<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<int> last = count(words[0]);
        for (int i = 1; i < words.size(); i++) {
            last = intersection(last, count(words[i]));
        }
        
        vector<string> result;
        for (int i = 0; i < 26; i++) {
            while (last[i] > 0) {
                result.push_back(string(1, 'a' + i));
                last[i]--;
            }
        }
        
        return result;
    }
    
private:
    vector<int> count(const string& str) {
        vector<int> frequency(26, 0);
        for (char c : str) {
            frequency[c - 'a']++;
        }
        return frequency;
    }
    
    vector<int> intersection(const vector<int>& a, const vector<int>& b) {
        vector<int> t(26, 0);
        for (int i = 0; i < 26; i++) {
            t[i] = min(a[i], b[i]);
        }
        return t;
    }
};
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