leetcode-problemset/leetcode-cn/originData/rearrange-k-substrings-to-form-target-string.json

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        "question": {
            "questionId": "3595",
            "questionFrontendId": "3365",
            "categoryTitle": "Algorithms",
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            "title": "Rearrange K Substrings to Form Target String",
            "titleSlug": "rearrange-k-substrings-to-form-target-string",
            "content": "<p>You are given two strings <code>s</code> and <code>t</code>, both of which are anagrams of each other, and an integer <code>k</code>.</p>\n\n<p>Your task is to determine whether it is possible to split the string <code>s</code> into <code>k</code> equal-sized substrings, rearrange the substrings, and concatenate them in <em>any order</em> to create a new string that matches the given string <code>t</code>.</p>\n\n<p>Return <code>true</code> if this is possible, otherwise, return <code>false</code>.</p>\n\n<p>An <strong>anagram</strong> is a word or phrase formed by rearranging the letters of a different word or phrase, using all the original letters exactly once.</p>\n\n<p>A <strong>substring</strong> is a contiguous <b>non-empty</b> sequence of characters within a string.</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;abcd&quot;, t = &quot;cdab&quot;, k = 2</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">true</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ul>\n\t<li>Split <code>s</code> into 2 substrings of length 2: <code>[&quot;ab&quot;, &quot;cd&quot;]</code>.</li>\n\t<li>Rearranging these substrings as <code>[&quot;cd&quot;, &quot;ab&quot;]</code>, and then concatenating them results in <code>&quot;cdab&quot;</code>, which matches <code>t</code>.</li>\n</ul>\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;aabbcc&quot;, t = &quot;bbaacc&quot;, k = 3</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">true</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ul>\n\t<li>Split <code>s</code> into 3 substrings of length 2: <code>[&quot;aa&quot;, &quot;bb&quot;, &quot;cc&quot;]</code>.</li>\n\t<li>Rearranging these substrings as <code>[&quot;bb&quot;, &quot;aa&quot;, &quot;cc&quot;]</code>, and then concatenating them results in <code>&quot;bbaacc&quot;</code>, which matches <code>t</code>.</li>\n</ul>\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;aabbcc&quot;, t = &quot;bbaacc&quot;, k = 2</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">false</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ul>\n\t<li>Split <code>s</code> into 2 substrings of length 3: <code>[&quot;aab&quot;, &quot;bcc&quot;]</code>.</li>\n\t<li>These substrings cannot be rearranged to form <code>t = &quot;bbaacc&quot;</code>, so the output is <code>false</code>.</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length == t.length &lt;= 2 * 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= s.length</code></li>\n\t<li><code>s.length</code> is divisible by <code>k</code>.</li>\n\t<li><code>s</code> and <code>t</code> consist only of lowercase English letters.</li>\n\t<li>The input is generated such that<!-- notionvc: 53e485fc-71ce-4032-aed1-f712dd3822ba --> <code>s</code> and <code>t</code> are anagrams of each other.</li>\n</ul>\n",
            "translatedTitle": "重排子字符串以形成目标字符串",
            "translatedContent": "<p>给你两个字符串 <code>s</code> 和 <code>t</code>（它们互为字母异位词），以及一个整数 <code>k</code>。</p>\n\n<p>你的任务是判断是否可以将字符串 <code>s</code> 分割成 <code>k</code> 个等长的子字符串，然后重新排列这些子字符串，并以任意顺序连接它们，使得最终得到的新字符串与给定的字符串 <code>t</code> 相匹配。</p>\n\n<p>如果可以做到，返回 <code>true</code>；否则，返回 <code>false</code>。</p>\n\n<p><strong>字母异位词&nbsp;</strong>是指由另一个单词或短语的所有字母重新排列形成的单词或短语，使用所有原始字母恰好一次。</p>\n\n<p><strong>子字符串&nbsp;</strong>是字符串中的一个连续&nbsp;<b>非空&nbsp;</b>字符序列。</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><strong>输入：</strong> <span class=\"example-io\">s = \"abcd\", t = \"cdab\", k = 2</span></p>\n\n<p><strong>输出：</strong> <span class=\"example-io\">true</span></p>\n\n<p><strong>解释：</strong></p>\n\n<ul>\n\t<li>将 <code>s</code> 分割成 2 个长度为 2 的子字符串：<code>[\"ab\", \"cd\"]</code>。</li>\n\t<li>重新排列这些子字符串为 <code>[\"cd\", \"ab\"]</code>，然后连接它们得到 <code>\"cdab\"</code>，与 <code>t</code> 相匹配。</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong> <span class=\"example-io\">s = \"aabbcc\", t = \"bbaacc\", k = 3</span></p>\n\n<p><strong>输出：</strong> <span class=\"example-io\">true</span></p>\n\n<p><strong>解释：</strong></p>\n\n<ul>\n\t<li>将 <code>s</code> 分割成 3 个长度为 2 的子字符串：<code>[\"aa\", \"bb\", \"cc\"]</code>。</li>\n\t<li>重新排列这些子字符串为 <code>[\"bb\", \"aa\", \"cc\"]</code>，然后连接它们得到 <code>\"bbaacc\"</code>，与 <code>t</code> 相匹配。</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong> <span class=\"example-io\">s = \"aabbcc\", t = \"bbaacc\", k = 2</span></p>\n\n<p><strong>输出：</strong> <span class=\"example-io\">false</span></p>\n\n<p><strong>解释：</strong></p>\n\n<ul>\n\t<li>将 <code>s</code> 分割成 2 个长度为 3 的子字符串：<code>[\"aab\", \"bcc\"]</code>。</li>\n\t<li>这些子字符串无法重新排列形成 <code>t = \"bbaacc\"</code>，所以输出 <code>false</code>。</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 == t.length &lt;= 2 * 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= s.length</code></li>\n\t<li><code>s.length</code> 能被 <code>k</code> 整除。</li>\n\t<li><code>s</code> 和 <code>t</code> 仅由小写英文字母组成。</li>\n\t<li>输入保证 <code>s</code> 和 <code>t</code> 互为字母异位词。</li>\n</ul>\n",
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\\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.11<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap \\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"http:\\/\\/www.grantjenks.com\\/docs\\/sortedcontainers\\/\\\" 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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