leetcode-problemset/leetcode-cn/originData/count-anagrams.json

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            "questionId": "2605",
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            "categoryTitle": "Algorithms",
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            "title": "Count Anagrams",
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            "content": "<p>You are given a string <code>s</code> containing one or more words. Every consecutive pair of words is separated by a single space <code>&#39; &#39;</code>.</p>\n\n<p>A string <code>t</code> is an <strong>anagram</strong> of string <code>s</code> if the <code>i<sup>th</sup></code> word of <code>t</code> is a <strong>permutation</strong> of the <code>i<sup>th</sup></code> word of <code>s</code>.</p>\n\n<ul>\n\t<li>For example, <code>&quot;acb dfe&quot;</code> is an anagram of <code>&quot;abc def&quot;</code>, but <code>&quot;def cab&quot;</code>&nbsp;and <code>&quot;adc bef&quot;</code> are not.</li>\n</ul>\n\n<p>Return <em>the number of <strong>distinct anagrams</strong> of </em><code>s</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;too hot&quot;\n<strong>Output:</strong> 18\n<strong>Explanation:</strong> Some of the anagrams of the given string are &quot;too hot&quot;, &quot;oot hot&quot;, &quot;oto toh&quot;, &quot;too toh&quot;, and &quot;too oht&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;aa&quot;\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> There is only one anagram possible for the given string.</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s</code> consists of lowercase English letters and spaces <code>&#39; &#39;</code>.</li>\n\t<li>There is single space between consecutive words.</li>\n</ul>\n",
            "translatedTitle": "统计同位异构字符串数目",
            "translatedContent": "<p>给你一个字符串&nbsp;<code>s</code>&nbsp;，它包含一个或者多个单词。单词之间用单个空格&nbsp;<code>' '</code>&nbsp;隔开。</p>\n\n<p>如果字符串 <code>t</code>&nbsp;中第 <code>i</code>&nbsp;个单词是 <code>s</code>&nbsp;中第 <code>i</code>&nbsp;个单词的一个&nbsp;<strong>排列</strong>&nbsp;，那么我们称字符串&nbsp;<code>t</code>&nbsp;是字符串&nbsp;<code>s</code>&nbsp;的同位异构字符串。</p>\n\n<ul>\n\t<li>比方说，<code>\"acb dfe\"</code>&nbsp;是&nbsp;<code>\"abc def\"</code>&nbsp;的同位异构字符串，但是&nbsp;<code>\"def cab\"</code>&nbsp;和&nbsp;<code>\"adc bef\"</code>&nbsp;不是。</li>\n</ul>\n\n<p>请你返回<em>&nbsp;</em><code>s</code>&nbsp;的同位异构字符串的数目，由于答案可能很大，请你将它对&nbsp;<code>10<sup>9</sup> + 7</code>&nbsp;<strong>取余</strong> 后返回。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>s = \"too hot\"\n<b>输出：</b>18\n<b>解释：</b>输入字符串的一些同位异构字符串为 \"too hot\" ，\"oot hot\" ，\"oto toh\" ，\"too toh\" 以及 \"too oht\" 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>s = \"aa\"\n<b>输出：</b>1\n<strong>解释：</strong>输入字符串只有一个同位异构字符串。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s</code> 只包含小写英文字母和空格&nbsp;<code>' '</code>&nbsp;。</li>\n\t<li>相邻单词之间由单个空格隔开。</li>\n</ul>\n",
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                "For each word, can you count the number of permutations possible if all characters are distinct?",
                "How to reduce overcounting when letters are repeated?",
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