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leetcode-problemset/leetcode-cn/originData/compare-strings-by-frequency-of-the-smallest-character.json
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{
"data": {
"question": {
"questionId": "1273",
"questionFrontendId": "1170",
"categoryTitle": "Algorithms",
"boundTopicId": 22197,
"title": "Compare Strings by Frequency of the Smallest Character",
"titleSlug": "compare-strings-by-frequency-of-the-smallest-character",
"content": "<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>\n\n<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>\n\n<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>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> queries = [&quot;cbd&quot;], words = [&quot;zaaaz&quot;]\n<strong>Output:</strong> [1]\n<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;).\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> queries = [&quot;bbb&quot;,&quot;cc&quot;], words = [&quot;a&quot;,&quot;aa&quot;,&quot;aaa&quot;,&quot;aaaa&quot;]\n<strong>Output:</strong> [1,2]\n<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;).\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= queries.length &lt;= 2000</code></li>\n\t<li><code>1 &lt;= words.length &lt;= 2000</code></li>\n\t<li><code>1 &lt;= queries[i].length, words[i].length &lt;= 10</code></li>\n\t<li><code>queries[i][j]</code>, <code>words[i][j]</code> consist of lowercase English letters.</li>\n</ul>\n",
"translatedTitle": "比较字符串最小字母出现频次",
"translatedContent": "<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>\n\n<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>\n\n<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>\n\n<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>\n\n<p> </p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入:</strong>queries = [\"cbd\"], words = [\"zaaaz\"]\n<strong>输出:</strong>[1]\n<strong>解释:</strong>查询 f(\"cbd\") = 1而 f(\"zaaaz\") = 3 所以 f(\"cbd\") < f(\"zaaaz\")。\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>queries = [\"bbb\",\"cc\"], words = [\"a\",\"aa\",\"aaa\",\"aaaa\"]\n<strong>输出:</strong>[1,2]\n<strong>解释:</strong>第一个查询 f(\"bbb\") < f(\"aaaa\"),第二个查询 f(\"aaa\") 和 f(\"aaaa\") 都 > f(\"cc\")。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 <= queries.length <= 2000</code></li>\n\t<li><code>1 <= words.length <= 2000</code></li>\n\t<li><code>1 <= queries[i].length, words[i].length <= 10</code></li>\n\t<li><code>queries[i][j]</code>、<code>words[i][j]</code> 都由小写英文字母组成</li>\n</ul>\n",
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"For each string from words calculate the leading count and store it in an array, then sort the integer array.",
"For each string from queries calculate the leading count \"p\" and in base of the sorted array calculated on the step 1 do a binary search to count the number of items greater than \"p\"."
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