Leetcode 49 Group anagrams

Providing not the fastest solution, but normalization of anagrams is fast enough, otherwise go with the hash function as alternative:

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        // Another possible alternative is to hash the string by 
        // assigning an integer to each letter
        // better if each integer is prime, shoudl definitely not be 
        // one, and multiply all the letter values
        // Store in HashMap<Long, List<String>> and at the end 
        // convert to List<List<String>>
        
        Map<String, List<String>> res = new HashMap<>();
        int n = strs.length;
        for (int i = 0; i < n; i++) {
            String s = strs[i];
            String norm = normalize(s);
            boolean added = false;
            if (res.containsKey(norm)) {
                res.get(norm).add(s);
            } else {
                res.put(norm, new ArrayList<>());
                res.get(norm).add(s);
            }
        }
        List<List<String>> lstRes = new ArrayList<>();
        for (String norm : res.keySet()) {
            lstRes.add(res.get(norm));
        }
        return lstRes;
        
        // alternative solution 2 (slow, map of map)
        // Map<Map<Character, Integer>, List<String>> res = 
        //     new HashMap<>();
        // int n = strs.length;
        // for (int i = 0; i < n; i++) {
        //     Map<Character, Integer> map = count(strs[i]);
        //     boolean added = false;
        //     for (Map<Character, Integer> m : res.keySet()) {
        //         if…