LeetCode 318. Maximum Product of Word Lengths-优快云博客

本文探讨如何通过位操作来检测两个仅包含小写字母'a'-'z'的单词是否共用字母，通过比较两个单词对应的位表示，若无共用字母，则位与运算结果为0。

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The tough part here is on how to detect whether two words sharing common letters.

As indicate in the question, each word only contains lower case letters 'a' - 'z', we can actually correlate this problem with "in a given string, detect whether there is any duplicates". The difference here is to detect duplicates between two strings.

thus, we can use idea that if two words contains no common letters, their & operation result should be 0.

Example:

['"abc", "de"] -- > [0x111, 0x11000], thus, 0x111 & 0x11000 -> 0.

// I originally thought about to use unordered_map<int, int> to store the getBits and string size. However, it is easy to make problems.
int getBits(string& str) {
    int bits = 0;
    for(char c : str) {
        bits |= (1 << (c - 'a'));
    }
    return bits;
}
int maxProduct(vector<string>& words) {
    int result= 0;
    int size = words.size();
    vector<int> m(size, 0);
    vector<int> n(size, 0);
    for(int i = 0; i < size; ++i) {
        m[i] = getBits(words[i]);
        n[i] = words[i].size();
    }
    // seems like there is no better method to get the max.....
    for(int i = 0;  i < size; ++i) {
        for(int j = i + 1; j < size; ++j) {
            // there is no common letters.
            if((m[i] & m[j]) == 0) { 
                result = max(n[i] * n[j], result);
            }
        }
    }
    return result;
}