leetcode-problemset/leetcode-cn/originData/string-transformation.json

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        "question": {
            "questionId": "3024",
            "questionFrontendId": "2851",
            "categoryTitle": "Algorithms",
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            "title": "String Transformation",
            "titleSlug": "string-transformation",
            "content": "<p>You are given two strings <code>s</code> and <code>t</code> of equal length <code>n</code>. You can perform the following operation on the string <code>s</code>:</p>\n\n<ul>\n\t<li>Remove a <strong>suffix</strong> of <code>s</code> of length <code>l</code> where <code>0 &lt; l &lt; n</code> and append it at the start of <code>s</code>.<br />\n\tFor example, let <code>s = &#39;abcd&#39;</code> then in one operation you can remove the suffix <code>&#39;cd&#39;</code> and append it in front of <code>s</code> making <code>s = &#39;cdab&#39;</code>.</li>\n</ul>\n\n<p>You are also given an integer <code>k</code>. Return <em>the number of ways in which </em><code>s</code> <em>can be transformed into </em><code>t</code><em> in <strong>exactly</strong> </em><code>k</code><em> operations.</em></p>\n\n<p>Since the answer can be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;abcd&quot;, t = &quot;cdab&quot;, k = 2\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> \nFirst way:\nIn first operation, choose suffix from index = 3, so resulting s = &quot;dabc&quot;.\nIn second operation, choose suffix from index = 3, so resulting s = &quot;cdab&quot;.\n\nSecond way:\nIn first operation, choose suffix from index = 1, so resulting s = &quot;bcda&quot;.\nIn second operation, choose suffix from index = 1, so resulting s = &quot;cdab&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;ababab&quot;, t = &quot;ababab&quot;, k = 1\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> \nFirst way:\nChoose suffix from index = 2, so resulting s = &quot;ababab&quot;.\n\nSecond way:\nChoose suffix from index = 4, so resulting s = &quot;ababab&quot;.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= s.length &lt;= 5 * 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= 10<sup>15</sup></code></li>\n\t<li><code>s.length == t.length</code></li>\n\t<li><code>s</code> and <code>t</code> consist of only lowercase English alphabets.</li>\n</ul>\n",
            "translatedTitle": "字符串转换",
            "translatedContent": "<p>给你两个长度都为 <code>n</code>&nbsp;的字符串&nbsp;<code>s</code> 和&nbsp;<code>t</code>&nbsp;。你可以对字符串 <code>s</code>&nbsp;执行以下操作：</p>\n\n<ul>\n\t<li>将 <code>s</code>&nbsp;长度为 <code>l</code>&nbsp;（<code>0 &lt; l &lt; n</code>）的 <strong>后缀字符串</strong>&nbsp;删除，并将它添加在 <code>s</code>&nbsp;的开头。<br />\n\t比方说，<code>s = 'abcd'</code>&nbsp;，那么一次操作中，你可以删除后缀&nbsp;<code>'cd'</code>&nbsp;，并将它添加到&nbsp;<code>s</code>&nbsp;的开头，得到&nbsp;<code>s = 'cdab'</code>&nbsp;。</li>\n</ul>\n\n<p>给你一个整数&nbsp;<code>k</code>&nbsp;，请你返回&nbsp;<strong>恰好</strong> <code>k</code>&nbsp;次操作将<em>&nbsp;</em><code>s</code> 变为<em>&nbsp;</em><code>t</code>&nbsp;的方案数。</p>\n\n<p>由于答案可能很大，返回答案对&nbsp;<code>10<sup>9</sup> + 7</code>&nbsp;<strong>取余</strong>&nbsp;后的结果。</p>\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 = \"abcd\", t = \"cdab\", k = 2\n<b>输出：</b>2\n<b>解释：</b>\n第一种方案：\n第一次操作，选择 index = 3 开始的后缀，得到 s = \"dabc\" 。\n第二次操作，选择 index = 3 开始的后缀，得到 s = \"cdab\" 。\n\n第二种方案：\n第一次操作，选择 index = 1 开始的后缀，得到 s = \"bcda\" 。\n第二次操作，选择 index = 1 开始的后缀，得到 s = \"cdab\" 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"ababab\", t = \"ababab\", k = 1\n<b>输出：</b>2\n<b>解释：</b>\n第一种方案：\n选择 index = 2 开始的后缀，得到 s = \"ababab\" 。\n\n第二种方案：\n选择 index = 4 开始的后缀，得到 s = \"ababab\" 。\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;= 5 * 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= 10<sup>15</sup></code></li>\n\t<li><code>s.length == t.length</code></li>\n\t<li><code>s</code> 和&nbsp;<code>t</code>&nbsp;都只包含小写英文字母。</li>\n</ul>\n",
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                "String <code>t</code> can be only constructed if it is a rotated version of string <code>s</code>.",
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