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leetcode-problemset/leetcode/originData/find-the-maximum-length-of-a-good-subsequence-ii.json
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"title": "Find the Maximum Length of a Good Subsequence II",
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"content": "<p>You are given an integer array <code>nums</code> and a <strong>non-negative</strong> integer <code>k</code>. A sequence of integers <code>seq</code> is called <strong>good</strong> if there are <strong>at most</strong> <code>k</code> indices <code>i</code> in the range <code>[0, seq.length - 2]</code> such that <code>seq[i] != seq[i + 1]</code>.</p>\n\n<p>Return the <strong>maximum</strong> possible length of a <strong>good</strong> <span data-keyword=\"subsequence-array\">subsequence</span> of <code>nums</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\">nums = [1,2,1,1,3], k = 2</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 maximum length subsequence is <code>[<u>1</u>,<u>2</u>,<u>1</u>,<u>1</u>,3]</code>.</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\">nums = [1,2,3,4,5,1], k = 0</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">2</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The maximum length subsequence is <code>[<u>1</u>,2,3,4,5,<u>1</u>]</code>.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>3</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= k &lt;= min(50, nums.length)</code></li>\n</ul>\n",
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"The absolute values in <code>nums</code> dont really matter. So we can remap the set of values to the range <code>[0, n - 1]</code>.",
"Let <code>dp[i][j]</code> be the length of the longest subsequence till index <code>j</code> with at most <code>i</code> positions such that <code>seq[i] != seq[i + 1]</code>.",
"For each value <code>x</code> from left to right, update <code>dp[i][x] = max(dp[i][x] + 1, dp[i - 1][y] + 1)</code>, where <code>y != x</code>."
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