Leetcode: 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.

 1 public class Solution {
 2     public int maxProduct(String[] words) {
 3         if (words==null || words.length==0) return 0;
 4         int max = 0;
 5         int[] bitmasks = new int[words.length]; 
 6         Arrays.sort(words, new Comparator<String>(){
 7             public int compare(String s1, String s2) {
 8                 return s2.length()-s1.length();
 9             }
10         });
11         
12         for (int i=0; i<words.length; i++) {
13             for (int j=0; j<words[i].length(); j++) {
14                 bitmasks[i] |= 1<<((int)(words[i].charAt(j) - ‘a‘));
15             }
16         }
17         
18         for (int i=0; i<words.length-1; i++) {
19             for (int j=i+1; j<words.length; j++) {
20                 if (words[i].length() * words[j].length() <= max) break; //pruning
21                 if ((bitmasks[i] & bitmasks[j]) != 0) continue;
22                 max = words[i].length() * words[j].length();
23                 break;//pruning
24             }
25         }
26         return max;
27     }
28 }