leetcode-problemset/leetcode-cn/originData/minimum-number-of-operations-to-make-word-k-periodic.json

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        "question": {
            "questionId": "3384",
            "questionFrontendId": "3137",
            "categoryTitle": "Algorithms",
            "boundTopicId": 2764713,
            "title": "Minimum Number of Operations to Make Word K-Periodic",
            "titleSlug": "minimum-number-of-operations-to-make-word-k-periodic",
            "content": "<p>You are given a string <code>word</code> of size <code>n</code>, and an integer <code>k</code> such that <code>k</code> divides <code>n</code>.</p>\n\n<p>In one operation, you can pick any two indices <code>i</code> and <code>j</code>, that are divisible by <code>k</code>, then replace the <span data-keyword=\"substring\">substring</span> of length <code>k</code> starting at <code>i</code> with the substring of length <code>k</code> starting at <code>j</code>. That is, replace the substring <code>word[i..i + k - 1]</code> with the substring <code>word[j..j + k - 1]</code>.<!-- notionvc: 49ac84f7-0724-452a-ab43-0c5e53f1db33 --></p>\n\n<p>Return <em>the <strong>minimum</strong> number of operations required to make</em> <code>word</code> <em><strong>k-periodic</strong></em>.</p>\n\n<p>We say that <code>word</code> is <strong>k-periodic</strong> if there is some string <code>s</code> of length <code>k</code> such that <code>word</code> can be obtained by concatenating <code>s</code> an arbitrary number of times. For example, if <code>word == &ldquo;ababab&rdquo;</code>, then <code>word</code> is 2-periodic for <code>s = &quot;ab&quot;</code>.</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\" style=\"\n    font-family: Menlo,sans-serif;\n    font-size: 0.85rem;\n\">word = &quot;leetcodeleet&quot;, k = 4</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\" style=\"\nfont-family: Menlo,sans-serif;\nfont-size: 0.85rem;\n\">1</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>We can obtain a 4-periodic string by picking i = 4 and j = 0. After this operation, word becomes equal to &quot;leetleetleet&quot;.</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\" style=\"\n    font-family: Menlo,sans-serif;\n    font-size: 0.85rem;\n\">word = &quot;</span>leetcoleet<span class=\"example-io\" style=\"\n    font-family: Menlo,sans-serif;\n    font-size: 0.85rem;\n\">&quot;, k = 2</span></p>\n\n<p><strong>Output:</strong> 3</p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>We can obtain a 2-periodic string by applying the operations in the table below.</p>\n\n<table border=\"1\" bordercolor=\"#ccc\" cellpadding=\"5\" cellspacing=\"0\" height=\"146\" style=\"border-collapse:collapse; text-align: center; vertical-align: middle;\">\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>i</th>\n\t\t\t<th>j</th>\n\t\t\t<th>word</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td style=\"padding: 5px 15px;\">0</td>\n\t\t\t<td style=\"padding: 5px 15px;\">2</td>\n\t\t\t<td style=\"padding: 5px 15px;\">etetcoleet</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td style=\"padding: 5px 15px;\">4</td>\n\t\t\t<td style=\"padding: 5px 15px;\">0</td>\n\t\t\t<td style=\"padding: 5px 15px;\">etetetleet</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td style=\"padding: 5px 15px;\">6</td>\n\t\t\t<td style=\"padding: 5px 15px;\">0</td>\n\t\t\t<td style=\"padding: 5px 15px;\">etetetetet</td>\n\t\t</tr>\n\t</tbody>\n</table>\n</div>\n\n<div id=\"gtx-trans\" style=\"position: absolute; left: 107px; top: 238.5px;\">\n<div class=\"gtx-trans-icon\">&nbsp;</div>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == word.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= word.length</code></li>\n\t<li><code>k</code> divides <code>word.length</code>.</li>\n\t<li><code>word</code> consists only of lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "K 周期字符串需要的最少操作次数",
            "translatedContent": "<p>给你一个长度为 <code>n</code> 的字符串 <code>word</code> 和一个整数 <code>k</code> ，其中 <code>k</code> 是 <code>n</code> 的因数。</p>\n\n<p>在一次操作中，你可以选择任意两个下标 <code>i</code> 和 <code>j</code>，其中 <code>0 &lt;= i, j &lt; n</code> ，且这两个下标都可以被 <code>k</code> 整除，然后用从 <code>j</code> 开始的长度为 <code>k</code> 的子串替换从 <code>i</code> 开始的长度为 <code>k</code> 的子串。也就是说，将子串 <code>word[i..i + k - 1]</code> 替换为子串 <code>word[j..j + k - 1]</code> 。</p>\n\n<p>返回使 <code>word</code> 成为 <strong>K 周期字符串</strong> 所需的<strong> 最少</strong> 操作次数。</p>\n\n<p>如果存在某个长度为 <code>k</code> 的字符串 <code>s</code>，使得 <code>word</code> 可以表示为任意次数连接 <code>s</code> ，则称字符串 <code>word</code> 是 <strong>K 周期字符串</strong> 。例如，如果 <code>word == \"ababab\"</code>，那么 <code>word</code> 就是 <code>s = \"ab\"</code> 时的 2 周期字符串 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\" style=\"\n    border-color: var(--border-tertiary);\n    border-left-width: 2px;\n    color: var(--text-secondary);\n    font-size: .875rem;\n    margin-bottom: 1rem;\n    margin-top: 1rem;\n    overflow: visible;\n    padding-left: 1rem;\n\">\n<p><strong>输入：</strong><span class=\"example-io\" style=\"\n    font-family: Menlo,sans-serif;\n    font-size: 0.85rem;\n\">word = \"leetcodeleet\", k = 4</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\" style=\"\nfont-family: Menlo,sans-serif;\nfont-size: 0.85rem;\n\">1</span></p>\n\n<p><strong>解释：</strong>可以选择 i = 4 和 j = 0 获得一个 4 周期字符串。这次操作后，word 变为 \"leetleetleet\" 。</p>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\" style=\"\n    border-color: var(--border-tertiary);\n    border-left-width: 2px;\n    color: var(--text-secondary);\n    font-size: .875rem;\n    margin-bottom: 1rem;\n    margin-top: 1rem;\n    overflow: visible;\n    padding-left: 1rem;\n\">\n<p><strong>输入：</strong><span class=\"example-io\" style=\"\n    font-family: Menlo,sans-serif;\n    font-size: 0.85rem;\n\">word = \"leetcoleet\", k = 2</span></p>\n\n<p><strong>输出：</strong>3</p>\n\n<p><strong>解释：</strong>可以执行以下操作获得一个 2 周期字符串。</p>\n\n<table border=\"1\" bordercolor=\"#ccc\" cellpadding=\"5\" cellspacing=\"0\" height=\"146\" style=\"border-collapse:collapse; text-align: center; vertical-align: middle;\">\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>i</th>\n\t\t\t<th>j</th>\n\t\t\t<th>word</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td style=\"padding: 5px 15px;\">0</td>\n\t\t\t<td style=\"padding: 5px 15px;\">2</td>\n\t\t\t<td style=\"padding: 5px 15px;\">etetcoleet</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td style=\"padding: 5px 15px;\">4</td>\n\t\t\t<td style=\"padding: 5px 15px;\">0</td>\n\t\t\t<td style=\"padding: 5px 15px;\">etetetleet</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td style=\"padding: 5px 15px;\">6</td>\n\t\t\t<td style=\"padding: 5px 15px;\">0</td>\n\t\t\t<td style=\"padding: 5px 15px;\">etetetetet</td>\n\t\t</tr>\n\t</tbody>\n</table>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == word.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= word.length</code></li>\n\t<li><code>k</code> 能整除 <code>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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