Maximum Product of Word

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.

Analyse: As seeing this problem, I first think of brute force. For each word in the array, find other words don't share common words with it. Time exceeded. 

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int result = 0;
        if(words.size() < 2) return result;
        
        for(int i = 0; i < words.size(); i++){
            for(int j = i + 1; j < words.size(); j++){
                if(!hasCommon(words[i], words[j])){
                    int temp = (words[i].length()) * (words[j].length());
                    result = max(result, temp);
                }
            }
        }
        return result;
    }
    
    bool hasCommon(string s1, string s2){
        if(s1.empty() || s2.empty()) return false;
        
        for(int i = 0; i < s1.length(); i++){
            for(int j = 0; j < s2.length(); j++){
                if(s2[j] == s1[i])
                    return true;
            }
        }
        return false;
    }
};

 Of course it will exceed time limit because the time complexity is O(n^4)....

Below I'll give a O(n^2) solution. 

class Solution {
public:
    int maxProduct(vector<string>& words) {
        if(words.size() < 2) return 0;
        
        int result = 0;
        vector<int> bits(words.size(), 0);
        for(int i = 0; i < words.size(); i++){
            for(int j = 0; j < words[i].size(); j++)
                bits[i] |= 1 << (words[i][j] - 'a');
        }
        
        for(int i = 0; i < words.size(); i++){
            for(int j = i; j < words.size(); j++){
                if(!(bits[j] & bits[i])){
                    int temp = words[i].length() * words[j].length(); 
                    result = max(result, temp);
                }
            }
        }
        return result;
    }
};