# LeetCode Maximum Product of Word Lengths

LeetCode 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(vector<string>& words) {
int ans = 0;
for (int i = 0; i < words.size(); ++i) {
for (int j = 0; j < words[i].size(); ++j) {
mask[i] |= 1 << (words[i][j] - 'a');
}
for (int k = 0; k < i; ++k) {
if ((mask[i] & mask[k]) == 0) // 注意 == 的优先级高于 &
ans = max(ans, int(words[i].size()*words[k].size()));
}
}
return ans;
}
};
```