Maximum Product of Word Lengths

Maximum Product of Word Lengths

Total Accepted: 750 Total Submissions: 2060 Difficulty: Medium

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.

把每个单词转换成整数，利用与的性质，如果两个数没有同为1的二进制比特位，则与结果为零。

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int words_size = words.size();
        vector<int> bit_map(words_size,0);
        for(int i=0;i<words_size;i++){
            int word_size = words[i].size();
            for(int j=0;j<word_size;j++){
                bit_map[i] |= 1<<(words[i][j]-'a');
            }
        }
        int max_len = 0;
        for(int i=0;i<words_size;i++){
            for(int j=i+1;j<words_size;j++){
                if((bit_map[i] & bit_map[j]) ==0 ){
                    int isize = words[i].size() ;
                    int jsize = words[j].size() ;
                    max_len = max(max_len , isize*jsize );
                }
            }
        }
        return max_len;
    }
};

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