You are given a cyclic array nums and an integer k.
Partition nums into at most k subarrays. As nums is cyclic, these subarrays may wrap around from the end of the array back to the beginning.
The range of a subarray is the difference between its maximum and minimum values. The score of a partition is the sum of subarray ranges.
\n\nReturn the maximum possible score among all cyclic partitions.
\n\n\n
Example 1:
\n\nInput: nums = [1,2,3,3], k = 2
\n\nOutput: 3
\n\nExplanation:
\n\nnums into [2, 3] and [3, 1] (wrapped around).[2, 3] is max(2, 3) - min(2, 3) = 3 - 2 = 1.[3, 1] is max(3, 1) - min(3, 1) = 3 - 1 = 2.1 + 2 = 3.Example 2:
\n\nInput: nums = [1,2,3,3], k = 1
\n\nOutput: 2
\n\nExplanation:
\n\nnums into [1, 2, 3, 3].[1, 2, 3, 3] is max(1, 2, 3, 3) - min(1, 2, 3, 3) = 3 - 1 = 2.Example 3:
\n\nInput: nums = [1,2,3,3], k = 4
\n\nOutput: 3
\n\nExplanation:
\n\nIdentical to Example 1, we partition nums into [2, 3] and [3, 1]. Note that nums may be partitioned into fewer than k subarrays.
\n
Constraints:
\n\n1 <= nums.length <= 10001 <= nums[i] <= 1091 <= k <= nums.length给你一个 循环 数组 nums 和一个整数 k。
将 nums 划分 为 最多 k 个子数组。由于 nums 是循环数组，这些子数组可以从数组末尾环绕回起点。
子数组的 范围 定义为其 最大值 与 最小值 的差值。划分的 得分 是所有子数组范围的总和。
\n\n返回所有循环划分方案中可能获得的 最大得分 。
\n\n子数组 是数组中的一个连续非空的元素序列。
\n\n\n\n
示例 1：
\n\n输入： nums = [1,2,3,3], k = 2
\n\n输出： 3
\n\n解释：
\n\nnums 划分为 [2, 3] 和 [3, 1]（环绕）。[2, 3] 的范围是 max(2, 3) - min(2, 3) = 3 - 2 = 1。[3, 1] 的范围是 max(3, 1) - min(3, 1) = 3 - 1 = 2。1 + 2 = 3。示例 2：
\n\n输入： nums = [1,2,3,3], k = 1
\n\n输出： 2
\n\n解释：
\n\nnums 划分为 [1, 2, 3, 3]。[1, 2, 3, 3] 的范围是 max(1, 2, 3, 3) - min(1, 2, 3, 3) = 3 - 1 = 2。2。示例 3：
\n\n输入： nums = [1,2,3,3], k = 4
\n\n输出： 3
\n\n解释：
\n\n与示例 1 相同，将 nums 划分为 [2, 3] 和 [3, 1]。注意，可以将 nums 划分为少于 k 个子数组。
\n\n
提示：
\n\n1 <= nums.length <= 10001 <= nums[i] <= 1091 <= k <= nums.length2 * k picks (each partition can supply a + and a -), so the problem becomes choosing which elements are + or -.",
"Model a partition by its max (+nums[i]) and min (-nums[i]) contributions; selecting an element as a max adds +nums[i], and selecting it as a min adds -nums[i].",
"Use a DP state (picks, balance) where picks is the total number of + and - chosen so far (<= 2 * k), and balance represents the difference between the counts of + and - currently; at each i, you conceptually take +, take -, or skip.",
"Handle cyclicity by limiting balance to the range [0, 2]. You can show that such a balance is always achievable."
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