# 318. Maximum Product of Word Lengths

#### Medium

***

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, return `0`.

 

**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.
```

 

**Constraints:**

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

```python
class Solution:
    def maxProduct(self, words: List[str]) -> int:
        bits = [0]*len(words)
        for index, word in enumerate(words):
            for char in list(word):
                # Here we are simply record occurence of each char in bit place
                # For Example "a" will set 0001 "b" will set 0010, each character will set its own place in 
                # 26 length bit array and when we perform AND operation we will get 0 if no character matches.
                bits[index] |= (1 << (ord(char) - ord('a')))
        product = 0
        for i in range(len(words)):
            for j in range(i, len(words)):
                if (bits[i] & bits[j]) == 0 and len(words[i])*len(words[j]) > product:
                    product = len(words[i])*len(words[j])
        return product
```