<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 compareMap(unordered_map<char, int> &mapA, unordered_map<char, int> &mapB){
        for(auto pr: mapB){
            if(mapA.find(pr.first)!=mapA.end())return false;
        }
        return true;
    }
    int maxProduct(vector<string>& words) {
        unordered_map<int,unordered_map<char, int>> mp;
        int ans = 0;
        for(int i = 0; i<words.size(); i++){
            unordered_map<char, int> charMap;
            for(auto ch: words[i])charMap[ch]++;
            for(auto it = mp.begin(); it!=mp.end(); it++){
                if(ans < words[it->first].length() * words[i].length() && compareMap(it->second, charMap)){
                    ans = (int)max(ans, (int)words[it->first].length() * (int)words[i].length());
                }
            }
            mp[i] = charMap;
        } 
        return ans;
    }
};
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