<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) {
          vector<string>res;
          map<char,int>sol;
          map<char,int>mapping;
          if(words.size()==0)
            return res;
          for(int i=0;i<words[0].size();i++){
            sol[words[0][i]]++;
          }
          for(int i=1;i<words.size();i++){
             mapping.clear();
            for(int j=0;j<(words[i].size());j++){
                mapping[words[i][j]]++;
            }
         for (auto& pair : sol) {
                char c = pair.first;
                sol[c] = min(sol[c], mapping[c]);
            }

          }

          for (auto& pair : sol) {
            char c = pair.first;
            int count = pair.second;
            while (count-- > 0) {
                res.push_back(string(1, c)); 
            }
        }
          
             return res;
    }
};
16000