public static int sherlockAndAnagrams(String s) {
        Map<String, Integer> occurrences = new HashMap<>();

        for (int i = 0; i < s.length(); i++) {
            for (int j = i; j < s.length(); j++) {
                String val = sortString(s.substring(i, j + 1));
                occurrences.put(val, occurrences.getOrDefault(val, 0) + 1);
            }
        }
        int anagramPairCount = 0;
        occurrences = removeSingleOccurrences(occurrences);
        for (Map.Entry<String, Integer> e : occurrences.entrySet()) {
            // Once occurrence ‘o’ of each frequency array is stored, total anagrams will be the sum of
            // o*(o-1)/2 for all different frequency arrays because if a particular substring has ‘o’ anagrams in
            // string total o*(o-1)/2 anagram pairs can be formed. Below is the implementation of above idea.

            // The division by two is needed to disregard different combinations between two elements:
            // [0, 1] and [1, 0]
            anagramPairCount += (e.getValue() * (e.getValue() - 1)) / 2;
        }
        return anagramPairCount;
    }

    private static Map<String, Integer> removeSingleOccurrences(Map<String, Integer> occurrences) {
        return occurrences
            .entrySet()
            .stream()
            .filter(e -> e.getValue() > 1)
            .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
    }

    private static String sortString(String str) {
        char[] charArray = str.toCharArray();
        Arrays.sort(charArray);
        return String.valueOf(charArray);
    }