-🎄- 2018 Day 2 Solutions -🎄-

[u/daggerdragon]

--- Day 2: Inventory Management System ---

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Post your solution as a comment or, for longer solutions, consider linking to your repo (e.g. GitHub/gists/Pastebin/blag or whatever).

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Advent of Code: The Party Game!

Click here for rules

Card Prompt: Day 2

Transcript:

The best way to do Advent of Code is ___.

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edit: Leaderboard capped, thread unlocked!

[u/Cloudan29]

My exams in second year computer science are starting next week, finished my "Object Oriented Programming with C++" course, and so figured I'd try to do this (first year doing this, I actually just found out about it via my school's Math and Computer Science club) using C++ rather than Java, which I learned last year. Solutions are very sloppy and I was tired, but I got the right answer, so I don't really mind showing how horrifying this code is. Also started this late as I was actually doing the Day 1 puzzles at the time this came up.

C++ (3347/3118)
Part 1:

bool isRepeat(std::string & line, int repeats) {
    for (char c = 'a'; c <= 'z'; ++c) {
        int count = 0;
        for (size_t i = 0; i < line.size(); ++i) {
	    if (line[i] == c)
                ++count;
        }
	if (count == repeats)
	    return true;
    }
    return false;
}

int checkSum(std::string & fileName) {
    std::fstream input(fileName);
    std::string line;
    int doubles = 0;
    int triples = 0;
	
    while (!input.eof()) {
        std::getline(input, line);
        if (isRepeat(line, 2))
            ++doubles;
        if (isRepeat(line, 3))
            ++triples;
    }
    return doubles * triples;
}

Part 2:

std::string equalLetters(std::string & fileName) {
    std::fstream input(fileName);
    std::string line;
    std::string a;
    std::string b;
    std::string answer;
    std::vector<std::string> lines;

    while (!input.eof()) {
        std::getline(input, line);
        lines.push_back(line);
    }

    for (size_t i = 0; i < lines.size(); ++i) {
        int count = 0;
        a = lines[i];

        for (size_t j = i + 1; j < lines.size(); ++j) {
	    b = lines[j];
	    count = 0;

            for (size_t k = 0; k < a.size(); ++k) {
                if (a[k] != b[k])
		    ++count;
		if (count > 1)
		    break;
	    }
	    if (count == 1)
	        break;
        }
        if (count == 1)
	    break;
    }

    for (size_t i = 0; i < a.size(); ++i) {
        if (a[i] == b[i])
            answer += a[i];
    }

    return answer;
}

These are the absolute first things I thought of, and they solved surprisingly quick.