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

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        "question": {
            "questionId": "3313",
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            "title": "Maximum Strength of K Disjoint Subarrays",
            "titleSlug": "maximum-strength-of-k-disjoint-subarrays",
            "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",
            "translatedTitle": "K 个不相交子数组的最大能量值",
            "translatedContent": "<p>给你一个长度为 <code>n</code>&nbsp;下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;和一个 <strong>正奇数</strong>&nbsp;整数&nbsp;<code>k</code>&nbsp;。</p>\n\n<p><code>x</code> 个子数组的能量值定义为&nbsp;<code>strength = sum[1] * x - sum[2] * (x - 1) + sum[3] * (x - 2) - sum[4] * (x - 3) + ... + sum[x] * 1</code> ，其中&nbsp;<code>sum[i]</code>&nbsp;是第 <code>i</code>&nbsp;个子数组的和。更正式的，能量值是满足&nbsp;<code>1 &lt;= i &lt;= x</code>&nbsp;的所有&nbsp;<code>i</code>&nbsp;对应的&nbsp;<code>(-1)<sup>i+1</sup> * sum[i] * (x - i + 1)</code>&nbsp;之和。</p>\n\n<p>你需要在 <code>nums</code>&nbsp;中选择 <code>k</code>&nbsp;个 <strong>不相交</strong><strong>子数组</strong>&nbsp;，使得&nbsp;<strong>能量值最大</strong>&nbsp;。</p>\n\n<p>请你返回可以得到的 <strong>最大</strong><strong>能量值</strong>&nbsp;。</p>\n\n<p><strong>注意</strong>，选出来的所有子数组&nbsp;<strong>不</strong>&nbsp;需要覆盖整个数组。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [1,2,3,-1,2], k = 3\n<b>输出：</b>22\n<b>解释：</b>选择 3 个子数组的最好方式是选择：nums[0..2] ，nums[3..3] 和 nums[4..4] 。能量值为 (1 + 2 + 3) * 3 - (-1) * 2 + 2 * 1 = 22 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [12,-2,-2,-2,-2], k = 5\n<b>输出：</b>64\n<b>解释：</b>唯一一种选 5 个不相交子数组的方案是：nums[0..0] ，nums[1..1] ，nums[2..2] ，nums[3..3] 和 nums[4..4] 。能量值为 12 * 5 - (-2) * 4 + (-2) * 3 - (-2) * 2 + (-2) * 1 = 64 。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [-1,-2,-3], k = 1\n<b>输出：</b>-1\n<b>解释：</b>选择 1 个子数组的最优方案是：nums[0..0] 。能量值为 -1 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</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> 是奇数。</li>\n</ul>\n",
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                "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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