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

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        "question": {
            "questionId": "3233",
            "questionFrontendId": "3003",
            "categoryTitle": "Algorithms",
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            "title": "Maximize the Number of Partitions After Operations",
            "titleSlug": "maximize-the-number-of-partitions-after-operations",
            "content": "<p>You are given a <strong>0-indexed</strong> string <code>s</code> and an integer <code>k</code>.</p>\n\n<p>You are to perform the following partitioning operations 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><strong>Before</strong> the operations, 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>Return <em>an integer denoting the <strong>maximum</strong> number of resulting partitions after the operations by optimally choosing at most one index to change.</em></p>\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;accca&quot;, k = 2\n<strong>Output:</strong> 3\n<strong>Explanation: </strong>In this example, to maximize the number of resulting partitions, s[2] can be changed to &#39;b&#39;.\ns becomes &quot;acbca&quot;.\nThe operations can now be performed as follows until s becomes empty:\n- Choose the longest prefix containing at most 2 distinct characters, &quot;<u>ac</u>bca&quot;.\n- Delete the prefix, and s becomes &quot;bca&quot;. The number of partitions is now 1.\n- Choose the longest prefix containing at most 2 distinct characters, &quot;<u>bc</u>a&quot;.\n- Delete the prefix, and s becomes &quot;a&quot;. The number of partitions is now 2.\n- Choose the longest prefix containing at most 2 distinct characters, &quot;<u>a</u>&quot;.\n- Delete the prefix, and s becomes empty. The number of partitions is now 3.\nHence, the answer is 3.\nIt can be shown that it is not possible to obtain more than 3 partitions.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;aabaab&quot;, k = 3\n<strong>Output:</strong> 1\n<strong>Explanation: </strong>In this example, to maximize the number of resulting partitions we can leave s as it is.\nThe operations can now be performed as follows until s becomes empty: \n- Choose the longest prefix containing at most 3 distinct characters, &quot;<u>aabaab</u>&quot;.\n- Delete the prefix, and s becomes empty. The number of partitions becomes 1. \nHence, the answer is 1. \nIt can be shown that it is not possible to obtain more than 1 partition.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;xxyz&quot;, k = 1\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> In this example, to maximize the number of resulting partitions, s[1] can be changed to &#39;a&#39;.\ns becomes &quot;xayz&quot;.\nThe operations can now be performed as follows until s becomes empty:\n- Choose the longest prefix containing at most 1 distinct character, &quot;<u>x</u>ayz&quot;.\n- Delete the prefix, and s becomes &quot;ayz&quot;. The number of partitions is now 1.\n- Choose the longest prefix containing at most 1 distinct character, &quot;<u>a</u>yz&quot;.\n- Delete the prefix, and s becomes &quot;yz&quot;. The number of partitions is now 2.\n- Choose the longest prefix containing at most 1 distinct character, &quot;<u>y</u>z&quot;.\n- Delete the prefix, and s becomes &quot;z&quot;. The number of partitions is now 3.\n- Choose the longest prefix containing at most 1 distinct character, &quot;<u>z</u>&quot;.\n- Delete the prefix, and s becomes empty. The number of partitions is now 4.\nHence, the answer is 4.\nIt can be shown that it is not possible to obtain more than 4 partitions.\n</pre>\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<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最大分割数量。</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>s = \"accca\", k = 2\n<strong>输出：</strong>3\n<strong>解释：</strong>在此示例中，为了最大化得到的分割数量，可以将 s[2] 改为 'b'。\ns 变为 \"acbca\"。\n按照以下方式执行操作，直到 s 变为空：\n- 选择最长且至多包含 2 个不同字符的前缀，\"<em><strong>ac</strong></em>bca\"。\n- 删除该前缀，s 变为 \"bca\"。现在分割数量为 1。\n- 选择最长且至多包含 2 个不同字符的前缀，\"<em><strong>bc</strong></em>a\"。\n- 删除该前缀，s 变为 \"a\"。现在分割数量为 2。\n- 选择最长且至多包含 2 个不同字符的前缀，\"<strong><em>a</em></strong>\"。\n- 删除该前缀，s 变为空。现在分割数量为 3。\n因此，答案是 3。\n可以证明，分割数量不可能超过 3。</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"aabaab\", k = 3\n<strong>输出：</strong>1\n<strong>解释：</strong>在此示例中，为了最大化得到的分割数量，可以保持 s 不变。\n按照以下方式执行操作，直到 s 变为空： \n- 选择最长且至多包含 3 个不同字符的前缀，\"<em><strong>aabaab</strong></em>\"。\n- 删除该前缀，s 变为空。现在分割数量为 1。\n因此，答案是 1。\n可以证明，分割数量不可能超过 1。</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"xxyz\", k = 1\n<strong>输出：</strong>4\n<strong>解释：</strong>在此示例中，为了最大化得到的分割数量，可以将 s[1] 改为 'a'。\ns 变为 \"xayz\"。\n按照以下方式执行操作，直到 s 变为空：\n- 选择最长且至多包含 1 个不同字符的前缀，\"<em><strong>x</strong></em>ayz\"。\n- 删除该前缀，s 变为 \"ayz\"。现在分割数量为 1。\n- 选择最长且至多包含 1 个不同字符的前缀，\"<em><strong>a</strong></em>yz\"。\n- 删除该前缀，s 变为 \"yz\"，现在分割数量为 2。\n- 选择最长且至多包含 1 个不同字符的前缀，\"<em><strong>y</strong></em>z\"。\n- 删除该前缀，s 变为 \"z\"。现在分割数量为 3。\n- 选择最且至多包含 1 个不同字符的前缀，\"<em>z</em>\"。\n- 删除该前缀，s 变为空。现在分割数量为 4。\n因此，答案是 4。\n可以证明，分割数量不可能超过 4。</pre>\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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