AdventOfCode2017 – Day 4

Hooray, another easy one.  After Day number 3, I could certainly use it.  This challenge involved taking a list of passphrases, and counting up the number of passphrases that had no duplicate words.   This seems simple, just a split, sort and application of std::unique.

Part 2 had me check for anagrams rather than duplicate words.  This was also easy, as I could map over a sort function to each word in the passphrase, and then check for uniqueness.

Let’s look at the code


#include <algorithm>
#include <iostream>
#include <string>

#include "algo.h"
#include "input.h"

std::string sortString(std::string s) {
    std::sort(s.begin(), s.end());
    return s;
}

bool isUnique(const std::string& passphrase) {
    auto words = algo::map(input::split(passphrase), sortString);
    std::sort(words.begin(), words.end());
    return words.end() == std::unique(words.begin(), words.end());
}

int main() {
    auto passPhrases = input::readMultiLineFile("input/input04.txt");
    auto count = std::count_if(passPhrases.begin(), passPhrases.end(), isUnique);
    std::cout << count << "\n";
    return 0;
}

Super straight forward.  I think the trickiest thing was std::unique, because I didn’t realize it returned the end iterator of the range.  But once I figured that out, this wasn’t so bad.

Stay tuned for day 5!

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