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.

Straight forward solution...

class Solution {
public:
    bool shareCommon(string s1, string s2) {
        sort(s1.begin(), s1.end());
        sort(s2.begin(), s2.end());
        int i1 = 0, i2 = 0;
        int n1 = s1.size(), n2 = s2.size();
        while ( i1 < n1 and i2 < n2 ) {
            if ( s1[i1] == s2[i2] ) return true;
            if ( s1[i1] > s2[i2] ) i2++;
            else i1++;
        }
        return false;
    }
    int maxProduct(vector<string>& words) {
        int n = words.size();
        int res = 0;
        for ( int i = 0; i < n-1; i++ ) {
            for ( int j = i+1; j < n; j++ ) {
                if ( not shareCommon(words[i], words[j]) ) res = max(res, int(words[i].size() * words[j].size()));
            }
        }
        return res;
    }
};

Bit-manipulation solution

class Solution {
public:
    int toInt(string s) {
        int n = s.size();
        int res = 0;
        for ( int i = 0; i < n; i++ ) res |= (1 << s[i]-'a');
        return res;
    }
    int maxProduct(vector<string>& words) {
        int n = words.size(), res = 0;
        vector<int> nums;
        for ( int i = 0; i < n; i++ ) nums.push_back(toInt(words[i]));
        for ( int i = 0; i < n-1; i++ ) {
            for ( int j = i+1; j < n; j++ ) {
                if ( (nums[i]&nums[j]) == 0 ) res = max(res, int(words[i].size()*words[j].size()));
            }
        }
        return res;
    }
};