<p>Let the function <code>f(s)</code> be the <strong>frequency of the lexicographically smallest character</strong> in a non-empty string <code>s</code>. For example, if <code>s = &quot;dcce&quot;</code> then <code>f(s) = 2</code> because the lexicographically smallest character is <code>&#39;c&#39;</code>, which has a frequency of 2.</p>

<p>You are given an array of strings <code>words</code> and another array of query strings <code>queries</code>. For each query <code>queries[i]</code>, count the <strong>number of words</strong> in <code>words</code> such that <code>f(queries[i])</code> &lt; <code>f(W)</code> for each <code>W</code> in <code>words</code>.</p>

<p>Return <em>an integer array </em><code>answer</code><em>, where each </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query</em>.</p>

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

<pre>
<strong>Input:</strong> queries = [&quot;cbd&quot;], words = [&quot;zaaaz&quot;]
<strong>Output:</strong> [1]
<strong>Explanation:</strong> On the first query we have f(&quot;cbd&quot;) = 1, f(&quot;zaaaz&quot;) = 3 so f(&quot;cbd&quot;) &lt; f(&quot;zaaaz&quot;).
</pre>

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

<pre>
<strong>Input:</strong> queries = [&quot;bbb&quot;,&quot;cc&quot;], words = [&quot;a&quot;,&quot;aa&quot;,&quot;aaa&quot;,&quot;aaaa&quot;]
<strong>Output:</strong> [1,2]
<strong>Explanation:</strong> On the first query only f(&quot;bbb&quot;) &lt; f(&quot;aaaa&quot;). On the second query both f(&quot;aaa&quot;) and f(&quot;aaaa&quot;) are both &gt; f(&quot;cc&quot;).
</pre>

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

<ul>
	<li><code>1 &lt;= queries.length &lt;= 2000</code></li>
	<li><code>1 &lt;= words.length &lt;= 2000</code></li>
	<li><code>1 &lt;= queries[i].length, words[i].length &lt;= 10</code></li>
	<li><code>queries[i][j]</code>, <code>words[i][j]</code> consist of lowercase English letters.</li>
</ul>

class Solution {
public:
    int lexicographically(string s)
    {
        vector<int> freqn(26,0);
        for(int i=0 ; i<s.size() ; i++)
        {
            freqn[s[i]-'a']++;
        }
        for(int i=0 ; i<26 ; i++)
        {
            if(freqn[i]!=0)
            {
                return freqn[i];
            }
        }
        return 0;
    }
    vector<int> numSmallerByFrequency(vector<string>& queries, vector<string>& words) {
    vector<int> querry(queries.size());
    vector<int> word(words.size());
    for(int i=0 ; i<querry.size() ; i++)
    {
        querry[i] = lexicographically(queries[i]);
    }
    for(int i=0 ; i<word.size() ; i++)
    {
        word[i] = lexicographically(words[i]);
    }
    vector<int> ans(querry.size(),0);
    int i=0,j=0;
    sort(word.begin(),word.end());
    while(i<querry.size())
    {
        int lo = 0;
        int hi = word.size()-1;
        int target = querry[i];
        while(lo<=hi)
        {
            int mid = lo + (hi - lo)/2;
            if(word[mid]>target)
            {
                int midn = mid;
                while(mid>=0 && word[mid]>target)
                {
                    midn = mid;
                    mid--;
                }
                ans[i] = word.size() - midn;
                break;
            }
            else
            {
                lo = mid + 1;
            }
        }
        i++;
    }
    return ans;
    }
};
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