Add solution for "Day 21: Keypad Conundrum", part 1

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2025-06-05 23:31:30 +02:00
parent c67bb9054a
commit 60798118ea
6 changed files with 212 additions and 0 deletions

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src/extra/KeypadRobot.cpp Normal file
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// Solutions to the Advent Of Code 2024.
// Copyright (C) 2025 Stefan Müller
//
// This program is free software: you can redistribute it and/or modify it under
// the terms of the GNU General Public License as published by the Free Software
// Foundation, either version 3 of the License, or (at your option) any later
// version.
//
// This program is distributed in the hope that it will be useful, but WITHOUT
// ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS
// FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.
//
// You should have received a copy of the GNU General Public License along with
// this program. If not, see <http://www.gnu.org/licenses/>.
#include <aoc/extra/KeypadRobot.hpp>
#include <sstream>
KeypadRobot::KeypadRobot(const std::map<char, Point2>&& keypad, const Point2&& forbidden)
: keypad_{ keypad }, forbidden_{ forbidden }
{
}
std::string KeypadRobot::calcInputKeys(const std::string& targetOutputKeys) const
{
std::ostringstream stream{};
Point2 position{ 0, 0 };
for (const char c : targetOutputKeys)
{
Point2 next = keypad_.at(c);
// This specific order of robot arm movements aims to reduce resulting combinations of 'A' and '<' for the
// second robot, which expand to more key presses starting with the third robot.
bool horizontalFirst{ (next.x < position.x && !(next.x == forbidden_.x && position.y == forbidden_.y)) ||
(position.x == forbidden_.x && next.y == forbidden_.y) };
if (horizontalFirst)
{
move(stream, next.x - position.x, '>', '<');
}
move(stream, next.y - position.y, 'v', '^');
if (!horizontalFirst)
{
move(stream, next.x - position.x, '>', '<');
}
stream << 'A';
position = next;
}
return stream.str();
}
void KeypadRobot::move(std::ostringstream& stream, const int delta, const char positive, const char negative) const
{
if (delta > 0)
{
for (int i{ 0 }; i < delta; i++)
{
stream << positive;
}
}
else
{
for (int i{ 0 }; i < -delta; i++)
{
stream << negative;
}
}
}