Add solution for "Day 5: Print Queue", part 2
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@ -13,6 +13,7 @@
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// You should have received a copy of the GNU General Public License along with
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// this program. If not, see <http://www.gnu.org/licenses/>.
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#include <algorithm>
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#include <sstream>
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#include <vector>
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@ -77,7 +78,8 @@ void PrintQueue::processUpdatePages(const std::string& line)
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std::stringstream stream{ line };
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std::string token;
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auto isCorrectOrder{ true };
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while (isCorrectOrder && std::getline(stream, token, ','))
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// We completely construct 'pages' for part 2, even if the ordering is not correct.
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while (std::getline(stream, token, ','))
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{
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size_t page = std::stoi(token);
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size_t i{ 0 };
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@ -86,14 +88,19 @@ void PrintQueue::processUpdatePages(const std::string& line)
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isCorrectOrder = !orderingRules_[page][pages[i]];
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i++;
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}
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if (isCorrectOrder)
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{
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pages.push_back(page);
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}
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pages.push_back(page);
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}
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if (isCorrectOrder)
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{
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part1 += pages[pages.size() / 2];
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}
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else
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{
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// This works because the input defines a complete ordering on the occurring
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// page numbers.
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std::sort(pages.begin(), pages.end(),
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[&](int const& a, int const& b) { return orderingRules_[a][b]; });
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part2 += pages[pages.size() / 2];
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}
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}
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@ -38,6 +38,12 @@ A simple [finite state machine](AdventOfCode2024/StringStateMachine.h) crawling
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For this puzzle I added a class for [points in two-dimensional space](AdventOfCode2024/Point2.h), so I can use these for simplifying directional computations. With that, the algorithm looks for all `X` and `A` for part 1 and 2, respectively, and tries to find the remaining characters, starting from that `X` or `A`.
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### Day 5: Print Queue
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:mag_right: Puzzle: <https://adventofcode.com/2024/day/5>, :white_check_mark: Solver: [`PrintQueue.cpp`](AdventOfCode2024/PrintQueue.cpp)
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My implementation uses an ordering matrix (a two-dimensional boolean array) to track which page combinations are ordered, and then queries that matrix for each ordered combination of pages in a single line. The same matrix can then also be used for a custom sort function for part 2.
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## License
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Copyright (C) 2024 Stefan Müller
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