Maximum Product of Word Lengths
<p>Given a string array <code>words</code>, return <em>the maximum value of</em> <code>length(word[i]) * length(word[j])</code> <em>where the two words do not share common letters</em>. If no such two words exist, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;abcw&quot;,&quot;baz&quot;,&quot;foo&quot;,&quot;bar&quot;,&quot;xtfn&quot;,&quot;abcdef&quot;]
<strong>Output:</strong> 16
<strong>Explanation:</strong> The two words can be &quot;abcw&quot;, &quot;xtfn&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;ab&quot;,&quot;abc&quot;,&quot;d&quot;,&quot;cd&quot;,&quot;bcd&quot;,&quot;abcd&quot;]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The two words can be &quot;ab&quot;, &quot;cd&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;aa&quot;,&quot;aaa&quot;,&quot;aaaa&quot;]
<strong>Output:</strong> 0
<strong>Explanation:</strong> No such pair of words.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= words.length &lt;= 1000</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 1000</code></li>
	<li><code>words[i]</code> consists only of lowercase English letters.</li>
</ul>
Bit Manipulation; Array; String
class Solution {
public:
    int maxProduct(vector<string>& words) {
        int n = words.size();
        int mask[n];
        memset(mask, 0, sizeof(mask));
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (char& c : words[i]) {
                mask[i] |= 1 << (c - 'a');
            }
            for (int j = 0; j < i; ++j) {
                if ((mask[i] & mask[j]) == 0) {
                    ans = max(ans, (int) (words[i].size() * words[j].size()));
                }
            }
        }
        return ans;
    }
};