Day 1 Part 1 and Part 2

.gitignore

@@ -1,2 +1,2 @@
 /target
-Days/*/target
+Days/*/*/target

.idea/adventofcode.iml

@@ -2,9 +2,11 @@
 <module type="EMPTY_MODULE" version="4">
   <component name="NewModuleRootManager">
     <content url="file://$MODULE_DIR$">
-      <sourceFolder url="file://$MODULE_DIR$/Days/One/src" isTestSource="false" />
+      <sourceFolder url="file://$MODULE_DIR$/Days/One/OnePart1/src" isTestSource="false" />
+      <sourceFolder url="file://$MODULE_DIR$/Days/One/OnePart2/src" isTestSource="false" />
       <sourceFolder url="file://$MODULE_DIR$/src" isTestSource="false" />
-      <excludeFolder url="file://$MODULE_DIR$/Days/One/target" />
+      <excludeFolder url="file://$MODULE_DIR$/Days/One/OnePart1/target" />
+      <excludeFolder url="file://$MODULE_DIR$/Days/One/OnePart2/target" />
       <excludeFolder url="file://$MODULE_DIR$/target" />
     </content>
     <orderEntry type="inheritedJdk" />

Days/One/OnePart1/src/main.rs

@@ -57,16 +57,44 @@
 //
 // Your actual left and right lists contain many location IDs. What is the total distance between your lists?
 
-use std::fs;
+use std::fs::File;
-
+use std::io::{BufRead, BufReader};
-fn compute(left: Vec<i64>, right: Vec<i64>) -> i64 {
+use std::path::Path;
 
+// Function to do the computing
+fn compute(mut left: Vec<i64>, mut right: Vec<i64>) -> i64 {
+    let mut sum: i64 = 0;
+    let mut idx = 0;
+    // Sort the lists
+    left.sort();
+    right.sort();
+    // Compare and sum the lists
+    for left_item in left.clone() {
+        let right_item = right[idx];
+        sum = sum + (left_item - right_item).abs();
+        idx += 1;
+    }
+    sum
 }
 
+// Function to load the lists
 fn load_lists() -> (Vec<i64>, Vec<i64>) {
-
+    let mut left: Vec<i64> = vec![0];
+    let mut right: Vec<i64> = vec![0];
+    // Load the inputs file in
+    let file = File::open(Path::new("./input")).unwrap();
+    let reader = BufReader::new(file);
+    for line in reader.lines() {
+        let line = line.unwrap();
+        let mut split = line.split("   ");
+        let collection = split.collect::<Vec<&str>>();
+        left.push(collection[0].parse::<i64>().unwrap());
+        right.push(collection[1].parse::<i64>().unwrap());
+    }
+    (left, right)
 }
 
 fn main() {
-    println!("Hello, world!");
+    let (left, right) = load_lists();
+    println!("{:?}", compute(left, right));
 }

Days/One/OnePart2/Cargo.lock

@@ -0,0 +1,7 @@
+# This file is automatically @generated by Cargo.
+# It is not intended for manual editing.
+version = 3
+
+[[package]]
+name = "OnePart2"
+version = "0.1.0"

Days/One/OnePart2/Cargo.toml

@@ -0,0 +1,6 @@
+[package]
+name = "OnePart2"
+version = "0.1.0"
+edition = "2021"
+
+[dependencies]

Days/One/OnePart2/src/main.rs

@@ -0,0 +1,129 @@
+// --- Day 1: Historian Hysteria ---
+// The Chief Historian is always present for the big Christmas sleigh launch, but nobody has seen
+// him in months! Last anyone heard, he was visiting locations that are historically significant to
+// the North Pole; a group of Senior Historians has asked you to accompany them as they check the
+// places they think he was most likely to visit.
+//
+// As each location is checked, they will mark it on their list with a star. They figure the Chief
+// Historian must be in one of the first fifty places they'll look, so in order to save Christmas,
+// you need to help them get fifty stars on their list before Santa takes off on December 25th.
+//
+// Collect stars by solving puzzles. Two puzzles will be made available on each day in the Advent
+// calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star.
+// Good luck!
+//
+// You haven't even left yet and the group of Elvish Senior Historians has already hit a problem:
+// their list of locations to check is currently empty. Eventually, someone decides that the best
+// place to check first would be the Chief Historian's office.
+//
+// Upon pouring into the office, everyone confirms that the Chief Historian is indeed nowhere to be
+// found. Instead, the Elves discover an assortment of notes and lists of historically significant
+// locations! This seems to be the planning the Chief Historian was doing before he left. Perhaps
+// these notes can be used to determine which locations to search?
+//
+// Throughout the Chief's office, the historically significant locations are listed not by name but
+// by a unique number called the location ID. To make sure they don't miss anything, The Historians
+// split into two groups, each searching the office and trying to create their own complete list of location IDs.
+//
+// There's just one problem: by holding the two lists up side by side (your puzzle input), it
+// quickly becomes clear that the lists aren't very similar.
+// Maybe you can help The Historians reconcile their lists?
+//
+// For example:
+//
+// 3   4
+// 4   3
+// 2   5
+// 1   3
+// 3   9
+// 3   3
+// Maybe the lists are only off by a small amount! To find out, pair up the numbers and measure
+// how far apart they are. Pair up the smallest number in the left list with the smallest number in
+// the right list, then the second-smallest left number with the second-smallest right number, and so on.
+//
+// Within each pair, figure out how far apart the two numbers are; you'll need to add up all of
+// those distances. For example, if you pair up a 3 from the left list with a 7 from the right list,
+// the distance apart is 4; if you pair up a 9 with a 3, the distance apart is 6.
+//
+// In the example list above, the pairs and distances would be as follows:
+//
+// The smallest number in the left list is 1, and the smallest number in the right list is 3. The
+// distance between them is 2.
+// The second-smallest number in the left list is 2, and the second-smallest number in the right
+// list is another 3. The distance between them is 1.
+// The third-smallest number in both lists is 3, so the distance between them is 0.
+// The next numbers to pair up are 3 and 4, a distance of 1.
+// The fifth-smallest numbers in each list are 3 and 5, a distance of 2.
+// Finally, the largest number in the left list is 4, while the largest number in the right list is
+// 9; these are a distance 5 apart.
+// To find the total distance between the left list and the right list, add up the distances between
+// all of the pairs you found. In the example above, this is 2 + 1 + 0 + 1 + 2 + 5, a total distance of 11!
+//
+// Your actual left and right lists contain many location IDs. What is the total distance between
+// your lists?
+//
+// --- Part Two ---
+// Your analysis only confirmed what everyone feared: the two lists of location IDs are indeed very
+// different.
+//
+// Or are they?
+//
+// The Historians can't agree on which group made the mistakes or how to read most of the Chief's
+// handwriting, but in the commotion you notice an interesting detail: a lot of location IDs appear
+// in both lists! Maybe the other numbers aren't location IDs at all but rather misinterpreted handwriting.
+//
+// This time, you'll need to figure out exactly how often each number from the left list appears in
+// the right list. Calculate a total similarity score by adding up each number in the left list
+// after multiplying it by the number of times that number appears in the right list.
+//
+// Here are the same example lists again:
+//
+// 3   4
+// 4   3
+// 2   5
+// 1   3
+// 3   9
+// 3   3
+// For these example lists, here is the process of finding the similarity score:
+//
+// The first number in the left list is 3. It appears in the right list three times, so the similarity score increases by 3 * 3 = 9.
+// The second number in the left list is 4. It appears in the right list once, so the similarity score increases by 4 * 1 = 4.
+// The third number in the left list is 2. It does not appear in the right list, so the similarity score does not increase (2 * 0 = 0).
+// The fourth number, 1, also does not appear in the right list.
+// The fifth number, 3, appears in the right list three times; the similarity score increases by 9.
+// The last number, 3, appears in the right list three times; the similarity score again increases by 9.
+// So, for these example lists, the similarity score at the end of this process is 31 (9 + 4 + 0 + 0 + 9 + 9).
+//
+// Once again consider your left and right lists. What is their similarity score?
+
+use std::fs::File;
+use std::io::{BufRead, BufReader};
+use std::path::Path;
+
+fn compute(mut left: Vec<i64>, mut right: Vec<i64>) -> i64 {
+    left.iter().fold(0, |sum, left_item| sum + (left_item * (right.iter().fold(0, |count, right_item| count + (left_item == right_item) as i64))))
+}
+
+fn load_lists() -> (Vec<i64>, Vec<i64>) {
+    let mut left: Vec<i64> = vec![0];
+    let mut right: Vec<i64> = vec![0];
+    let file = File::open(Path::new("./input")).unwrap();
+    let reader = BufReader::new(file);
+    for line in reader.lines() {
+        let line = line.unwrap();
+        let mut split = line.split("   ");
+        let collection = split.collect::<Vec<&str>>();
+        left.push(collection[0].parse::<i64>().unwrap());
+        right.push(collection[1].parse::<i64>().unwrap());
+    }
+    (left, right)
+}
+
+fn main() {
+    let (left, right) = load_lists();
+    use std::time::Instant;
+    let now = Instant::now();
+    println!("{:?}", compute(left, right));
+    let elapsed = now.elapsed();
+    println!("Elapsed: {:?} seconds.", elapsed.as_micros());
+}