You are given an integer array nums and an integer k.
Split the array into some number of non-empty subarrays. The cost of a split is the sum of the importance value of each subarray in the split.
\n\nLet trimmed(subarray) be the version of the subarray where all numbers which appear only once are removed.
trimmed([3,1,2,4,3,4]) = [3,4,3,4].The importance value of a subarray is k + trimmed(subarray).length.
[1,2,3,3,3,4,4], then trimmed([1,2,3,3,3,4,4]) = [3,3,3,4,4].The importance value of this subarray will be k + 5.Return the minimum possible cost of a split of nums.
A subarray is a contiguous non-empty sequence of elements within an array.
\n\n\n
Example 1:
\n\n\nInput: nums = [1,2,1,2,1,3,3], k = 2\nOutput: 8\nExplanation: 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\n\n
Example 2:
\n\n\nInput: nums = [1,2,1,2,1], k = 2\nOutput: 6\nExplanation: 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\n\n
Example 3:
\n\n\nInput: nums = [1,2,1,2,1], k = 5\nOutput: 10\nExplanation: 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\n\n
\n
Constraints:
\n\n1 <= nums.length <= 10000 <= nums[i] < nums.length1 <= k <= 109\n\n", "translatedTitle": "拆分数组的最小代价", "translatedContent": "
给你一个整数数组 nums 和一个整数 k 。
将数组拆分成一些非空子数组。拆分的 代价 是每个子数组中的 重要性 之和。
\n\n令 trimmed(subarray) 作为子数组的一个特征，其中所有仅出现一次的数字将会被移除。
trimmed([3,1,2,4,3,4]) = [3,4,3,4] 。子数组的 重要性 定义为 k + trimmed(subarray).length 。
[1,2,3,3,3,4,4] ，trimmed([1,2,3,3,3,4,4]) = [3,3,3,4,4] 。这个子数组的重要性就是 k + 5 。找出并返回拆分 nums 的所有可行方案中的最小代价。
子数组 是数组的一个连续 非空 元素序列。
\n\n\n\n
示例 1：
\n\n\n输入：nums = [1,2,1,2,1,3,3], k = 2\n输出：8\n解释：将 nums 拆分成两个子数组：[1,2], [1,2,1,3,3]\n[1,2] 的重要性是 2 + (0) = 2 。\n[1,2,1,3,3] 的重要性是 2 + (2 + 2) = 6 。\n拆分的代价是 2 + 6 = 8 ，可以证明这是所有可行的拆分方案中的最小代价。\n\n\n
示例 2：
\n\n\n输入：nums = [1,2,1,2,1], k = 2\n输出：6\n解释：将 nums 拆分成两个子数组：[1,2], [1,2,1] 。\n[1,2] 的重要性是 2 + (0) = 2 。\n[1,2,1] 的重要性是 2 + (2) = 4 。\n拆分的代价是 2 + 4 = 6 ，可以证明这是所有可行的拆分方案中的最小代价。\n\n\n
示例 3：
\n\n\n输入：nums = [1,2,1,2,1], k = 5\n输出：10\n解释：将 nums 拆分成一个子数组：[1,2,1,2,1].\n[1,2,1,2,1] 的重要性是 5 + (3 + 2) = 10 。\n拆分的代价是 10 ，可以证明这是所有可行的拆分方案中的最小代价。\n\n\n
\n\n
提示：
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