def sherlockAndAnagrams(s)
    substrings = Hash.new { |h,k| h[k] = 0 }
    (0..s.length - 1).each do |i|
        (i..s.length - 1).each do |j|
            sorted = s[i..j].chars.sort.join
            substrings[sorted] += 1
        end
    end
        
    substrings.values.
        filter { |f| f > 1 }.
        map { |k| k * (k - 1) / 2 }.
        sum
end

int sherlockAndAnagrams(std::string s) {
    std::map<std::map<char,int>, int> unique_strings;
    int result = 0;
    for (size_t i = 0; i<s.size(); ++i){
        std::map<char,int> submap;
        for (size_t j = i; j<s.size(); ++j) {
            submap[s[j]]++;
            unique_strings[submap]++;
        }
    }
    for (auto& [substr, count]: unique_strings) {
        result += (count - 1)* count / 2;
    }
    return result;
}

fun sherlockAndAnagrams(s: String): Int {
    var windowedSize = s.length - 1 // define possible windowed steps for that string
    var listOfWindows : List<List<Char>> = listOf()
    var anagramsCounter = 0
    
    // itterate based on possible windowed steps
    while(windowedSize > 0){
        // get all windowed sets
        listOfWindows = s.toList().windowed(size = windowedSize, step = 1) { it ->
            it.sortedBy { it }
        }
        // count all anagrams
        listOfWindows.forEachIndexed{ index, window ->
            for(j in (index + 1) .. listOfWindows.lastIndex) {
                if(window == listOfWindows[j])  
                    anagramsCounter++
            }
        }
        windowedSize-- // reduce the windowed size
    }
    return anagramsCounter
}

import itertools
def sherlockAndAnagrams(s):
    return sum([
    len([
        pair
        for pair in itertools.combinations([s[i@i+l] for i in range(len(s) - l + 1)], 2)
        if sorted(pair[0]) == sorted(pair[1])
    ])
    for l in range(1, len(s))
    ])