<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>

class Solution {
public:
    bool valid(vector<bool> a, vector<bool> b) {
        for (int i=0;i<26;i++)
            if (a[i] && b[i])
                return false;

        return true;
    }
    int max (int a, int b) {
        return a > b ? a:b;
    }
    int maxProduct(vector<string>& words) {
        int ans = 0;
        vector<vector<bool>> data(words.size(),vector<bool>(26,false));
        for (int i=0;i<words.size();i++)
        {
            for (int j=0;j<words[i].size();j++)
                data[i][words[i][j]-'a'] = true;
        }
        for (int i=0;i<words.size();i++)
        {
            for (int j=i+1;j<words.size();j++)
            {
                if(valid(data[i],data[j]))
                    ans = max(ans, words[i].size()*words[j].size());
            }
        }

        return ans;
    }
};
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