its a program you'd run on irb, that asks for the expenses and the factor you need to group to find the magic number (2 or 3) and it prints out the number to enter at the end.

Here's the way I solved it:

class AdventAccount

  def initialize(expenses, factor=2)
    @expenses = expenses
    @factor = factor
  end

  def perform
    filter_big_nums
    winning_number(do_si_do)
  end

  private

  def filter_big_nums
    @expenses.reject! { |expense| expense >= 2020 }
  end

  def find_your_partner
    @expenses.combination(@factor).to_a()
  end

  def winning_set(group)
    group.sum == 2020
  end

  def do_si_do
    find_your_partner.each do |group|
      puts "Checking group #{group}"
      return group if winning_set(group)
    end
  end

  def winning_number(group)
    puts "the winning number is #{group.reduce(:*)}"
  end

end

puts "Starting up the Advent Account checker"
puts "Submit expenses now:"
expenses = gets.chomp.split(",").map(&:to_i)
puts "Thanks! Now, what factor should we search the expenses with?"
factor = gets.chomp.to_i
test = AdventAccount.new(expenses, factor)

test.perform

Ryan Palo • Dec 2 '20

Ooh filtering out numbers > 2020 is a really nice touch. I hadn’t thought of that 😁

ttyago • Dec 1 '21

Hi! A little bit late, I've start with AOC this year and waiting for tomorrow I've try the past event this is my solution in python for 2020 - DAY 1:

# PART01 - Find the two number that sum == T then multiply each other
# PART02 - Find the triplets that sum == T  then multiply each other

from itertools import combinations
from math import prod

I = list([int(x) for x in open('input.txt', 'r').readlines()])
T = 2020

# PART01
I2 = [x for x in combinations(I,2) if sum(x) == T]
R_P01 = prod([x for x in I2[0]])

# PART02
I3 = [x for x in combinations(I,3) if sum(x) == T]
R_P02 = prod([x for x in I3[0]])

Very good content here Ryan!

Neil Gall • Dec 3 '20

I wasn't even going to do AoC this year after a bit of burnout last year. But I have no self-control. I have done day 1 in Haskell, Rust and C to compare performance. Rust wins by a hair.

github.com/neilgall/advent-of-code...

Harry Gibson • Dec 8 '20

Only just heard of this, have some catching up to do! Here's my day 1 - similar to a couple of others already posted. Weirdly the part 1 problem came up in an interview just a couple of weeks ago!

def part_one(nums, target):
    seen = set()
    possible = False
    for num in nums:
        required = target - num
        if required in seen:
            possible = True
            break
        seen.add(num)
    if possible:
        return num, required
    return -1

def part_two(nums, target):
    seen = set()
    possible = False
    for num in nums:
        for num_2 in nums:
            req_num = target - num - num_2
            if req_num in seen:
                possible = True
                break
            seen.add(num_2)
        else:
            continue
        break
    if possible:
        return num, num_2, req_num
    return -1

with open("input.txt") as input:
    nums = [int(line) for line in input]
    res = part_one(nums, 2020)
    if res != -1:
        x,y = res
        print(f"Part 1 numbers are {x}, {y}, product is {x*y}")
    else:
        print(f"Part 1 cannot be solved from this dataset")

    res_2 = part_two(nums, 2020)
    if res_2 != -1:
        x,y,z = res_2
        print(f"Part 2 numbers are {x}, {y}, {z} product is {x*y*z}")
    else:
        print(f"Part 2 cannot be solved from this dataset")

Benoit Ranque • Dec 21 '20 • Edited

Bit late to the party, here is my iterator focused solution in rust. input is a Vec, aka the parsed input data

    if let Some((a, b, c)) = input.iter().find_map(|a| {
        if let Some((b, c)) = input.iter().find_map(|b| {
            if let Some(c) = input.iter().find(|c| a + b + *c == 2020) {
                Some((b, c))
            } else {
                None
            }
        }) {
            Some((a, b, c))
        } else {
            None
        }
    }) {
        println!("A: {}, B: {}, C: {} result: {}", a, b, c, a * b * c);
    } else {
        println!("Could not find numbers fullfilling the requirement")
    }

Edit: solution 2, same concept, but more concise

    match input.iter().find_map(|a| {
        input.iter().find_map(|b| {
            input.iter().find_map(|c| {
                if a + b + c == 2020 {
                    Some((a, b, c))
                } else {
                    None
                }
            })
        })
    }) {
        Some((a, b, c)) => println!("A: {}, B: {}, C: {} result: {}", a, b, c, a * b * c),
        None => println!("Could not find numbers fullfilling the requirement"),
    }

Mr. Bartucz • Dec 3 '20

Here are my solutions in Scratch. Enjoy! scratch.mit.edu/studios/28204945/

Saad-py • Mar 5 '21 • Edited

Not anything difficult or complex in my code
It is in rust

use std::fs;
fn main() {
let file = fs::read_to_string("D:\Coding\AdventOfCode\day_onw\src\Data.txt").expect("Not read").to_string();
let sum = 2020;
let nums = file.lines();
let mut result = Vec::new();
let mut vector = Vec::new();
for i in file.lines() {vector.push(i.parse::().unwrap())}
for i in nums {
let num = i.parse::().unwrap();
for a in file.lines() {
let num2 = a.parse::().unwrap();
if vector.contains(&(sum - (num + num2))) {
result.push(num);
result.push(num2);
result.push((sum - (num + num2)));
}
}
}
print!("{}",result[0]*result[1]*result[2]);
}

Dirk Fraanje (the Netherlands) • Dec 7 '20

My solution in C#: (The timer is just for fun ..)

    public static void Execute()
    {
        var timer = new Stopwatch();
        timer.Start();

        for (int i = 0; i < input.Count; i++)
        {
            for (int i2 = i + 1; i2 < input.Count; i2++)
            {
                var valueneeded = 2020 - input[i] - input[i2];
                if (input.Contains(valueneeded))
                {
                    timer.Stop();
                    Console.WriteLine();
                    Console.WriteLine($"Answer: {input[i] * input[i2] * valueneeded}");
                    Console.WriteLine($"Executed in: {timer.ElapsedMilliseconds} milliseconds, {timer.ElapsedTicks}  Ticks");
                }

            }
        }
    }

Thibaut Patel • Dec 9 '20

My JavaScript walkthrough:

Emil Privér • Dec 1 '20

Hello, I solves it like this github.com/emilpriver/Advent-of-co...