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leetcode-problemset/leetcode-cn/problem (Chinese)/比较字符串最小字母出现频次 [compare-strings-by-frequency-of-the-smallest-character].html
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<p>定义一个函数 <code>f(s)</code>,统计 <code>s</code>  中<strong>(按字典序比较)最小字母的出现频次</strong> ,其中 <code>s</code> 是一个非空字符串。</p>
<p>例如,若 <code>s = "dcce"</code>,那么 <code>f(s) = 2</code>,因为字典序最小字母是 <code>"c"</code>它出现了 2 次。</p>
<p>现在,给你两个字符串数组待查表 <code>queries</code> 和词汇表 <code>words</code> 。对于每次查询 <code>queries[i]</code> ,需统计 <code>words</code> 中满足 <code>f(queries[i])</code> < <code>f(W)</code> 的<strong> 词的数目</strong> <code>W</code> 表示词汇表 <code>words</code> 中的每个词。</p>
<p>请你返回一个整数数组 <code>answer</code> 作为答案,其中每个 <code>answer[i]</code> 是第 <code>i</code> 次查询的结果。</p>
<p> </p>
<p><strong>示例 1</strong></p>
<pre>
<strong>输入:</strong>queries = ["cbd"], words = ["zaaaz"]
<strong>输出:</strong>[1]
<strong>解释:</strong>查询 f("cbd") = 1而 f("zaaaz") = 3 所以 f("cbd") < f("zaaaz")
</pre>
<p><strong>示例 2</strong></p>
<pre>
<strong>输入:</strong>queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
<strong>输出:</strong>[1,2]
<strong>解释:</strong>第一个查询 f("bbb") < f("aaaa")第二个查询 f("aaa") f("aaaa") > f("cc")。
</pre>
<p> </p>
<p><strong>提示:</strong></p>
<ul>
<li><code>1 <= queries.length <= 2000</code></li>
<li><code>1 <= words.length <= 2000</code></li>
<li><code>1 <= queries[i].length, words[i].length <= 10</code></li>
<li><code>queries[i][j]</code><code>words[i][j]</code> 都由小写英文字母组成</li>
</ul>