# 318. Maximum Product of Word Lengths

## 1. Question

Given a string array words, return the maximum value of length(word[i]) length(word[j])where the two words do not share common letters.* If no such two words exist, return0.

## 2. Examples

Example 1:

Input: words = ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16
Explanation: The two words can be "abcw", "xtfn".


Example 2:

Input: words = ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4
Explanation: The two words can be "ab", "cd".


Example 3:

Input: words = ["a","aa","aaa","aaaa"]
Output: 0
Explanation: No such pair of words.


## 3. Constraints

• 2 <= words.length <= 1000
• 1 <= words[i].length <= 1000
• words[i] consists only of lowercase English letters.

## 5. Solutions

class Solution {
public int maxProduct(String[] words) {
int len = words.length;
int[] arr = new int[len];

for (int i = 0; i < len; i++) {
char[] ch = words[i].toCharArray();

for (char c : ch) {
// 此处是一个异或运算，用二进制标志是否已经有此字母。
// c - 'a' 计算出要位移的个数，将1进行左移，并与对应的int值进行或运算。
// ac --> 00000...00101
arr[i] |= 1 << (c - 'a');
}
}

int res = 0;
for (int i = 0; i < len; i++) {
for (int j = i + 1; j < len; j++) {
// 与运算
if ((arr[i] & arr[j]) == 0) {
res = Math.max(res, words[i].length() * words[j].length());
}
}
}
return res;
}
}