leetcode-problemset/leetcode-cn/originData/maximize-the-number-of-partitions-after-operations.json

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        "question": {
            "questionId": "3233",
            "questionFrontendId": "3003",
            "categoryTitle": "Algorithms",
            "boundTopicId": 2592229,
            "title": "Maximize the Number of Partitions After Operations",
            "titleSlug": "maximize-the-number-of-partitions-after-operations",
            "content": "<p>You are given a string <code>s</code> and an integer <code>k</code>.</p>\n\n<p>First, you are allowed to change <strong>at most</strong> <strong>one</strong> index in <code>s</code> to another lowercase English letter.</p>\n\n<p>After that, do the following partitioning operation until <code>s</code> is <strong>empty</strong>:</p>\n\n<ul>\n\t<li>Choose the <strong>longest</strong> <strong>prefix</strong> of <code>s</code> containing at most <code>k</code> <strong>distinct</strong> characters.</li>\n\t<li><strong>Delete</strong> the prefix from <code>s</code> and increase the number of partitions by one. The remaining characters (if any) in <code>s</code> maintain their initial order.</li>\n</ul>\n\n<p>Return an integer denoting the <strong>maximum</strong> number of resulting partitions after the operations by optimally choosing at most one index to change.</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;accca&quot;, k = 2</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">3</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The optimal way is to change <code>s[2]</code> to something other than a and c, for example, b. then it becomes <code>&quot;acbca&quot;</code>.</p>\n\n<p>Then we perform the operations:</p>\n\n<ol>\n\t<li>The longest prefix containing at most 2 distinct characters is <code>&quot;ac&quot;</code>, we remove it and <code>s</code> becomes <code>&quot;bca&quot;</code>.</li>\n\t<li>Now The longest prefix containing at most 2 distinct characters is <code>&quot;bc&quot;</code>, so we remove it and <code>s</code> becomes <code>&quot;a&quot;</code>.</li>\n\t<li>Finally, we remove <code>&quot;a&quot;</code> and <code>s</code> becomes empty, so the procedure ends.</li>\n</ol>\n\n<p>Doing the operations, the string is divided into 3 partitions, so the answer is 3.</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\">s = &quot;aabaab&quot;, k = 3</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">1</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>Initially&nbsp;<code>s</code>&nbsp;contains 2 distinct characters, so whichever character we change, it will contain at most 3 distinct characters, so the longest prefix with at most 3 distinct characters would always be all of it, therefore the answer is 1.</p>\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;xxyz&quot;, k = 1</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">4</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The optimal way is to change&nbsp;<code>s[0]</code>&nbsp;or&nbsp;<code>s[1]</code>&nbsp;to something other than characters in&nbsp;<code>s</code>, for example, to change&nbsp;<code>s[0]</code>&nbsp;to&nbsp;<code>w</code>.</p>\n\n<p>Then&nbsp;<code>s</code>&nbsp;becomes <code>&quot;wxyz&quot;</code>, which consists of 4 distinct characters, so as <code>k</code> is 1, it will divide into 4 partitions.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>s</code> consists only of lowercase English letters.</li>\n\t<li><code>1 &lt;= k &lt;= 26</code></li>\n</ul>\n",
            "translatedTitle": "执行操作后的最大分割数量",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始的字符串&nbsp;<code>s</code>&nbsp;和一个整数&nbsp;<code>k</code>。</p>\n\n<p>你需要执行以下分割操作，直到字符串&nbsp;<code>s&nbsp;</code>变为&nbsp;<strong>空</strong>：</p>\n\n<ul>\n\t<li>选择&nbsp;<code>s</code>&nbsp;的最长&nbsp;<strong>前缀</strong>，该前缀最多包含&nbsp;<code>k&nbsp;</code>个&nbsp;<strong>不同&nbsp;</strong>字符。</li>\n\t<li><strong>删除&nbsp;</strong>这个前缀，并将分割数量加一。如果有剩余字符，它们在&nbsp;<code>s</code>&nbsp;中保持原来的顺序。</li>\n</ul>\n\n<p>执行操作之 <strong>前</strong> ，你可以将&nbsp;<code>s</code>&nbsp;中&nbsp;<strong>至多一处 </strong>下标的对应字符更改为另一个小写英文字母。</p>\n\n<p>在最优选择情形下改变至多一处下标对应字符后，用整数表示并返回操作结束时得到的 <strong>最大</strong> 分割数量。</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 = \"accca\", k = 2</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">3</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>最好的方式是把&nbsp;<code>s[2]</code>&nbsp;变为除了 a 和 c 之外的东西，比如&nbsp;b。然后它变成了&nbsp;<code>\"acbca\"</code>。</p>\n\n<p>然后我们执行以下操作：</p>\n\n<ol>\n\t<li>最多包含 2 个不同字符的最长前缀是 <code>\"ac\"</code>，我们删除它然后&nbsp;<code>s</code> 变为&nbsp;<code>\"bca\"</code>。</li>\n\t<li>现在最多包含 2 个不同字符的最长前缀是&nbsp;<code>\"bc\"</code>，所以我们删除它然后&nbsp;<code>s</code> 变为&nbsp;<code>\"a\"</code>。</li>\n\t<li>最后，我们删除&nbsp;<code>\"a\"</code>&nbsp;并且&nbsp;<code>s</code>&nbsp;变成空串，所以该过程结束。</li>\n</ol>\n\n<p>进行操作时，字符串被分成 3 个部分，所以答案是 3。</p>\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 = \"aabaab\", k = 3</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">1</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>一开始&nbsp;<code>s</code>&nbsp;包含 2 个不同的字符，所以无论我们改变哪个，&nbsp;它最多包含 3 个不同字符，因此最多包含 3 个不同字符的最长前缀始终是所有字符，因此答案是 1。</p>\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 = \"xxyz\", k = 1</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>4</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>最好的方式是将&nbsp;<code>s[0]</code>&nbsp;或&nbsp;<code>s[1]</code>&nbsp;变为&nbsp;<code>s</code>&nbsp;中字符以外的东西，例如将&nbsp;<code>s[0]</code>&nbsp;变为&nbsp;<code>w</code>。</p>\n\n<p>然后&nbsp;<code>s</code>&nbsp;变为&nbsp;<code>\"wxyz\"</code>，包含 4 个不同的字符，所以当&nbsp;<code>k</code>&nbsp;为 1，它将分为 4 个部分。</p>\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 &lt;= 10<sup>4</sup></code></li>\n\t<li><code>s</code>&nbsp;只包含小写英文字母。</li>\n\t<li><code>1 &lt;= k &lt;= 26</code></li>\n</ul>\n",
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            "hints": [
                "For each position, try to brute-force the replacements.",
                "To speed up the brute-force solution, we can precompute the following (without changing any index) using prefix sums and binary search:<ul>\r\n<li><code>pref[i]</code>: The number of resulting partitions from the operations by performing the operations on <code>s[0:i]</code>.</li>\r\n<li><code>suff[i]</code>: The number of resulting partitions from the operations by performing the operations on <code>s[i:n - 1]</code>, where <code>n == s.length</code>.</li>\r\n<li><code>partition_start[i]</code>: The start index of the partition containing the <code>i<sup>th</sup></code> index after performing the operations.</li>\r\n</ul>",
                "Now, for a position <code>i</code>, we can try all possible <code>25</code> replacements:<br />\r\nFor a replacement, using prefix sums and binary search, we need to find the rightmost index, <code>r</code>, such that the number of distinct characters in the range <code>[partition_start[i], r]</code> is at most <code>k</code>.<br />\r\nThere are <code>2</code> cases:<ul>\r\n<li><code>r >= i</code>: the number of resulting partitions in this case is <code>1 + pref[partition_start[i] - 1] + suff[r + 1]</code>.</li>\r\n<li>Otherwise, we need to find the rightmost index <code>r<sub>2</sub></code> such that the number of distinct characters in the range <code>[r:r<sub>2</sub>]</code> is at most <code>k</code>. The answer in this case is <code>2 + pref[partition_start[i] - 1] + suff[r<sub>2</sub> + 1]</code></li>\r\n</ul>",
                "The answer is the maximum among all replacements."
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