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leetcode-problemset/leetcode/originData/maximize-the-number-of-partitions-after-operations.json
2024-01-09 10:57:06 +08:00

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"title": "Maximize the Number of Partitions After Operations",
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"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",
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"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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