<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> minFreq(26, INT_MAX);  // Frequency map for minimum occurrences of each character
        
        // Populate minFreq using the first word
        vector<int> freq(26, 0);  // Frequency map for the first word
        
        // Calculate frequency for the first word
        for (int i = 0; i < words[0].size(); ++i) {
            freq[words[0][i] - 'a']++;
        }
        
        // Update minFreq based on the first word
        for (int i = 0; i < 26; ++i) {
            minFreq[i] = freq[i];
        }
        
        // For the rest of the words, update the minimum frequency of each character
        for (int i = 1; i < words.size(); ++i) {
            vector<int> freq(26, 0);  // Temporary frequency map for the current word
            
            for (int j = 0; j < words[i].size(); ++j) {
                freq[words[i][j] - 'a']++;
            }
            
            // Update the global minFreq to keep track of the minimum count of each character
            for (int j = 0; j < 26; ++j) {
                minFreq[j] = min(minFreq[j], freq[j]);
            }
        }
        
        // Collect the result based on minFreq
        vector<string> result;
        for (int i = 0; i < 26; ++i) {
            while (minFreq[i] > 0) {
                result.push_back(string(1, i + 'a'));  // Add character 'a' + i to the result
                minFreq[i]--;
            }
        }
        
        return result;
    }
};

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