Hackerrank Sherlock and Anagrams

sherlockanagrams.js

// I opened the page one day and wrote the solution on another. I had forgotten it was in the dict/hashmap section...

const sum = (acc, next) => acc + next;
const matches = (val, i, arr) => {
    let count = 0;
    for (++i; i < arr.length; i++) {
        if (val == arr[i]) count++;
    }
    return count;
};

// Complete the sherlockAndAnagrams function below.
function sherlockAndAnagrams(s) {
  let strs = [];

  // get substring permutations
  for(let chunkSize = 1; chunkSize < s.length; chunkSize++) {
    let temp = [];

    // make array of sorted chunkSize substrings
    for(let i = 0; i <= s.length - chunkSize; i++) {
      temp.push(s.substr(i, chunkSize).split('').sort().join(''));
    }
    // count pairs
    temp = temp.map(matches)
    // add total
    strs.push(temp.reduce(sum));
  }

  return strs.reduce(sum)
}

It was 2 years ago and I don't really remember writing it, so I couldn't tell you what my thought process was at the time. I can try to work through what I may have been thinking:

• The problem statement is "Two strings are anagrams of each other if the letters of one string can be rearranged to form the other string. Given a string, find the number of pairs of substrings of the string that are anagrams of each other."
  So we've got 2 things to do: generate all the substrings, and determine which of them are anagrams
• As the comment on line 16 notes, I started with 'get substring permutations'. I have a window of chunkSize (substring length) that increases from 1, and I slide the window along the array (of characters, aka a string) saving each as I go.
• The .sort() on ln 22 is in anticipation of the other part of the problem. Sorted arrays or strings are a) easier to compare and b) meet the anagram requirement when identical. Consider the pair 'abb' and 'bba' from the sample data. When sorted, they are 'abb' and 'abb'; identical strings.
• on ln 25 im counting the pairs for the current chunk size. At this point temp is a list of sorted substrings of the current chunk size. matches() maps each substring to the number of matching substrings i've stored (in other words, the number of pairs). So if temp is ['a', 'b', 'b', 'a'], it maps to [1, 1, 0, 0]
• ln 27 just adds those pair counts together (again, for only the current chunk size). So [1, 1, 0, 0] reduces to 2
• at ln 30 i'm left with a list of pair counts for each chunk size, so again those number are added together to leave me with the total pair count

Hopefully that helps. Happy to elaborate as needed, when/if I have the free time to do so. It's worth noting that, as far as I can tell, I made no attempt at achieving efficiency. It looks like I got something that works and just left it at that. Readability may also be a concern.