<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 maxProduct(vector<string>& words) {
        
        vector < pair < int, vector <int> > > freq;

        for(string s : words) {
            vector <int> count(26, 0);

            for( char c : s) {
                count[c-'a']++;
            }

            freq.push_back({s.size(), count} );
        }

        sort(freq.begin(), freq.end(), 
            [&](auto v, auto w){ return v.first > w.first; 
        });

        int ans = 0;
        int n = freq.size();

        for(int i=0; i<n; i++) {
            for(int j=i+1; j<n; j++) {
                bool ok = true;

                for(int k=0; k<26; k++) {
                    if(freq[i].second[k] != 0 && freq[j].second[k] != 0) {
                        ok = false;
                        break;
                    }
                }

                if(ok) {
                    ans = max(freq[i].first * freq[j].first , ans);
                }
            }
        }

        return ans;



    }
};
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