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.

class Solution {
    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;  
    }
}