Given a dictionary and a list of letters find all valid words that can be built with the letters

Question

The brute force way can solve the problem in O(n!), basically calculating all the permutations and checking the results in a dictionary. I am looking for ways to improve the com

Answer 1

Assume that letters only contains letters from a to z.

Use an integer array to count the number of occurrence of a character in letters.

For each word in the dictionary, check if there is a specific character in the word that appears more than allowed, if not, add this word into result.

    List findValidWords(List dict, char letters[]){
        int []avail = new int[26];
        for(char c : letters){
            int index = c - 'a';
            avail[index]++;
        }
        List result = new ArrayList();
        for(String word: dict){
            int []count = new int[26];
            boolean ok = true;
            for(char c : word.toCharArray()){
                int index = c - 'a';
                count[index]++;
                if(count[index] > avail[index]){
                    ok = false;
                    break;
                }
            }
            if(ok){
                result.add(word);
            }
        }
        return result;
    }

So we can see that the time complexity is O(m*k) with m is number of word in the dictionary and k is the maximum total of characters in a word