leetcode-problemset/leetcode-cn/originData/minimum-operations-to-make-character-frequencies-equal.json

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        "question": {
            "questionId": "3638",
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            "categoryTitle": "Algorithms",
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            "title": "Minimum Operations to Make Character Frequencies Equal",
            "titleSlug": "minimum-operations-to-make-character-frequencies-equal",
            "content": "<p>You are given a string <code>s</code>.</p>\n\n<p>A string <code>t</code> is called <strong>good</strong> if all characters of <code>t</code> occur the same number of times.</p>\n\n<p>You can perform the following operations <strong>any number of times</strong>:</p>\n\n<ul>\n\t<li>Delete a character from <code>s</code>.</li>\n\t<li>Insert a character in <code>s</code>.</li>\n\t<li>Change a character in <code>s</code> to its next letter in the alphabet.</li>\n</ul>\n\n<p><strong>Note</strong> that you cannot change <code>&#39;z&#39;</code> to <code>&#39;a&#39;</code> using the third operation.</p>\n\n<p>Return<em> </em>the <strong>minimum</strong> number of operations required to make <code>s</code> <strong>good</strong>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">s = &quot;acab&quot;</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">1</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>We can make <code>s</code> good by deleting one occurrence of character <code>&#39;a&#39;</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">s = &quot;wddw&quot;</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">0</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>We do not need to perform any operations since <code>s</code> is initially good.</p>\n</div>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">s = &quot;aaabc&quot;</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">2</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>We can make <code>s</code> good by applying these operations:</p>\n\n<ul>\n\t<li>Change one occurrence of <code>&#39;a&#39;</code> to <code>&#39;b&#39;</code></li>\n\t<li>Insert one occurrence of <code>&#39;c&#39;</code> into <code>s</code></li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>3 &lt;= s.length &lt;= 2&nbsp;* 10<sup>4</sup></code></li>\n\t<li><code>s</code> contains only lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "使字符频率相等的最少操作次数",
            "translatedContent": "<p>给你一个字符串&nbsp;<code>s</code>&nbsp;。</p>\n\n<p>如果字符串 <code>t</code>&nbsp;中的字符出现次数相等，那么我们称&nbsp;<code>t</code>&nbsp;为 <strong>好的</strong>&nbsp;。</p>\n\n<p>你可以执行以下操作 <strong>任意次</strong>&nbsp;：</p>\n\n<ul>\n\t<li>从&nbsp;<code>s</code>&nbsp;中删除一个字符。</li>\n\t<li>往&nbsp;<code>s</code>&nbsp;中添加一个字符。</li>\n\t<li>将&nbsp;<code>s</code>&nbsp;中一个字母变成字母表中下一个字母。</li>\n</ul>\n\n<p><b>注意</b>&nbsp;，第三个操作不能将&nbsp;<code>'z'</code>&nbsp;变为&nbsp;<code>'a'</code>&nbsp;。</p>\n\n<p>请你返回将 <code>s</code>&nbsp;变 <strong>好</strong>&nbsp;的 <strong>最少</strong>&nbsp;操作次数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>s = \"acab\"</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>1</span></p>\n\n<p><b>解释：</b></p>\n\n<p>删掉一个字符&nbsp;<code>'a'</code>&nbsp;，<code>s</code>&nbsp;变为好的。</p>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>s = \"wddw\"</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>0</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p><code>s</code>&nbsp;一开始就是好的，所以不需要执行任何操作。</p>\n</div>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>s = \"aaabc\"</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>2</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>通过以下操作，将&nbsp;<code>s</code>&nbsp;变好：</p>\n\n<ul>\n\t<li>将一个&nbsp;<code>'a'</code>&nbsp;变为&nbsp;<code>'b'</code>&nbsp;。</li>\n\t<li>往 <code>s</code>&nbsp;中插入一个&nbsp;<code>'c'</code>&nbsp;。</li>\n</ul>\n</div>\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;= 2&nbsp;* 10<sup>4</sup></code></li>\n\t<li><code>s</code>&nbsp;只包含小写英文字母。</li>\n</ul>\n",
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                "The order of the letters in the string is irrelevant.",
                "Compute an occurrence array <code>occ</code> where <code>occ[x]</code> is the number of occurrences of the <code>x<supth</sup></code> character of the alphabet. How do the described operations change <code>occ</code>?",
                "We have three types of operations: increase any <code>occ[x]</code> by 1, decrease any <code>occ[x]</code> by 1, or decrease any <code>occ[x]</code> by 1 and simultaneously increase <code>occ[x + 1]</code> by 1 at the same time. To make <code>s</code> good, we need to make <code>occ</code> good. <code>occ</code> is good if and only if every <code>occ[x]</code> equals either 0 or some constant <code>c</code>.",
                "If you know the value of <code>c</code>, how can you calculate the minimum operations required to make <code>occ</code> good?",
                "Observation 1: It is never optimal to apply the third type of operation (simultaneous decrease and increase) on two continuous elements <code>occ[x]</code> and <code>occ[x + 1]</code>. Instead, we can decrease <code>occ[x]</code> by 1 then increase <code>occ[x + 2]</code> by 1 to achieve the same effect.",
                "Observation 2: It is never optimal to increase an element of <code>occ</code> then decrease it, or vice versa.",
                "Use dynamic programming where <code>dp[i]</code> is the minimum number of operations required to make <code>occ[0..i]</code> good. You will need to use the above observations to come up with the transitions."
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