class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        List<List<String>> result = new ArrayList<>();


        Map<String, List<String>> map = new HashMap<>();
        // iterating over each string from giver array of strings
        for(String str : strs)
        {
            // convert the current string to charArray, and store hash values
            // suppose str is : "cat"
            int[] count = new int[26];
            for(char ch : str.toCharArray())
            {
                count[ch - 'a']++;
            }
            // count[0]=1 --> for 'a', count[2]=1 --> for 'c', count[19]=1 --> for 't'

            // convert count array into string
            StringBuilder sb = new StringBuilder();
            for(int index : count)
            {
                // adding hashes in between the count of characters
                sb.append("#").append(index);
            }
            // sb: "#1#0#1#0#0#0#0#0#0#0#0#0#0#0#0#0#0#0#0#1#0#0#0#0#0#0"

            // convert string builder to string
            String key = sb.toString();
            // key = "#1#0#1#0#0#0#0#0#0#0#0#0#0#0#0#0#0#0#0#1#0#0#0#0#0#0"
            if(!map.containsKey(key))
                map.put(key, new ArrayList<String>());

            // Always add the string to the list, whether the key is new or not.
            // so suppose this key : {cat, act}, adding current string as anagram
            map.get(key).add(str);
        }

        result.addAll(map.values());
        return result;
    }
}