Solution for counting number of lines to anagrams in a list, done recursively

count-anagrams.js

'use strict';

/**

* Count the number of anagrams in a list of words.

*

* An anagram is a word made up of the same letters as another word in the array (but not

* necessarily in the same order).

*

* I like to think of it like this: Draw a line connecting each pair of words that contain exactly

* the same letters. For example, with this list of words:

*

* TAR RAT BOB ART

*

* You would end up with 3 connecting lines (TAR---RAT, TAR---ART, RAT---ART). So the result is 3.

*

* @param {string[]} wordsList - The list of words

*

* @return {number} The number of words which are anagrams of other words

*/

const isAnagram = (a, b) => a.length === b.length
  && a
    .toLowerCase()
    .split('')
    .every(character => b.toLowerCase().includes(character));

const countAnagrams = (wordsList) => {
  // base case
  if (wordsList.length < 2) {
    return 0;
  }

  const lastElement = wordsList.pop();

  let lineCount = 0;

  for (let i = 0; i < wordsList.length; i++) {
    if (isAnagram(lastElement, wordsList[i])) {
      lineCount++;
    }
  }

  return lineCount + countAnagrams(wordsList);
};

// helper tests
// console.log('true:', isAnagram('Pot', 'Top'));

// console.log('false:', isAnagram('Foo', 'Tom'));

console.log('Zero:', countAnagrams([]));

console.log('Zero:', countAnagrams(['Top']));

console.log('Zero:', countAnagrams(['Top', 'Spot']));

console.log('One:', countAnagrams(['Top', 'Pot']));

console.log('Two:', countAnagrams(['Top', 'Pot', 'Spot', 'Stop', 'Foo']));

console.log('Three:', countAnagrams(['Top', 'Pot', 'Opt']));

console.log(
  'Nine:',
  countAnagrams(['Top', 'Pot', 'Opt', 'Spot', 'Stop', 'Tops', 'Opts']),
);