<p>You are given an array of strings <code>words</code> and a string <code>chars</code>.</p>

<p>A string is <strong>good</strong> if it can be formed by characters from <code>chars</code> (each character can only be used once).</p>

<p>Return <em>the sum of lengths of all good strings in words</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;cat&quot;,&quot;bt&quot;,&quot;hat&quot;,&quot;tree&quot;], chars = &quot;atach&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> The strings that can be formed are &quot;cat&quot; and &quot;hat&quot; so the answer is 3 + 3 = 6.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;hello&quot;,&quot;world&quot;,&quot;leetcode&quot;], chars = &quot;welldonehoneyr&quot;
<strong>Output:</strong> 10
<strong>Explanation:</strong> The strings that can be formed are &quot;hello&quot; and &quot;world&quot; so the answer is 5 + 5 = 10.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 1000</code></li>
	<li><code>1 &lt;= words[i].length, chars.length &lt;= 100</code></li>
	<li><code>words[i]</code> and <code>chars</code> consist of lowercase English letters.</li>
</ul>

class Solution {
public:
    int countCharacters(const std::vector<std::string>& words, const std::string& chars) {
        // Build frequency map for chars
        std::unordered_map<char, int> charFreq;
        for (char c : chars) {
            charFreq[c]++;
        }

        int ans = 0;
        
        for (const std::string& word : words) {
            std::unordered_map<char, int> wordFreq;
            bool canForm = true;

            // Count the frequency of each character in the current word
            for (char c : word) {
                wordFreq[c]++;
            }

            // Check if the word can be formed with the characters in chars
            for (const auto& [charInWord, countInWord] : wordFreq) {
                if (charFreq[charInWord] < countInWord) {
                    canForm = false;
                    break;
                }
            }

            // If the word can be formed, add its length to the answer
            if (canForm) {
                ans += word.length();
            }
        }

        return ans;
    }
};
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