<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(vector<string>& words, string chars) {
        unordered_map<char,int> hash1, hash2;
        int len=0;
        for(char ch : chars)
            hash1[ch]++;
        
        for(string word : words){
            hash2.clear();
            bool flag1=true;
            for(char ch : word){
                if(hash1.find(ch) == hash1.end()){
                    flag1=false;
                    break;
                }
                else
                    hash2[ch]++;
            }
            bool flag=false;
            if(flag1){
                for(auto it : hash2){
                    if(hash1.find(it.first) != hash1.end() && it.second <= hash1[it.first])
                        flag=true;
                    else{
                        flag=false;
                        break;
                    }
                }
            }
            if(flag)
                len += word.size();
        }
        return len;

    }
};
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