30. Substring with Concatenation of All Words

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

2 pointer one stand the first letter, the other go through

    public IList<int> FindSubstring(string s, string[] words) {
        var res = new List<int>();
        if(words.Count()==0) return res;
        var hashtable = new Dictionary<string,int>();
        foreach(var word in words)
        {
            if(hashtable.ContainsKey(word)) hashtable[word] +=1;
            else hashtable.Add(word,1);
        }
        int wordSize = words[0].Length;
        
        for(int j =0;j<= s.Length - words.Count()*wordSize;j++)
        {
            var exist = new Dictionary<string,int>();
            int i=j;
            int count =0;
            while(i<=(s.Length - wordSize))
            {
                 string newWord = s.Substring(i,wordSize);
                 if(hashtable.ContainsKey(newWord))
                 {
                     
                     if(exist.ContainsKey(newWord))
                     {
                         if(exist[newWord]<hashtable[newWord])
                         {
                             exist[newWord]++;
                             count++;
                             i += wordSize;
                             if(count ==words.Count())
                             {
                                 res.Add(j);
                                 break;
                             }
                         }
                         else  break;
                     }
                     else
                     {
                         exist.Add(newWord,1);
                         count++;
                         if(count ==words.Count())
                        {
                                 res.Add(j);
                                 break;
                        }
                         i += wordSize;
                     }
                 }
                 else break;
            }
        }
        return res;
    }