<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:
  int count(string s1,string s2){
    int n1=s1.length();
    int n2=s2.length();

   vector<bool> flag(26,false);
   
    for(int i=0;i<s1.size();i++){
        flag[s1[i]-'a']=true;
    }
    for(int i=0;i<s2.size();i++){
        if(flag[s2[i]-'a']){
             return 0;
        }
    }
    return n1*n2;
  }
  int ans=0;
    int maxProduct(vector<string>& words) {
        for ( int i=0;i<words.size()-1;i++){
           
           for(int j=i+1;j<words.size();j++){
            if(words[i].size()*words[j].size()>ans){
           ans=max(ans,count(words[i],words[j]));}
           }
        }
      return ans;
    
    }
};
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