I wanted to practice functional programming (FP) without using any library but using vanilla JS only. So I took a problem from Advent of Code (the 2nd part of Day 4):

https://adventofcode.com/2017/day/4

You can only access the 2nd part if you type in the solution for the 1st part: 466 or look at my solution for part 1: Advent Of Code 2017 Day 4 (part 1) in Functional Programming (FP)

--- Part Two --- For added security, yet another system policy has been put in place. Now, a valid passphrase must contain no two words that are anagrams of each other - that is, a passphrase is invalid if any word's letters can be rearranged to form any other word in the passphrase.

For example:

• abcde fghij is a valid passphrase.
• abcde xyz ecdab is not valid - the letters from the third word can be rearranged to form the first word.
• a ab abc abd abf abj is a valid passphrase, because all letters need to be used when forming another word.
• iiii oiii ooii oooi oooo is valid.
• oiii ioii iioi iiio is not valid - any of these words can be rearranged to form any other word. Under this new system policy, how many passphrases are valid?

My solution in FP:

const INPUT =
`pphsv ojtou brvhsj cer ntfhlra udeh ccgtyzc zoyzmh jum lugbnk
spjb xkkak anuvk ejoklh nyerw bsjp zxuq vcwitnd xxtjmjg zfgq xkpf
...
juo pmiyoh xxk myphio ogfyf dovlmwm moevao qqxidn`;

const get = input => input.split('\n');
const countDuplicate = words => words.reduce((acc, word) => {
  return Object.assign(acc, {[word]: (acc[word] || 0) + 1});
}, {});
const onlyUniqueWords = phrases => {
  const words = phrases.split(' ');
  const duplicateWords = countDuplicate(words);
  return !Object.values(duplicateWords).some(w => w > 1);
};
const sortWords = words => words.map(word =>  word.split('').sort().join(''));
const noAnagrams = phrases => {
  const words = phrases.split(' ');
  const sortedWords = sortWords(words).sort().join(' ');
  return onlyUniqueWords(sortedWords);
};
const phrasesWithNoAnagrams = get(INPUT)
.filter(noAnagrams);
console.log("solution ", phrasesWithNoAnagrams.length);

Is there a better way to write it in FP with pure JavaScript, i.e. no additional FP library? Any other improvement suggestions are welcomed.

I wanted to practice functional programming (FP) without using any library but using vanilla JavaScript only. So I took a problem from Advent of Code (the 2nd part of Day 4)

You can only access the 2nd part if you type in the solution for the 1st part: 466 or look at my solution for part 1

--- Part Two --- For added security, yet another system policy has been put in place. Now, a valid passphrase must contain no two words that are anagrams of each other - that is, a passphrase is invalid if any word's letters can be rearranged to form any other word in the passphrase.

For example:

• abcde fghij is a valid passphrase.
• abcde xyz ecdab is not valid - the letters from the third word can be rearranged to form the first word.
• a ab abc abd abf abj is a valid passphrase, because all letters need to be used when forming another word.
• iiii oiii ooii oooi oooo is valid.
• oiii ioii iioi iiio is not valid - any of these words can be rearranged to form any other word. Under this new system policy, how many passphrases are valid?

My solution in FP:

const INPUT =
`pphsv ojtou brvhsj cer ntfhlra udeh ccgtyzc zoyzmh jum lugbnk
spjb xkkak anuvk ejoklh nyerw bsjp zxuq vcwitnd xxtjmjg zfgq xkpf
...
juo pmiyoh xxk myphio ogfyf dovlmwm moevao qqxidn`;

const get = input => input.split('\n');
const countDuplicate = words => words.reduce((acc, word) => {
  return Object.assign(acc, {[word]: (acc[word] || 0) + 1});
}, {});
const onlyUniqueWords = phrases => {
  const words = phrases.split(' ');
  const duplicateWords = countDuplicate(words);
  return !Object.values(duplicateWords).some(w => w > 1);
};
const sortWords = words => words.map(word =>  word.split('').sort().join(''));
const noAnagrams = phrases => {
  const words = phrases.split(' ');
  const sortedWords = sortWords(words).sort().join(' ');
  return onlyUniqueWords(sortedWords);
};
const phrasesWithNoAnagrams = get(INPUT)
.filter(noAnagrams);
console.log("solution ", phrasesWithNoAnagrams.length);

Is there a better way to write it in FP with pure JavaScript, i.e. no additional FP library? Any other improvement suggestions are welcomed.