leetcode-problemset/leetcode-cn/originData/minimum-changes-to-make-k-semi-palindromes.json

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        "question": {
            "questionId": "2879",
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            "categoryTitle": "Algorithms",
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            "title": "Minimum Changes to Make K Semi-palindromes",
            "titleSlug": "minimum-changes-to-make-k-semi-palindromes",
            "content": "<p>Given a string <code>s</code> and an integer <code>k</code>, partition <code>s</code> into <code>k</code> <strong><span data-keyword=\"substring-nonempty\">substrings</span></strong> such that the letter changes needed to make each substring a <strong>semi-palindrome</strong>&nbsp;are minimized.</p>\n\n<p>Return the <em><strong>minimum</strong> number of letter changes</em> required<em>.</em></p>\n\n<p>A <strong>semi-palindrome</strong> is a special type of string that can be divided into <strong><span data-keyword=\"palindrome\">palindromes</span></strong> based on a repeating pattern. To check if a string is a semi-palindrome:</p>\n\n<ol>\n\t<li>Choose a positive divisor <code>d</code> of the string&#39;s length. <code>d</code> can range from <code>1</code> up to, but not including, the string&#39;s length. For a string of length <code>1</code>, it does not have a valid divisor as per this definition, since the only divisor is its length, which is not allowed.</li>\n\t<li>For a given divisor <code>d</code>, divide the string into groups where each group contains characters from the string that follow a repeating pattern of length <code>d</code>. Specifically, the first group consists of characters at positions <code>1</code>, <code>1 + d</code>, <code>1 + 2d</code>, and so on; the second group includes characters at positions <code>2</code>, <code>2 + d</code>, <code>2 + 2d</code>, etc.</li>\n\t<li>The string is considered a semi-palindrome if each of these groups forms a palindrome.</li>\n</ol>\n\n<p>Consider the string <code>&quot;abcabc&quot;</code>:</p>\n\n<ul>\n\t<li>The length of <code>&quot;abcabc&quot;</code> is <code>6</code>. Valid divisors are <code>1</code>, <code>2</code>, and <code>3</code>.</li>\n\t<li>For <code>d = 1</code>: The entire string <code>&quot;abcabc&quot;</code> forms one group. Not a palindrome.</li>\n\t<li>For <code>d = 2</code>:\n\t<ul>\n\t\t<li>Group 1 (positions <code>1, 3, 5</code>): <code>&quot;acb&quot;</code></li>\n\t\t<li>Group 2 (positions <code>2, 4, 6</code>): <code>&quot;bac&quot;</code></li>\n\t\t<li>Neither group forms a palindrome.</li>\n\t</ul>\n\t</li>\n\t<li>For <code>d = 3</code>:\n\t<ul>\n\t\t<li>Group 1 (positions <code>1, 4</code>): <code>&quot;aa&quot;</code></li>\n\t\t<li>Group 2 (positions <code>2, 5</code>): <code>&quot;bb&quot;</code></li>\n\t\t<li>Group 3 (positions <code>3, 6</code>): <code>&quot;cc&quot;</code></li>\n\t\t<li>All groups form palindromes. Therefore, <code>&quot;abcabc&quot;</code> is a semi-palindrome.</li>\n\t</ul>\n\t</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1: </strong></p>\n\n<div class=\"example-block\" style=\"border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;\">\n<p><strong>Input: </strong> <span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> s = &quot;abcac&quot;, k = 2 </span></p>\n\n<p><strong>Output: </strong> <span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> 1 </span></p>\n\n<p><strong>Explanation: </strong> Divide <code>s</code> into <code>&quot;ab&quot;</code> and <code>&quot;cac&quot;</code>. <code>&quot;cac&quot;</code> is already semi-palindrome. Change <code>&quot;ab&quot;</code> to <code>&quot;aa&quot;</code>, it becomes semi-palindrome with <code>d = 1</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 2: </strong></p>\n\n<div class=\"example-block\" style=\"border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;\">\n<p><strong>Input: </strong> <span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> s = &quot;abcdef&quot;, k = 2 </span></p>\n\n<p><strong>Output: </strong> <span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> 2 </span></p>\n\n<p><strong>Explanation: </strong> Divide <code>s</code> into substrings <code>&quot;abc&quot;</code> and <code>&quot;def&quot;</code>. Each&nbsp;needs one change to become semi-palindrome.</p>\n</div>\n\n<p><strong class=\"example\">Example 3: </strong></p>\n\n<div class=\"example-block\" style=\"border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;\">\n<p><strong>Input: </strong> <span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> s = &quot;aabbaa&quot;, k = 3 </span></p>\n\n<p><strong>Output: </strong> <span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> 0 </span></p>\n\n<p><strong>Explanation: </strong> Divide <code>s</code> into substrings <code>&quot;aa&quot;</code>, <code>&quot;bb&quot;</code> and <code>&quot;aa&quot;</code>.&nbsp;All are already semi-palindromes.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= s.length &lt;= 200</code></li>\n\t<li><code>1 &lt;= k &lt;= s.length / 2</code></li>\n\t<li><code>s</code> contains only lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "得到 K 个半回文串的最少修改次数",
            "translatedContent": "<p>给你一个字符串&nbsp;<code>s</code>&nbsp;和一个整数&nbsp;<code>k</code>&nbsp;，请你将&nbsp;<code>s</code> 分成&nbsp;<code>k</code>&nbsp;个<strong>&nbsp;子字符串</strong>&nbsp;，使得每个 <strong>子字符串</strong>&nbsp;变成&nbsp;<strong>半回文串</strong>&nbsp;需要修改的字符数目最少。</p>\n\n<p>请你返回一个整数，表示需要修改的 <strong>最少</strong>&nbsp;字符数目。</p>\n\n<p><strong>注意：</strong></p>\n\n<ul>\n\t<li>如果一个字符串从左往右和从右往左读是一样的，那么它是一个 <strong>回文串</strong>&nbsp;。</li>\n\t<li>如果长度为 <code>len</code>&nbsp;的字符串存在一个满足&nbsp;<code>1 &lt;= d &lt; len</code>&nbsp;的正整数&nbsp;<code>d</code>&nbsp;，<code>len % d == 0</code>&nbsp;成立且所有对 <code>d</code>&nbsp;做除法余数相同的下标对应的字符连起来得到的字符串都是 <strong>回文串</strong>&nbsp;，那么我们说这个字符串是 <strong>半回文串</strong>&nbsp;。比方说&nbsp;<code>\"aa\"</code>&nbsp;，<code>\"aba\"</code> ，<code>\"adbgad\"</code>&nbsp;和&nbsp;<code>\"abab\"</code>&nbsp;都是 <strong>半回文串</strong>&nbsp;，而&nbsp;<code>\"a\"</code>&nbsp;，<code>\"ab\"</code>&nbsp;和&nbsp;<code>\"abca\"</code>&nbsp;不是。</li>\n\t<li><strong>子字符串</strong>&nbsp;指的是一个字符串中一段连续的字符序列。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"abcac\", k = 2\n<b>输出：</b>1\n<b>解释：</b>我们可以将 s 分成子字符串 \"ab\" 和 \"cac\" 。子字符串 \"cac\" 已经是半回文串。如果我们将 \"ab\" 变成 \"aa\" ，它也会变成一个 d = 1 的半回文串。\n该方案是将 s 分成 2 个子字符串的前提下，得到 2 个半回文子字符串需要的最少修改次数。所以答案为 1 。</pre>\n\n<p><strong class=\"example\">示例 2:</strong></p>\n\n<pre>\n<b>输入：</b>s = \"abcdef\", k = 2\n<b>输出：</b>2\n<b>解释：</b>我们可以将 s 分成子字符串 \"abc\" 和 \"def\" 。子字符串 \"abc\" 和 \"def\" 都需要修改一个字符得到半回文串，所以我们总共需要 2 次字符修改使所有子字符串变成半回文串。\n该方案是将 s 分成 2 个子字符串的前提下，得到 2 个半回文子字符串需要的最少修改次数。所以答案为 2 。</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"aabbaa\", k = 3\n<b>输出：</b>0\n<b>解释：</b>我们可以将 s 分成子字符串 \"aa\" ，\"bb\" 和 \"aa\" 。\n字符串 \"aa\" 和 \"bb\" 都已经是半回文串了。所以答案为 0 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= s.length &lt;= 200</code></li>\n\t<li><code>1 &lt;= k &lt;= s.length / 2</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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                "Define <code>dp[i][j]</code> as the minimum count of letter changes needed to split the suffix of string <code>s</code> starting from <code>s[i]</code> into <code>j</code> valid parts.",
                "We have <code>dp[i][j] = min(dp[x + 1][j - 1] + v[i][x])</code>. Here <code>v[i][x]</code> is the minimum number of letter changes to change substring <code>s[i..x]</code> into semi-palindrome.",
                "<code>v[i][j]</code> can be calculated separately by <b>brute-force</b>. We can create a table of <code>v[i][j]</code> independently to improve the complexity. Also note that semi-palindrome’s length is at least <code>2</code>."
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<code>use-after-free<\\/code> \\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/troydhanson.github.io\\/uthash\\/\\\" target=\\\"_blank\\\">uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n<p><b>1. \\u5f80\\u54c8\\u5e0c\\u8868\\u4e2d\\u6dfb\\u52a0\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n<pre>\\r\\nstruct hash_entry {\\r\\n    int id;            \\/* we'll use this field as the key *\\/\\r\\n    char name[10];\\r\\n    UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry 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target=\\\"_blank\\\">sortedcontainers<\\/a> \\u5e93\\u3002<\\/p>\"],\"scala\":[\"Scala\",\"<p>\\u7248\\u672c\\uff1a<code>Scala 3.3.1<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"<p>\\u7248\\u672c\\uff1a<code>Kotlin 1.9.0<\\/code><\\/p>\\r\\n\\r\\n<p>\\u6211\\u4eec\\u4f7f\\u7528\\u7684\\u662f JetBrains \\u63d0\\u4f9b\\u7684 experimental compiler\\u3002\\u5982\\u679c\\u60a8\\u8ba4\\u4e3a\\u60a8\\u9047\\u5230\\u4e86\\u7f16\\u8bd1\\u5668\\u76f8\\u5173\\u7684\\u95ee\\u9898\\uff0c\\u8bf7\\u5411\\u6211\\u4eec\\u53cd\\u9988<\\/p>\"],\"rust\":[\"Rust\",\"<p>\\u7248\\u672c\\uff1a<code>rust 1.79.0<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 crates.io \\u7684 <a href=\\\"https:\\/\\/crates.io\\/crates\\/rand\\\" target=\\\"_blank\\\">rand<\\/a> \\u548c <a href=\\\"https:\\/\\/crates.io\\/crates\\/regex\\\" target=\\\"_blank\\\">regex<\\/a><\\/p>\"],\"php\":[\"PHP\",\"<p><code>PHP 8.2<\\/code>.<\\/p>\\r\\n\\r\\n<p>With bcmath module.<\\/p>\"],\"typescript\":[\"TypeScript\",\"<p>TypeScript 5.1.6<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2022<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/blob\\/v5\\/README.md\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.4.0<\\/a>\\uff0c<a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/v4.2.3\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.3<\\/a> \\u4ee5\\u53ca <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/deque\\/tree\\/v1.0.4\\\" target=\\\"_blank\\\"> datastructures-js\\/deque@1.0.4<\\/a>\\u3002<\\/p>\"],\"racket\":[\"Racket\",\"<p><a href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.15<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 26\"],\"elixir\":[\"Elixir\",\"Elixir 1.17 with Erlang\\/OTP 26\"],\"dart\":[\"Dart\",\"<p>Dart 3.2\\u3002\\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/pub.dev\\/packages\\/collection\\/versions\\/1.18.0\\\" target=\\\"_blank\\\">collection<\\/a> \\u5305<\\/p>\\r\\n\\r\\n<p>\\u60a8\\u7684\\u4ee3\\u7801\\u5c06\\u4f1a\\u88ab\\u4e0d\\u7f16\\u8bd1\\u76f4\\u63a5\\u8fd0\\u884c<\\/p>\"],\"cangjie\":[\"Cangjie\",\"<p>\\u7248\\u672c\\uff1a0.53.11 (cjnative)<\\/p>\\r\\n\\r\\n<p>\\u7f16\\u8bd1\\u53c2\\u6570\\uff1a<code>-O2 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