No `sort()` anagram tester

index.js

// Takes two strings.
// Returns true if they're anagrams, false if not.

function isAnagram (a, b) {
  const remove = (string, pattern) => string.replace(new RegExp(pattern, 'ig'), '');

  // Discard whitespace; anagrams can be different numbers of words
  let lettersA = remove(a, '\\s');
  let lettersB = remove(b, '\\s');

  while (lettersA.length && lettersA.length === lettersB.length) {
    const letter = lettersA[0];

    lettersA = remove(lettersA, letter);
    lettersB = remove(lettersB, letter);
  }

  return lettersA === lettersB;
}

Fun coding exercise suggested by Lara Schenck on Twitter: https://twitter.com/laras126/status/929780884835745792.

How my solution works:

1. Remove all whitespace. Anagrams can be different numbers of words.
2. Loop!
   1. If either string is empty or they're not equal length, exit the while loop.
   2. Remove all instances of the first letter in lettersA from both strings. Repeat the loop.
3. Compare strings. If the words are anagrams, both will be '', an empty string.