<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) {
        unordered_map<char,int> letter_freq;
        for (char letter='a'; letter<='z'; letter++) {
            letter_freq[letter]=INT_MAX;
        }
        for (int i=0; i<words.size(); i++) {
            unordered_map<char,int> curr_letter_freq;
            for (char letter='a'; letter<='z'; letter++) {
                curr_letter_freq[letter]=0;
            }
            for (int j=0; j<words[i].size(); j++) {
                curr_letter_freq[words[i][j]]++;
            }
            for (auto it:curr_letter_freq) {
                if (it.second<letter_freq[it.first]) {
                    letter_freq[it.first]=it.second;
                }
            }
        }
        vector<string> ans;
        for (auto it:letter_freq) {
            if (it.second!=INT_MAX) {
                string letter;
                letter.push_back(it.first);
                int freq=it.second;
                while (freq--) {
                    ans.push_back(letter);
                }
            }
        }
        return ans;
    }
};
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