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leetcode-problemset/leetcode-cn/originData/number-of-sets-of-k-non-overlapping-line-segments.json

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"questionId": "1725",
"questionFrontendId": "1621",
"categoryTitle": "Algorithms",
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"title": "Number of Sets of K Non-Overlapping Line Segments",
"titleSlug": "number-of-sets-of-k-non-overlapping-line-segments",
"content": "<p>Given <code>n</code> points on a 1-D plane, where the <code>i<sup>th</sup></code> point (from <code>0</code> to <code>n-1</code>) is at <code>x = i</code>, find the number of ways we can draw <strong>exactly</strong> <code>k</code> <strong>non-overlapping</strong> line segments such that each segment covers two or more points. The endpoints of each segment must have <strong>integral coordinates</strong>. The <code>k</code> line segments <strong>do not</strong> have to cover all <code>n</code> points, and they are <strong>allowed</strong> to share endpoints.</p>\n\n<p>Return <em>the number of ways we can draw </em><code>k</code><em> non-overlapping line segments</em><em>.</em> Since this number can be huge, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/09/07/ex1.png\" style=\"width: 179px; height: 222px;\" />\n<pre>\n<strong>Input:</strong> n = 4, k = 2\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> The two line segments are shown in red and blue.\nThe image above shows the 5 different ways {(0,2),(2,3)}, {(0,1),(1,3)}, {(0,1),(2,3)}, {(1,2),(2,3)}, {(0,1),(1,2)}.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 3, k = 1\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> The 3 ways are {(0,1)}, {(0,2)}, {(1,2)}.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 30, k = 7\n<strong>Output:</strong> 796297179\n<strong>Explanation:</strong> The total number of possible ways to draw 7 line segments is 3796297200. Taking this number modulo 10<sup>9</sup> + 7 gives us 796297179.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 1000</code></li>\n\t<li><code>1 &lt;= k &lt;= n-1</code></li>\n</ul>\n",
"translatedTitle": "大小为 K 的不重叠线段的数目",
"translatedContent": "<p>给你一维空间的 <code>n</code> 个点,其中第 <code>i</code> 个点(编号从 <code>0</code> 到 <code>n-1</code>)位于 <code>x = i</code> 处,请你找到 <strong>恰好</strong> <code>k</code> <strong>个不重叠</strong> 线段且每个线段至少覆盖两个点的方案数。线段的两个端点必须都是 <strong>整数坐标</strong> 。这 <code>k</code> 个线段不需要全部覆盖全部 <code>n</code> 个点,且它们的端点 <strong>可以 </strong>重合。</p>\n\n<p>请你返回 <code>k</code> 个不重叠线段的方案数。由于答案可能很大,请将结果对 <code>10<sup>9</sup> + 7</code> <strong>取余</strong> 后返回。</p>\n\n<p> </p>\n\n<p><strong>示例 1</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/10/17/ex1.png\" style=\"width: 179px; height: 222px;\" />\n<pre>\n<b>输入:</b>n = 4, k = 2\n<b>输出:</b>5\n<strong>解释:\n</strong>如图所示,两个线段分别用红色和蓝色标出。\n上图展示了 5 种不同的方案 {(0,2),(2,3)}{(0,1),(1,3)}{(0,1),(2,3)}{(1,2),(2,3)}{(0,1),(1,2)} 。</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<b>输入:</b>n = 3, k = 1\n<b>输出:</b>3\n<strong>解释:</strong>总共有 3 种不同的方案 {(0,1)}, {(0,2)}, {(1,2)} 。\n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre>\n<b>输入:</b>n = 30, k = 7\n<b>输出:</b>796297179\n<strong>解释:</strong>画 7 条线段的总方案数为 3796297200 种。将这个数对 10<sup>9</sup> + 7 取余得到 796297179 。\n</pre>\n\n<p><strong>示例 4</strong></p>\n\n<pre>\n<b>输入:</b>n = 5, k = 3\n<b>输出:</b>7\n</pre>\n\n<p><strong>示例 5</strong></p>\n\n<pre>\n<b>输入:</b>n = 3, k = 2\n<b>输出:</b>1</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>2 <= n <= 1000</code></li>\n\t<li><code>1 <= k <= n-1</code></li>\n</ul>\n",
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"Try to use dynamic programming where the current index and remaining number of line segments to form can describe any intermediate state.",
"To make the computation of each state in constant time, we could add another flag to the state that indicates whether or not we are in the middle of placing a line (placed start point but no endpoint)."
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