<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>

/*
store caracters of chars in a set
iterate over each word in words
iterate each caracter of word
*/

class Solution {
public:
    int countCharacters(vector<string>& words, string chars) {
        unordered_map<char, int>mp;
        unordered_map<char, int>repeated;
        int ans = 0, c = 0;
        string w = "";
        for(auto x : chars){
            mp[x]++;
        }
        
        
        for(int i = 0; i < words.size(); ++i){
            w = words[i];
            c = 0;
            repeated = mp;
            for(int y = 0; y < w.length(); ++y){
                if(repeated.find(w[y]) == repeated.end()) break;
                c++;
                repeated[w[y]]--;
                if(repeated[w[y]] == 0)repeated.erase(w[y]);
            }
            (c == w.length()) ? ans += c : 0;
        }
        
        return ans;
    }
};
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