[LeetCode 318] 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.
Solution:
Use bit manipulation to mark if two words has shared character, then iterate to check max product

public int maxProduct(String[] words) {
        int len = words.length;
        if(len <=1 ) return 0;
        int[] mask = new int[len];
        for(int i=0;i<len;i++) {
            for(int j=0;j<words[i].length();j++) {
                mask[i] |= 1 << (words[i].charAt(j)-'a');
            }
        }
        int max = 0;
        for(int i=0;i<len;i++) {
            for(int j=i+1;j<len;j++) {
                if((mask[i] & mask[j]) == 0) {
                    max = Math.max(max, words[i].length() * words[j].length());
                }
            }
        }
        return max;
    }