mapped occurrence of each number in column 2

days/day01.cpp

@@ -4,10 +4,195 @@
 
 #include "day01.h"
 
+#include <fstream>
 #include <iostream>
 #include <ostream>
+#include <unordered_map>
+#include <vector>
+#include <bits/fs_fwd.h>
+#include <bits/fs_path.h>
 
-void day01::Run()
+/*
+--- 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 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?
+*/
+
+void day01::Part1()
 {
-	std::cout << "Day 01" << std::endl;
+	std::vector<int> col1;
+	std::vector<int> col2;
+	int sum = 0;
+
+	std::ifstream input("input/day01.txt");
+
+	if (!input)
+	{
+		std::cerr << "Failed to open input file." << std::endl;
+		std::cout << "Current path is " << std::filesystem::current_path() << '\n';
+
+		return;
+	}
+
+	std::string delimiter = "   ";
+	auto delimiterLength = delimiter.length();
+
+	std::string line;
+	while (std::getline(input, line))
+	{
+		auto delimiterPos = line.find(delimiter);
+		col1.emplace_back(std::stoi(line.substr(0, delimiterPos)));
+		col2.emplace_back(std::stoi(line.substr(delimiterPos + delimiterLength, line.length())));
+	}
+
+	std::sort(col1.begin(), col1.end());
+	std::sort(col2.begin(), col2.end());
+
+
+	for (int i = 0; i < col1.size(); ++i)
+	{
+		sum += std::abs(col1[i] - col2[i]);
+	}
+}
+
+/*
+--- 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?
+
+ */
+
+void day01::Part2()
+{
+	std::vector<int> col1;
+	std::vector<int> col2;
+	std::unordered_map<int, int> occurrence_map;
+	int sum = 0;
+
+	std::ifstream input("input/day01.txt");
+	//std::ifstream input("input/day01short.txt");
+
+	if (!input)
+	{
+		std::cerr << "Failed to open input file." << std::endl;
+		std::cout << "Current path is " << std::filesystem::current_path() << '\n';
+
+		return;
+	}
+
+	std::string delimiter = "   ";
+	auto delimiterLength = delimiter.length();
+
+	std::string line;
+	while (std::getline(input, line))
+	{
+		auto delimiterPos = line.find(delimiter);
+		col1.emplace_back(std::stoi(line.substr(0, delimiterPos)));
+		col2.emplace_back(std::stoi(line.substr(delimiterPos + delimiterLength, line.length())));
+	}
+
+	for (auto number: col2)
+	{
+		if (occurrence_map.contains(number))
+		{
+			occurrence_map[number]++;
+		} else
+		{
+			occurrence_map[number] = 1;
+		}
+	}
+
+	for (auto number: col1)
+	{
+		sum += number * occurrence_map[number];
+	}
+	printf("Sum of all occurrence pairs is: %d\n", sum);
 }

days/day01.h

@@ -9,7 +9,8 @@
 class day01
 {
 public:
-	static void Run();
+	static void Part1();
+	static void Part2();
 };
 
 

input/day01short.txt

@@ -0,0 +1,3 @@
+66845   37619
+94793   99076
+76946   36179

main.cpp

@@ -1,9 +1,13 @@
+#include <chrono>
 #include <iostream>
 
 #include "days/day01.h"
 
 int main()
 {
-day01 day01;
-	day01.Run();
+	const auto startTime = std::chrono::high_resolution_clock::now();
+	day01::Part2();
+	const auto endTime = std::chrono::high_resolution_clock::now();
+	const std::chrono::duration<double> diff = endTime - startTime;
+	std::cout << "Execution time: " << diff.count() * 1e6 << " μs" << std::endl;
 }