Challenge_Day_1.md

Day 1 - Report Repair

Background

Source of the daily challenge comes from the Advent of Code for 2020 produced by Eric Wastl.

Data Source

The source of data in the expense report comes from: https://adventofcode.com/2020/day/1/input

library(dplyr)

## 
## Attaching package: 'dplyr'

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

Data management

Read in data from the advent of code website:

d <- read.csv('Data/input_Day1.txt', header = F, col.names = 'expense', colClasses = 'numeric')
head(d)

##   expense
## 1    1509
## 2    1857
## 3    1736
## 4    1815
## 5    1576
## 6    1970

Part 1

Find two numbers in the expense report that sum to 2020 and multiply the two numbers together.

Analysis

Find two numbers that add to 2020

Print summary statistics for the dataset

summary(d)

##     expense    
##  Min.   :  48  
##  1st Qu.:1508  
##  Median :1722  
##  Mean   :1618  
##  3rd Qu.:1836  
##  Max.   :2010

Based on the minimum value of 48, we can remove all expenses greater than 1972 since the sum would exceed 2020.

cd <- d %>%
  filter(expense <= 1972)

Loop through possible expense summations to see which two expenses add to 2020.

exp_list <- sort(cd$expense)

i <- j <- c()

for(i in exp_list){
  j <- 2020-i
  if(j %in% exp_list){
    print(c(i,j))
    break()
  }
}

## [1]  251 1769

Expenses of 251 and 1769 add to 2020.

Multiply these two numbers together

exp_mult <- i*j

The final result is:

print(exp_mult)

## [1] 444019

Part 2

Find three numbers in the expense report that sum to 2020 and multiply the three numbers together.

Analysis

Find three numbers that add to 2020

Loop through possible expense summations to see which three expenses add to 2020.

i <- j <- k <- c()

for(i in exp_list){
  exp_list2 <- exp_list[exp_list != i]
  for(j in exp_list2){
    k <- 2020 - (i + j)
    if(k %in% exp_list2){
      print(c(i,j,k)) 
      break()
    }
  }
break()}

## [1]   48  383 1589

Expenses of 48, 383, and 1589 add to 2020.

Multiply these three numbers together

exp_mult <- i*j*k

The final result is:

print(exp_mult)

## [1] 29212176