leetcode-problemset/leetcode-cn/originData/remove-letter-to-equalize-frequency.json

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            "questionId": "2532",
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            "categoryTitle": "Algorithms",
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            "title": "Remove Letter To Equalize Frequency",
            "titleSlug": "remove-letter-to-equalize-frequency",
            "content": "<p>You are given a <strong>0-indexed</strong> string <code>word</code>, consisting of lowercase English letters. You need to select <strong>one</strong> index and <strong>remove</strong> the letter at that index from <code>word</code> so that the <strong>frequency</strong> of every letter present in <code>word</code> is equal.</p>\n\n<p>Return<em> </em><code>true</code><em> if it is possible to remove one letter so that the frequency of all letters in </em><code>word</code><em> are equal, and </em><code>false</code><em> otherwise</em>.</p>\n\n<p><strong>Note:</strong></p>\n\n<ul>\n\t<li>The <b>frequency</b> of a letter <code>x</code> is the number of times it occurs in the string.</li>\n\t<li>You <strong>must</strong> remove exactly one letter and cannot choose to do nothing.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> word = &quot;abcc&quot;\n<strong>Output:</strong> true\n<strong>Explanation:</strong> Select index 3 and delete it: word becomes &quot;abc&quot; and each character has a frequency of 1.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> word = &quot;aazz&quot;\n<strong>Output:</strong> false\n<strong>Explanation:</strong> We must delete a character, so either the frequency of &quot;a&quot; is 1 and the frequency of &quot;z&quot; is 2, or vice versa. It is impossible to make all present letters have equal frequency.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= word.length &lt;= 100</code></li>\n\t<li><code>word</code> consists of lowercase English letters only.</li>\n</ul>\n",
            "translatedTitle": "删除字符使频率相同",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的字符串&nbsp;<code>word</code>&nbsp;，字符串只包含小写英文字母。你需要选择 <strong>一个</strong>&nbsp;下标并 <strong>删除</strong>&nbsp;下标处的字符，使得 <code>word</code>&nbsp;中剩余每个字母出现 <strong>频率</strong>&nbsp;相同。</p>\n\n<p>如果删除一个字母后，<code>word</code>&nbsp;中剩余所有字母的出现频率都相同，那么返回 <code>true</code>&nbsp;，否则返回 <code>false</code>&nbsp;。</p>\n\n<p><strong>注意：</strong></p>\n\n<ul>\n\t<li>字母&nbsp;<code>x</code>&nbsp;的 <strong>频率</strong><strong>&nbsp;</strong>是这个字母在字符串中出现的次数。</li>\n\t<li>你 <strong>必须</strong>&nbsp;恰好删除一个字母，不能一个字母都不删除。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>word = \"abcc\"\n<b>输出：</b>true\n<b>解释：</b>选择下标 3 并删除该字母：word 变成 \"abc\" 且每个字母出现频率都为 1 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>word = \"aazz\"\n<b>输出：</b>false\n<b>解释：</b>我们必须删除一个字母，所以要么 \"a\" 的频率变为 1 且 \"z\" 的频率为 2 ，要么两个字母频率反过来。所以不可能让剩余所有字母出现频率相同。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= word.length &lt;= 100</code></li>\n\t<li><code>word</code>&nbsp;只包含小写英文字母。</li>\n</ul>\n",
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                "Brute force all letters that could be removed.",
                "Use a frequency array of size 26."
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