We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Strings
  4. Sherlock and Anagrams
  5. Discussions

Sherlock and Anagrams

  • carlos_contreras
     + 1 comment

    WOW!! I really liked your solution. I made a shorter version based on your code:

    public static int sherlockAndAnagrams(string s)
{
   int counter = 0;

   for (int x = 1; x <= s.Length - 1; x++)
   {
      Dictionary<string, int> lista = new Dictionary<string, int>();

      for (int y = 0; y <= s.Length - x; y++)
      {
         var cadena = new string(s.Substring(y, x).OrderBy(c => c).ToArray());
         if (!lista.ContainsKey(cadena))
            lista.Add(cadena, 1);
         else
            lista[cadena] += 1;
      }

      lista.Values.Where(w => w > 1).ToList().ForEach(v =>
      {
         counter += v * (v - 1) / 2;
      });
   }

   return counter;
}