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leetcode-problemset/leetcode/originData/minimum-number-of-operations-to-make-word-k-periodic.json
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"title": "Minimum Number of Operations to Make Word K-Periodic",
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"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",
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"hints": [
"Calculate the frequency of each substring of length <code>k</code> that starts at an index that is divisible by <code>k</code>.",
"The period of the final string will be the substring with the highest frequency."
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