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

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            "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 an array of integers <code>nums</code> with length <code>n</code>, and a positive <strong>odd</strong> integer <code>k</code>.</p>\n\n<p>Select exactly <b><code>k</code></b> disjoint <span data-keyword=\"subarray-nonempty\">subarrays</span> <b><code>sub<sub>1</sub>, sub<sub>2</sub>, ..., sub<sub>k</sub></code></b> from <code>nums</code> such that the last element of <code>sub<sub>i</sub></code> appears before the first element of <code>sub<sub>{i+1}</sub></code> for all <code>1 &lt;= i &lt;= k-1</code>. The goal is to maximize their combined strength.</p>\n\n<p>The strength of the selected subarrays is defined as:</p>\n\n<p><code>strength = k * sum(sub<sub>1</sub>)- (k - 1) * sum(sub<sub>2</sub>) + (k - 2) * sum(sub<sub>3</sub>) - ... - 2 * sum(sub<sub>{k-1}</sub>) + sum(sub<sub>k</sub>)</code></p>\n\n<p>where <b><code>sum(sub<sub>i</sub>)</code></b> is the sum of the elements in the <code>i</code>-th subarray.</p>\n\n<p>Return the <strong>maximum</strong> possible strength that can be obtained from selecting exactly <b><code>k</code></b> disjoint subarrays from <code>nums</code>.</p>\n\n<p><strong>Note</strong> that the chosen 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<p><strong>Input:</strong> <span class=\"example-io\">nums = [1,2,3,-1,2], k = 3</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">22</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The best possible way to select 3 subarrays is: nums[0..2], nums[3..3], and nums[4..4]. The strength is calculated as follows:</p>\n\n<p><code>strength = 3 * (1 + 2 + 3) - 2 * (-1) + 2 = 22</code></p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [12,-2,-2,-2,-2], k = 5</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">64</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>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 calculated as follows:</p>\n\n<p><code>strength = 5 * 12 - 4 * (-2) + 3 * (-2) - 2 * (-2) + (-2) = 64</code></p>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [-1,-2,-3], k = </span>1</p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">-1</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The best possible way to select 1 subarray is: nums[0..0]. The strength is -1.</p>\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>.",
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target=\\\"_blank\\\">sortedcontainers<\\/a> \\u5e93\\u3002<\\/p>\"],\"scala\":[\"Scala\",\"<p>\\u7248\\u672c\\uff1a<code>Scala 3.3.1<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"<p>\\u7248\\u672c\\uff1a<code>Kotlin 1.9.0<\\/code><\\/p>\\r\\n\\r\\n<p>\\u6211\\u4eec\\u4f7f\\u7528\\u7684\\u662f JetBrains \\u63d0\\u4f9b\\u7684 experimental compiler\\u3002\\u5982\\u679c\\u60a8\\u8ba4\\u4e3a\\u60a8\\u9047\\u5230\\u4e86\\u7f16\\u8bd1\\u5668\\u76f8\\u5173\\u7684\\u95ee\\u9898\\uff0c\\u8bf7\\u5411\\u6211\\u4eec\\u53cd\\u9988<\\/p>\"],\"rust\":[\"Rust\",\"<p>\\u7248\\u672c\\uff1a<code>rust 1.79.0<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 crates.io \\u7684 <a href=\\\"https:\\/\\/crates.io\\/crates\\/rand\\\" target=\\\"_blank\\\">rand<\\/a> \\u548c <a href=\\\"https:\\/\\/crates.io\\/crates\\/regex\\\" target=\\\"_blank\\\">regex<\\/a><\\/p>\"],\"php\":[\"PHP\",\"<p><code>PHP 8.2<\\/code>.<\\/p>\\r\\n\\r\\n<p>With bcmath module.<\\/p>\"],\"typescript\":[\"TypeScript\",\"<p>TypeScript 5.1.6<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2022<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/blob\\/v5\\/README.md\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.4.0<\\/a>\\uff0c<a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/v4.2.3\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.3<\\/a> \\u4ee5\\u53ca <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/deque\\/tree\\/v1.0.4\\\" target=\\\"_blank\\\"> datastructures-js\\/deque@1.0.4<\\/a>\\u3002<\\/p>\"],\"racket\":[\"Racket\",\"<p><a href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.15<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 26\"],\"elixir\":[\"Elixir\",\"Elixir 1.17 with Erlang\\/OTP 26\"],\"dart\":[\"Dart\",\"<p>Dart 3.2\\u3002\\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/pub.dev\\/packages\\/collection\\/versions\\/1.18.0\\\" target=\\\"_blank\\\">collection<\\/a> \\u5305<\\/p>\\r\\n\\r\\n<p>\\u60a8\\u7684\\u4ee3\\u7801\\u5c06\\u4f1a\\u88ab\\u4e0d\\u7f16\\u8bd1\\u76f4\\u63a5\\u8fd0\\u884c<\\/p>\"],\"cangjie\":[\"Cangjie\",\"<p>\\u7248\\u672c\\uff1a0.53.11 (cjnative)<\\/p>\\r\\n\\r\\n<p>\\u7f16\\u8bd1\\u53c2\\u6570\\uff1a<code>-O2 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