192 lines
7.1 KiB
C++
192 lines
7.1 KiB
C++
//
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// Created by stedd on 01.12.24.
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//
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#include "day01.h"
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#include <fstream>
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#include <iostream>
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#include <ostream>
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#include <unordered_map>
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#include <vector>
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#include <bits/fs_fwd.h>
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#include <bits/fs_path.h>
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/*
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--- Day 1: Historian Hysteria ---
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The Chief Historian is always present for the big Christmas sleigh launch, but nobody has seen him in months! Last
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anyone heard, he was visiting locations that are historically significant to the North Pole; a group of Senior
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Historians has asked you to accompany them as they check the places they think he was most likely to visit.
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As each location is checked, they will mark it on their list with a star. They figure the Chief Historian must be in
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one of the first fifty places they'll look, so in order to save Christmas, you need to help them get fifty stars on
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their list before Santa takes off on December 25th.
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Collect stars by solving puzzles. Two puzzles will be made available on each day in the Advent calendar; the second
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puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!
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You haven't even left yet and the group of Elvish Senior Historians has already hit a problem: their list of locations
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to check is currently empty. Eventually, someone decides that the best place to check first would be the Chief
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Historian's office.
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Upon pouring into the office, everyone confirms that the Chief Historian is indeed nowhere to be found. Instead,
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the Elves discover an assortment of notes and lists of historically significant locations! This seems to be the
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planning the Chief Historian was doing before he left. Perhaps these notes can be used to determine which locations
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to search?
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Throughout the Chief's office, the historically significant locations are listed not by name but by a unique number
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called the location ID. To make sure they don't miss anything, The Historians split into two groups, each searching
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the office and trying to create their own complete list of location IDs.
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There's just one problem: by holding the two lists up side by side (your puzzle input), it quickly becomes clear
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that the lists aren't very similar. Maybe you can help The Historians reconcile their lists?
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For example:
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3 4
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4 3
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2 5
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1 3
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3 9
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3 3
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Maybe the lists are only off by a small amount! To find out, pair up the numbers and measure how far apart they are.
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Pair up the smallest number in the left list with the smallest number in the right list, then the second-smallest left
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number with the second-smallest right number, and so on.
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Within each pair, figure out how far apart the two numbers are; you'll need to add up all of those distances.
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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
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up a 9 with a 3, the distance apart is 6.
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In the example list above, the pairs and distances would be as follows:
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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.
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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.
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The third-smallest number in both lists is 3, so the distance between them is 0.
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The next numbers to pair up are 3 and 4, a distance of 1.
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The fifth-smallest numbers in each list are 3 and 5, a distance of 2.
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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.
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To find the total distance between the left list and the right list, add up the distances between all the pairs
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you found. In the example above, this is 2 + 1 + 0 + 1 + 2 + 5, a total distance of 11!
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Your actual left and right lists contain many location IDs. What is the total distance between your lists?
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*/
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void day01::Part1()
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{
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std::vector<int> col1;
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std::vector<int> col2;
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int sum = 0;
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std::ifstream input("input/day01.txt");
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if (!input)
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{
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std::cerr << "Failed to open input file." << std::endl;
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std::cout << "Current path is " << std::filesystem::current_path() << '\n';
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return;
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}
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std::string delimiter = " ";
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auto delimiterLength = delimiter.length();
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std::string line;
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while (std::getline(input, line))
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{
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auto delimiterPos = line.find(delimiter);
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col1.emplace_back(std::stoi(line.substr(0, delimiterPos)));
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col2.emplace_back(std::stoi(line.substr(delimiterPos + delimiterLength, line.length())));
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}
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std::sort(col1.begin(), col1.end());
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std::sort(col2.begin(), col2.end());
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for (int i = 0; i < col1.size(); ++i)
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{
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sum += std::abs(col1[i] - col2[i]);
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}
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}
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/*
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--- Part Two ---
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Your analysis only confirmed what everyone feared: the two lists of location IDs are indeed very different.
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Or are they?
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The Historians can't agree on which group made the mistakes or how to read most of the Chief's handwriting,
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but in the commotion you notice an interesting detail: a lot of location IDs appear in both lists! Maybe the other
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numbers aren't location IDs at all but rather misinterpreted handwriting.
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This time, you'll need to figure out exactly how often each number from the left list appears in the right
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list. Calculate a total similarity score by adding up each number in the left list after multiplying it by the number
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of times that number appears in the right list.
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Here are the same example lists again:
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3 4
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4 3
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2 5
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1 3
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3 9
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3 3
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For these example lists, here is the process of finding the similarity score:
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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.
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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.
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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).
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The fourth number, 1, also does not appear in the right list.
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The fifth number, 3, appears in the right list three times; the similarity score increases by 9.
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The last number, 3, appears in the right list three times; the similarity score again increases by 9.
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So, for these example lists, the similarity score at the end of this process is 31 (9 + 4 + 0 + 0 + 9 + 9).
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Once again consider your left and right lists. What is their similarity score?
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*/
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void day01::Part2()
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{
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std::vector<int> col1;
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std::vector<int> col2;
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std::unordered_map<int, int> occurrence_map;
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int sum = 0;
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std::ifstream input("input/day01.txt");
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//std::ifstream input("input/day01short.txt");
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if (!input)
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{
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std::cerr << "Failed to open input file." << std::endl;
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std::cout << "Current path is " << std::filesystem::current_path() << '\n';
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return;
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}
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std::string delimiter = " ";
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auto delimiterLength = delimiter.length();
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std::string line;
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while (std::getline(input, line))
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{
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auto delimiterPos = line.find(delimiter);
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col1.emplace_back(std::stoi(line.substr(0, delimiterPos)));
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col2.emplace_back(std::stoi(line.substr(delimiterPos + delimiterLength, line.length())));
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}
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for (auto number: col2)
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{
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occurrence_map[number]++;
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}
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for (auto number: col1)
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{
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sum += number * occurrence_map[number];
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}
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printf("Sum of all occurrence pairs is: %d\n", sum);
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}
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