Maximum Product of Word Lengths

Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.

Example 1:
Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be "abcw", "xtfn".

Example 2:
Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be "ab", "cd".

Example 3:
Given ["a", "aa", "aaa", "aaaa"]
Return 0
No such pair of words.

Bit Map

    public int maxProduct(String[] words) {
        int res = 0;
        if(words.length == 0) return 0;
        List<boolean[]> bml = new ArrayList();
        for(String w : words){
            boolean [] bitmap = new boolean[26];
            char [] wl = w.toCharArray();
            for(char c : wl){
                bitmap [c - 'a'] = true;
            }
            bml.add(bitmap);
        }

        for(int i = 0 ; i < bml.size() - 1; i ++){
            for(int j = i + 1; j < bml.size(); j ++){
                boolean canProduct = true;
                for(int k = 0 ; k < 26 ; k++){
                    if(bml.get(i)[k] && bml.get(j)[k]) {canProduct = false; break;}
                }
                if(canProduct) res = Math.max(res, words[i].length() * words[j].length());
            }
        }

        return res;
    }