Not going to do part 2 tonight, will do in the morning

Days/Two/TwoPart1/Cargo.toml

@@ -4,3 +4,8 @@ version = "0.1.0"
 edition = "2021"
 
 [dependencies]
+
+[profile.release]
+codegen-units = 1
+lto = "fat"
+panic = "abort"

Days/Two/TwoPart1/src/main.rs

@@ -1,3 +1,111 @@
-fn main() {
-    println!("Hello, world!");
+// --- Day 2: Red-Nosed Reports ---
+// Fortunately, the first location The Historians want to search isn't a long walk from the Chief
+// Historian's office.
+//
+// While the Red-Nosed Reindeer nuclear fusion/fission plant appears to contain no sign of the Chief
+// Historian, the engineers there run up to you as soon as they see you. Apparently, they still talk
+// about the time Rudolph was saved through molecular synthesis from a single electron.
+//
+// They're quick to add that - since you're already here - they'd really appreciate your help
+// analyzing some unusual data from the Red-Nosed reactor. You turn to check if The Historians are
+// waiting for you, but they seem to have already divided into groups that are currently searching
+// every corner of the facility. You offer to help with the unusual data.
+//
+// The unusual data (your puzzle input) consists of many reports, one report per line. Each report
+// is a list of numbers called levels that are separated by spaces. For example:
+//
+// 7 6 4 2 1
+// 1 2 7 8 9
+// 9 7 6 2 1
+// 1 3 2 4 5
+// 8 6 4 4 1
+// 1 3 6 7 9
+// This example data contains six reports each containing five levels.
+//
+// The engineers are trying to figure out which reports are safe. The Red-Nosed reactor safety
+// systems can only tolerate levels that are either gradually increasing or gradually decreasing.
+// So, a report only counts as safe if both of the following are true:
+//
+// The levels are either all increasing or all decreasing.
+// Any two adjacent levels differ by at least one and at most three.
+// In the example above, the reports can be found safe or unsafe by checking those rules:
+//
+// 7 6 4 2 1: Safe because the levels are all decreasing by 1 or 2.
+// 1 2 7 8 9: Unsafe because 2 7 is an increase of 5.
+// 9 7 6 2 1: Unsafe because 6 2 is a decrease of 4.
+// 1 3 2 4 5: Unsafe because 1 3 is increasing but 3 2 is decreasing.
+// 8 6 4 4 1: Unsafe because 4 4 is neither an increase or a decrease.
+// 1 3 6 7 9: Safe because the levels are all increasing by 1, 2, or 3.
+// So, in this example, 2 reports are safe.
+//
+// Analyze the unusual data from the engineers. How many reports are safe?
+
+use std::fs::File;
+use std::io::{BufRead, BufReader};
+use std::path::Path;
+
+fn compute(all_reports: Vec<Vec<i32>>) -> i32 {
+    let mut sum = 0;
+    for report in all_reports {
+        let mut going_up = false;
+        let mut first_check = true;
+        let mut is_good = true;
+        for (idx, level) in report.iter().enumerate() {
+            if idx == report.len() - 1 { continue; }
+            let difference = level - report.clone().get(idx + 1).unwrap();
+            if difference.abs() >= 1 && difference.abs() <= 3 {
+                if first_check {
+                    first_check = false;
+                    // If positive
+                    if difference > 0 {
+                        going_up = true;
+                    } else {
+                        going_up = false;
+                    }
+                } else {
+                    if (difference > 0 && going_up) || (difference < 0 && !going_up) {
+                        continue;
+                    } else {
+                        is_good = false;
+                    }
+                }
+            } else {
+                is_good = false;
+            }
+        }
+        if is_good {
+            sum += 1;
+        }
+    }
+    sum
+}
+
+fn load_input() -> Vec<Vec<i32>> {
+    let mut all_reports = vec![vec![]];
+    let file = File::open(Path::new("./input")).unwrap();
+    let reader = BufReader::new(file);
+    all_reports.remove(0);
+    for line in reader.lines() {
+        let line = line.unwrap();
+        let split = line.split_whitespace().collect::<Vec<_>>();
+        let mut temp_report: Vec<i32> = vec![];
+        for level in split {
+            temp_report.push(level.parse::<i32>().unwrap());
+        }
+        all_reports.push(temp_report);
+    }
+
+    all_reports
+}
+
+fn main() {
+    // Load in the input file
+    let all_reports = load_input();
+    // Setup a timing function for the lolz (gotta go fast)
+    use std::time::Instant;
+    let now = Instant::now();
+    let result = compute(all_reports);
+    let elapsed = now.elapsed();
+    println!("Result: {:?}", result);
+    println!("Elapsed: {:.2?} microseconds", elapsed.as_micros());
 }

Days/Two/TwoPart2/src/main.rs

@@ -1,3 +1,134 @@
-fn main() {
-    println!("Hello, world!");
+// --- Day 2: Red-Nosed Reports ---
+// Fortunately, the first location The Historians want to search isn't a long walk from the Chief
+// Historian's office.
+//
+// While the Red-Nosed Reindeer nuclear fusion/fission plant appears to contain no sign of the Chief
+// Historian, the engineers there run up to you as soon as they see you. Apparently, they still talk
+// about the time Rudolph was saved through molecular synthesis from a single electron.
+//
+// They're quick to add that - since you're already here - they'd really appreciate your help
+// analyzing some unusual data from the Red-Nosed reactor. You turn to check if The Historians are
+// waiting for you, but they seem to have already divided into groups that are currently searching
+// every corner of the facility. You offer to help with the unusual data.
+//
+// The unusual data (your puzzle input) consists of many reports, one report per line. Each report
+// is a list of numbers called levels that are separated by spaces. For example:
+//
+// 7 6 4 2 1
+// 1 2 7 8 9
+// 9 7 6 2 1
+// 1 3 2 4 5
+// 8 6 4 4 1
+// 1 3 6 7 9
+// This example data contains six reports each containing five levels.
+//
+// The engineers are trying to figure out which reports are safe. The Red-Nosed reactor safety
+// systems can only tolerate levels that are either gradually increasing or gradually decreasing.
+// So, a report only counts as safe if both of the following are true:
+//
+// The levels are either all increasing or all decreasing.
+// Any two adjacent levels differ by at least one and at most three.
+// In the example above, the reports can be found safe or unsafe by checking those rules:
+//
+// 7 6 4 2 1: Safe because the levels are all decreasing by 1 or 2.
+// 1 2 7 8 9: Unsafe because 2 7 is an increase of 5.
+// 9 7 6 2 1: Unsafe because 6 2 is a decrease of 4.
+// 1 3 2 4 5: Unsafe because 1 3 is increasing but 3 2 is decreasing.
+// 8 6 4 4 1: Unsafe because 4 4 is neither an increase or a decrease.
+// 1 3 6 7 9: Safe because the levels are all increasing by 1, 2, or 3.
+// So, in this example, 2 reports are safe.
+//
+// Analyze the unusual data from the engineers. How many reports are safe?
+
+// --- Part Two ---
+// The engineers are surprised by the low number of safe reports until they realize they forgot to
+// tell you about the Problem Dampener.
+//
+// The Problem Dampener is a reactor-mounted module that lets the reactor safety systems tolerate
+// a single bad level in what would otherwise be a safe report. It's like the bad level never happened!
+//
+// Now, the same rules apply as before, except if removing a single level from an unsafe report
+// would make it safe, the report instead counts as safe.
+//
+// More of the above example's reports are now safe:
+//
+// 7 6 4 2 1: Safe without removing any level.
+// 1 2 7 8 9: Unsafe regardless of which level is removed.
+// 9 7 6 2 1: Unsafe regardless of which level is removed.
+// 1 3 2 4 5: Safe by removing the second level, 3.
+// 8 6 4 4 1: Safe by removing the third level, 4.
+// 1 3 6 7 9: Safe without removing any level.
+// Thanks to the Problem Dampener, 4 reports are actually safe!
+//
+// Update your analysis by handling situations where the Problem Dampener can remove a single level
+// from unsafe reports. How many reports are now safe?
+
+use std::fs::File;
+use std::io::{BufRead, BufReader};
+use std::path::Path;
+
+fn compute(all_reports: Vec<Vec<i32>>) -> i32 {
+    let mut sum = 0;
+    for report in all_reports {
+        let mut going_up = false;
+        let mut first_check = true;
+        let mut is_good = true;
+        for (idx, level) in report.iter().enumerate() {
+            if idx == report.len() - 1 { continue; }
+            let difference = level - report.clone().get(idx + 1).unwrap();
+            if difference.abs() >= 1 && difference.abs() <= 3 {
+                if first_check {
+                    first_check = false;
+                    // If positive
+                    if difference > 0 {
+                        going_up = true;
+                    } else {
+                        going_up = false;
+                    }
+                } else {
+                    if (difference > 0 && going_up) || (difference < 0 && !going_up) {
+                        continue;
+                    } else {
+                        is_good = false;
+                    }
+                }
+            } else {
+                is_good = false;
+            }
+        }
+        if is_good {
+            sum += 1;
+        }
+    }
+    sum
+}
+
+fn load_input() -> Vec<Vec<i32>> {
+    let mut all_reports = vec![vec![]];
+    let file = File::open(Path::new("./input")).unwrap();
+    let reader = BufReader::new(file);
+    all_reports.remove(0);
+    for line in reader.lines() {
+        let line = line.unwrap();
+        let split = line.split_whitespace().collect::<Vec<_>>();
+        let mut temp_report: Vec<i32> = vec![];
+        for level in split {
+            temp_report.push(level.parse::<i32>().unwrap());
+        }
+        all_reports.push(temp_report);
+    }
+
+    all_reports
+}
+
+fn main() {
+    // Load in the input file
+    let all_reports = load_input();
+    // Setup a timing function for the lolz (gotta go fast)
+    use std::time::Instant;
+    let now = Instant::now();
+    let result = compute(all_reports);
+    let elapsed = now.elapsed();
+    println!("Result: {:?}", result);
+    println!("Elapsed: {:.2?} microseconds", elapsed.as_micros());
 }