leetcode-problemset/leetcode/originData/minimum-cost-to-split-an-array.json

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            "title": "Minimum Cost to Split an Array",
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            "content": "<p>You are given an integer array <code>nums</code> and an integer <code>k</code>.</p>\n\n<p>Split the array into some number of non-empty subarrays. The <strong>cost</strong> of a split is the sum of the <strong>importance value</strong> of each subarray in the split.</p>\n\n<p>Let <code>trimmed(subarray)</code> be the version of the subarray where all numbers which appear only once are removed.</p>\n\n<ul>\n\t<li>For example, <code>trimmed([3,1,2,4,3,4]) = [3,4,3,4].</code></li>\n</ul>\n\n<p>The <strong>importance value</strong> of a subarray is <code>k + trimmed(subarray).length</code>.</p>\n\n<ul>\n\t<li>For example, if a subarray is <code>[1,2,3,3,3,4,4]</code>, then <font face=\"monospace\">trimmed(</font><code>[1,2,3,3,3,4,4]) = [3,3,3,4,4].</code>The importance value of this subarray will be <code>k + 5</code>.</li>\n</ul>\n\n<p>Return <em>the minimum possible cost of a split of </em><code>nums</code>.</p>\n\n<p>A <strong>subarray</strong> is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,1,2,1,3,3], k = 2\n<strong>Output:</strong> 8\n<strong>Explanation:</strong> We split nums to have two subarrays: [1,2], [1,2,1,3,3].\nThe importance value of [1,2] is 2 + (0) = 2.\nThe importance value of [1,2,1,3,3] is 2 + (2 + 2) = 6.\nThe cost of the split is 2 + 6 = 8. It can be shown that this is the minimum possible cost among all the possible splits.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,1,2,1], k = 2\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> We split nums to have two subarrays: [1,2], [1,2,1].\nThe importance value of [1,2] is 2 + (0) = 2.\nThe importance value of [1,2,1] is 2 + (2) = 4.\nThe cost of the split is 2 + 4 = 6. It can be shown that this is the minimum possible cost among all the possible splits.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,1,2,1], k = 5\n<strong>Output:</strong> 10\n<strong>Explanation:</strong> We split nums to have one subarray: [1,2,1,2,1].\nThe importance value of [1,2,1,2,1] is 5 + (3 + 2) = 10.\nThe cost of the split is 10. It can be shown that this is the minimum possible cost among all the possible splits.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>0 &lt;= nums[i] &lt; nums.length</code></li>\n\t<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n<style type=\"text/css\">.spoilerbutton {display:block; border:dashed; padding: 0px 0px; margin:10px 0px; font-size:150%; font-weight: bold; color:#000000; background-color:cyan; outline:0; \n}\n.spoiler {overflow:hidden;}\n.spoiler > div {-webkit-transition: all 0s ease;-moz-transition: margin 0s ease;-o-transition: all 0s ease;transition: margin 0s ease;}\n.spoilerbutton[value=\"Show Message\"] + .spoiler > div {margin-top:-500%;}\n.spoilerbutton[value=\"Hide Message\"] + .spoiler {padding:5px;}\n</style>\n",
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                    "code": "class Solution(object):\n    def minCost(self, nums, k):\n        \"\"\"\n        :type nums: List[int]\n        :type k: int\n        :rtype: int\n        \"\"\"\n        ",
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                    "code": "public class Solution {\n    public int MinCost(int[] nums, int k) {\n        \n    }\n}",
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                    "code": "/**\n * @param {number[]} nums\n * @param {number} k\n * @return {number}\n */\nvar minCost = function(nums, k) {\n    \n};",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer[] $nums\n     * @param Integer $k\n     * @return Integer\n     */\n    function minCost($nums, $k) {\n        \n    }\n}",
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                    "code": "# @param {Integer[]} nums\n# @param {Integer} k\n# @return {Integer}\ndef min_cost(nums, k)\n    \nend",
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                "Let's denote dp[r] = minimum cost to partition the first r elements of nums. What would be the transitions of such dynamic programming?",
                "dp[r] = min(dp[l] + importance(nums[l..r])) over all 0 <= l < r. This already gives us an O(n^3) approach, as importance can be calculated in linear time, and there are a total of O(n^2) transitions.",
                "Can you think of a way to compute multiple importance values of related subarrays faster?",
                "importance(nums[l-1..r]) is either importance(nums[l..r]) if a new unique element is added, importance(nums[l..r]) + 1 if an old element that appeared at least twice is added, or importance(nums[l..r]) + 2, if a previously unique element is duplicated. This allows us to compute importance(nums[l..r]) for all 0 <= l < r in O(n) by keeping a frequency table and decreasing l from r-1 down to 0."
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