understanding solution of Valid Anagram in javascript

Question

This is from LeetCode - Valid Anagram

Given two strings s and t , write a function to determine if t is an anagram of s.
Example 1:
Input: s = "anagram", t = "nagaram"
Output: true
Example 2:

You may assume the string contains only lowercase alphabets.

Follow up: What if the inputs contain unicode characters? How would you adapt your solution to such case?

I don't understand these code below

• result1[s.charCodeAt(i) - 97]++; --> what does ++ mean?
• result2.length = 26; --> what does 26 stand for?
• result2.fill(0); --> why does fill it with 0?

Please advise!

var isAnagram = function(s,t) {
    if (s.length !== t.length)
        result false;
    const result1 = [];
    result1.length = 26;
    result1.fill(0);

    const result2 = [];
    result2.length = 26;
    result2.fill(0);

    for (let i = 0; i < s.length; i++) {
        result1[s.charCodeAt(i) - 97]++;
        result2[t.charCodeAt(i) - 97]++;
    }

    for (let i = 0; i < result1.length; i++) {
        if (result1[i] !== result2[i]) {
            return false;
        }
    }
    return true;
};

Answer 1

That's somewhat poorly written code, let's improve it a bit so that your question are automatically gone:

var isAnagram = function(string1, string2) {
    if (string1.length !== string2.length)
        return false; // here was a typo: result false

    const alphabetLength = 26;
    // returns an array of zeros (empty counters), each one per alphabet letter
    const getAlphabetCounters = function() {
        const counters = [];
        counters.length = alphabetLength;
        counters.fill(0);
        return counters;
    }
    const countCharacter = function(c, counters) {
        // zero for a, 25 for z
        const position = c.charCodeAt(0) - 97;
        counters[position]++;
    }

    const counters1 = getAlphabetCounters();
    const counters2 = getAlphabetCounters();

    for (let i = 0; i < string1.length; i++) {
        countCharacter(string1[i], counters1);
        countCharacter(string2[i], counters2);
    }

    for (let i = 0; i < counters1.length; i++) {
        if (counters1[i] !== counters2[i]) {
            return false;
        }
    }
    return true;
};

But it would be probably a better idea to use a "decremental" approach like this:

var isAnagram = function(string1, string2) {
    if (string1.length !== string2.length)
        return false;

    let letters1 = string1.split('');
    let letters2 = string2.split('');
    for (let letter of letters1) {
        let position = letters2.indexOf(letter);
        if(position == -1)
            return false;
        letters2.splice(position, 1);
    }
    return true;
};

or if one cares about performance for long strings, sorting letters in those and direct comparison would be the way to go.