I found it a lot easier to think about sorting substrings and using a map so that the key for anograms are the same : E.g. ab & ba both have key ab after sorting

 // Complete the sherlockAndAnagrams function below.
static int sherlockAndAnagrams(String s) {
    HashMap<String, Integer> map = new HashMap<String, Integer>();

    // Keep track of how many anagrams we've seen
    int totalCount = 0;

    // Generate all substrings (N^2)
    for(int i = 0 ; i < s.length(); i++)
    {
        for(int j=i+1 ; j <= s.length(); j++)
        {
            String currentSubString = s.substring(i,j);

            // Sort all strings E.g. ab & ba both == ab now
            char[] chars = currentSubString.toCharArray();
            Arrays.sort(chars);
            currentSubString = String.valueOf(chars);

            // If sorted substring has been seen before
            if(map.containsKey(currentSubString)) 
            {
                // Check how many times we've seen it and add that amount to the count
                int value = map.get(currentSubString);
                totalCount=totalCount+value;

                // Increment the times we've seen the string
                map.put(currentSubString, value+1);
            } 
            else 
            {
                // Never seen it before = insert and set to 1 to indiciate we've now seen it
                map.put(currentSubString, 1);
            }
        }
    }
    return totalCount;
}