leetcode-problemset/leetcode-cn/originData/minimum-time-to-revert-word-to-initial-state-i.json

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            "title": "Minimum Time to Revert Word to Initial State I",
            "titleSlug": "minimum-time-to-revert-word-to-initial-state-i",
            "content": "<p>You are given a <strong>0-indexed</strong> string <code>word</code> and an integer <code>k</code>.</p>\n\n<p>At every second, you must perform the following operations:</p>\n\n<ul>\n\t<li>Remove the first <code>k</code> characters of <code>word</code>.</li>\n\t<li>Add any <code>k</code> characters to the end of <code>word</code>.</li>\n</ul>\n\n<p><strong>Note</strong> that you do not necessarily need to add the same characters that you removed. However, you must perform <strong>both</strong> operations at every second.</p>\n\n<p>Return <em>the <strong>minimum</strong> time greater than zero required for</em> <code>word</code> <em>to revert to its <strong>initial</strong> state</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> word = &quot;abacaba&quot;, k = 3\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> At the 1st second, we remove characters &quot;aba&quot; from the prefix of word, and add characters &quot;bac&quot; to the end of word. Thus, word becomes equal to &quot;cababac&quot;.\nAt the 2nd second, we remove characters &quot;cab&quot; from the prefix of word, and add &quot;aba&quot; to the end of word. Thus, word becomes equal to &quot;abacaba&quot; and reverts to its initial state.\nIt can be shown that 2 seconds is the minimum time greater than zero required for word to revert to its initial state.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> word = &quot;abacaba&quot;, k = 4\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> At the 1st second, we remove characters &quot;abac&quot; from the prefix of word, and add characters &quot;caba&quot; to the end of word. Thus, word becomes equal to &quot;abacaba&quot; and reverts to its initial state.\nIt can be shown that 1 second is the minimum time greater than zero required for word to revert to its initial state.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> word = &quot;abcbabcd&quot;, k = 2\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> At every second, we will remove the first 2 characters of word, and add the same characters to the end of word.\nAfter 4 seconds, word becomes equal to &quot;abcbabcd&quot; and reverts to its initial state.\nIt can be shown that 4 seconds is the minimum time greater than zero required for word to revert to its initial state.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= word.length &lt;= 50 </code></li>\n\t<li><code>1 &lt;= k &lt;= word.length</code></li>\n\t<li><code>word</code> consists only of lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "将单词恢复初始状态所需的最短时间 I",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始的字符串 <code>word</code> 和一个整数 <code>k</code> 。</p>\n\n<p>在每一秒，你必须执行以下操作：</p>\n\n<ul>\n\t<li>移除 <code>word</code> 的前 <code>k</code> 个字符。</li>\n\t<li>在 <code>word</code> 的末尾添加 <code>k</code> 个任意字符。</li>\n</ul>\n\n<p><strong>注意 </strong>添加的字符不必和移除的字符相同。但是，必须在每一秒钟都执行 <strong>两种 </strong>操作。</p>\n\n<p>返回将 <code>word</code> 恢复到其 <strong>初始 </strong>状态所需的 <strong>最短 </strong>时间（该时间必须大于零）。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>word = \"abacaba\", k = 3\n<strong>输出：</strong>2\n<strong>解释：</strong>\n第 1 秒，移除 word 的前缀 \"aba\"，并在末尾添加 \"bac\" 。因此，word 变为 \"cababac\"。\n第 2 秒，移除 word 的前缀 \"cab\"，并在末尾添加 \"aba\" 。因此，word 变为 \"abacaba\" 并恢复到始状态。\n可以证明，2 秒是 word 恢复到其初始状态所需的最短时间。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>word = \"abacaba\", k = 4\n<strong>输出：</strong>1\n<strong>解释：\n</strong>第 1 秒，移除 word 的前缀 \"abac\"，并在末尾添加 \"caba\" 。因此，word 变为 \"abacaba\" 并恢复到初始状态。\n可以证明，1 秒是 word 恢复到其初始状态所需的最短时间。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>word = \"abcbabcd\", k = 2\n<strong>输出：</strong>4\n<strong>解释：</strong>\n每一秒，我们都移除 word 的前 2 个字符，并在 word 末尾添加相同的字符。\n4 秒后，word 变为 \"abcbabcd\" 并恢复到初始状态。\n可以证明，4 秒是 word 恢复到其初始状态所需的最短时间。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= word.length &lt;= 50</code></li>\n\t<li><code>1 &lt;= k &lt;= word.length</code></li>\n\t<li><code>word</code>仅由小写英文字母组成。</li>\n</ul>\n",
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