<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>& w) {
        int maxi = 0; int prod,fl; 
        int n = w.size();
        vector<unordered_map<int,int>> m;
        for(int i=0; i<n; i++){
            unordered_map<int,int> m1; 
            for(int t=0; t<w[i].size(); t++) m1[w[i][t]]++;
            m.push_back(m1);
        }
        for(int i=0; i<n; i++){
            for(int j=i+1; j<n; j++){
                fl = 0; prod = 1;
                for(auto it:m[i]){
                    if(m[j].find(it.first) != m[j].end()){ fl = 1; break; }
                }
                if(fl == 0){
                    prod = w[i].size() * w[j].size();
                    maxi = max(maxi , prod);
                }
            }
        }
        return maxi;
    }
};
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