Compare Strings by Frequency of the Smallest Character
<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>
Array; Hash Table; String; Binary Search; Sorting
class Solution {
public:
    vector<int> numSmallerByFrequency(vector<string>& queries, vector<string>& words) {
        auto f = [](string s) {
            int cnt[26] = {0};
            for (char c : s) {
                cnt[c - 'a']++;
            }
            for (int x : cnt) {
                if (x) {
                    return x;
                }
            }
            return 0;
        };
        int n = words.size();
        int nums[n];
        for (int i = 0; i < n; i++) {
            nums[i] = f(words[i]);
        }
        sort(nums, nums + n);
        vector<int> ans;
        for (auto& q : queries) {
            int x = f(q);
            ans.push_back(n - (upper_bound(nums, nums + n, x) - nums));
        }
        return ans;
    }
};