leetcode-problemset/leetcode/originData/maximum-strength-of-k-disjoint-subarrays.json

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            "title": "Maximum Strength of K Disjoint Subarrays",
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            "content": "<p>You are given a <strong>0-indexed</strong> array of integers <code>nums</code> of length <code>n</code>, and a <strong>positive</strong> <strong>odd</strong> integer <code>k</code>.</p>\n\n<p>The strength of <code>x</code> subarrays is defined as <code>strength = sum[1] * x - sum[2] * (x - 1) + sum[3] * (x - 2) - sum[4] * (x - 3) + ... + sum[x] * 1</code> where <code>sum[i]</code> is the sum of the elements in the <code>i<sup>th</sup></code> subarray. Formally, strength is sum of <code>(-1)<sup>i+1</sup> * sum[i] * (x - i + 1)</code> over all <code>i</code>&#39;s such that <code>1 &lt;= i &lt;= x</code>.</p>\n\n<p>You need to select <code>k</code> <strong>disjoint <span data-keyword=\"subarray-nonempty\">subarrays</span></strong> from <code>nums</code>, such that their <strong>strength</strong> is <strong>maximum</strong>.</p>\n\n<p>Return <em>the <strong>maximum</strong> possible <strong>strength</strong> that can be obtained</em>.</p>\n\n<p><strong>Note</strong> that the selected subarrays <strong>don&#39;t</strong> need to cover the entire array.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3,-1,2], k = 3\n<strong>Output:</strong> 22\n<strong>Explanation:</strong> The best possible way to select 3 subarrays is: nums[0..2], nums[3..3], and nums[4..4]. The strength is (1 + 2 + 3) * 3 - (-1) * 2 + 2 * 1 = 22.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [12,-2,-2,-2,-2], k = 5\n<strong>Output:</strong> 64\n<strong>Explanation:</strong> The only possible way to select 5 disjoint subarrays is: nums[0..0], nums[1..1], nums[2..2], nums[3..3], and nums[4..4]. The strength is 12 * 5 - (-2) * 4 + (-2) * 3 - (-2) * 2 + (-2) * 1 = 64.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [-1,-2,-3], k = 1\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> The best possible way to select 1 subarray is: nums[0..0]. The strength is -1.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= n</code></li>\n\t<li><code>1 &lt;= n * k &lt;= 10<sup>6</sup></code></li>\n\t<li><code>k</code> is odd.</li>\n</ul>\n",
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                    "lang": "C++",
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                    "code": "class Solution {\npublic:\n    long long maximumStrength(vector<int>& nums, int k) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def maximumStrength(self, nums, k):\n        \"\"\"\n        :type nums: List[int]\n        :type k: int\n        :rtype: int\n        \"\"\"\n        ",
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                    "code": "class Solution:\n    def maximumStrength(self, nums: List[int], k: int) -> int:\n        ",
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                    "lang": "C",
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                    "code": "long long maximumStrength(int* nums, int numsSize, int k) {\n    \n}",
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                    "code": "public class Solution {\n    public long MaximumStrength(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 maximumStrength = function(nums, k) {\n    \n};",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer[] $nums\n     * @param Integer $k\n     * @return Integer\n     */\n    function maximumStrength($nums, $k) {\n        \n    }\n}",
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                    "code": "class Solution {\n    func maximumStrength(_ nums: [Int], _ k: Int) -> Int {\n        \n    }\n}",
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                    "code": "class Solution {\n    fun maximumStrength(nums: IntArray, k: Int): Long {\n        \n    }\n}",
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                    "code": "func maximumStrength(nums []int, k int) int64 {\n    \n}",
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                    "lang": "Ruby",
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                    "code": "# @param {Integer[]} nums\n# @param {Integer} k\n# @return {Integer}\ndef maximum_strength(nums, k)\n    \nend",
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                    "code": "object Solution {\n    def maximumStrength(nums: Array[Int], k: Int): Long = {\n        \n    }\n}",
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                    "lang": "Rust",
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                    "code": "impl Solution {\n    pub fn maximum_strength(nums: Vec<i32>, k: i32) -> i64 {\n        \n    }\n}",
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                    "lang": "Racket",
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                    "code": "(define/contract (maximum-strength nums k)\n  (-> (listof exact-integer?) exact-integer? exact-integer?)\n  )",
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                    "lang": "Erlang",
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                    "code": "-spec maximum_strength(Nums :: [integer()], K :: integer()) -> integer().\nmaximum_strength(Nums, K) ->\n  .",
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                    "code": "defmodule Solution do\n  @spec maximum_strength(nums :: [integer], k :: integer) :: integer\n  def maximum_strength(nums, k) do\n    \n  end\nend",
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            "hints": [
                "Let <code>dp[i][j][x == 0/1]</code> be the maximum strength to select <code>j</code> disjoint subarrays from the original array’s suffix (<code>nums[i..(n - 1)]</code>), x denotes whether we select the element or not.",
                "Initially <code>dp[n][0][0] == 0</code>.",
                "We have \r\n<code>dp[i][j][1] = nums[i] * get(j) + max(dp[i + 1][j - 1][0], dp[i + 1][j][1])</code> where <code>get(j) = j</code> if <code>j</code> is odd, otherwise <code>-j</code>.",
                "We can select <code>nums[i]</code> as a separate subarray or select at least <code>nums[i]</code> and <code>nums[i + 1]</code> as the first subarray.\r\n<code>dp[i][j][0] = max(dp[i + 1][j][0], dp[i][j][1])</code>.",
                "The answer is <code>dp[0][k][0]</code>."
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