<p>Given a string array <code>words</code>, return <em>an array of all characters that show up in all strings within the </em><code>words</code><em> (including duplicates)</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> words = ["bella","label","roller"]
<strong>Output:</strong> ["e","l","l"]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> words = ["cool","lock","cook"]
<strong>Output:</strong> ["c","o"]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 100</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 100</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    vector<string> commonChars(vector<string>& words) {
        // char, vector<pair<int>>
        int n= words.size();
        unordered_map<char, vector<pair<int,int>>> dict; // {index, freq}
        for(int j=0;j<n;j++)
        {
            string str= words[j];
            vector<int> freq(26,0);
            for(auto ch: str)
            {
                freq[ch-'a']++;
            }
            for(int i=0;i<26;i++)
            {
                char ch= (i+'a');
                dict[ch].push_back({j,freq[i]});
            }
        }
        vector<string> ans;
        for(int i=0;i<26;i++)
        {
            char ch= (i+'a');
            // cout<<ch<<endl;
            if(dict[ch].size() < n)
            continue;
            int min_f=INT_MAX;
            // find min frequency across all words
            for(auto pr: dict[ch])
            min_f= min(min_f, pr.second);
            string s;
            s+=ch;
            // cout<<s<<endl;
            while(min_f--)
            ans.push_back(s);
        }
        return ans;
    }
};
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