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# 力扣题库（完整版）
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> 最后更新日期： **2023.01.23**
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> 最后更新日期： **2023.02.02**
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>
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> 使用脚本前请务必仔细完整阅读本 `README.md` 文件
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leetcode-cn/originData/count-distinct-numbers-on-board.json
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leetcode-cn/originData/count-distinct-numbers-on-board.json
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leetcode-cn/originData/count-increasing-quadruplets.json
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leetcode-cn/originData/count-increasing-quadruplets.json
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leetcode-cn/originData/put-marbles-in-bags.json
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leetcode-cn/originData/put-marbles-in-bags.json
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<p>你有 <code>k</code> 个背包。给你一个下标从 <strong>0</strong> 开始的整数数组 <code>weights</code> ，其中 <code>weights[i]</code> 是第 <code>i</code> 个珠子的重量。同时给你整数 <code>k</code> 。</p>
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<p>请你按照如下规则将所有的珠子放进 <code>k</code> 个背包。</p>
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<ul>
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<li>没有背包是空的。</li>
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<li>如果第 <code>i</code> 个珠子和第 <code>j</code> 个珠子在同一个背包里，那么下标在 <code>i</code> 到 <code>j</code> 之间的所有珠子都必须在这同一个背包中。</li>
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<li>如果一个背包有下标从 <code>i</code> 到 <code>j</code> 的所有珠子，那么这个背包的价格是 <code>weights[i] + weights[j]</code> 。</li>
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</ul>
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<p>一个珠子分配方案的 <strong>分数</strong> 是所有 <code>k</code> 个背包的价格之和。</p>
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<p>请你返回所有分配方案中，<strong>最大分数</strong> 与 <strong>最小分数</strong> 的 <strong>差值</strong> 为多少。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>weights = [1,3,5,1], k = 2
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<b>输出：</b>4
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<b>解释：</b>
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分配方案 [1],[3,5,1] 得到最小得分 (1+1) + (3+1) = 6 。
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分配方案 [1,3],[5,1] 得到最大得分 (1+3) + (5+1) = 10 。
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所以差值为 10 - 6 = 4 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>weights = [1, 3], k = 2
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<b>输出：</b>0
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<b>解释：</b>唯一的分配方案为 [1],[3] 。
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最大最小得分相等，所以返回 0 。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 <= k <= weights.length <= 10<sup>5</sup></code></li>
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<li><code>1 <= weights[i] <= 10<sup>9</sup></code></li>
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</ul>
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<p>现在有一个正凸多边形，其上共有 <code>n</code> 个顶点。顶点按顺时针方向从 <code>0</code> 到 <code>n - 1</code> 依次编号。每个顶点上 <strong>正好有一只猴子</strong> 。下图中是一个 6 个顶点的凸多边形。</p>
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<p><img alt="" src="https://assets.leetcode.com/uploads/2023/01/22/hexagon.jpg" style="width: 300px; height: 293px;"></p>
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<p>每个猴子同时移动到相邻的顶点。顶点 <code>i</code> 的相邻顶点可以是：</p>
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<ul>
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<li>顺时针方向的顶点 <code>(i + 1) % n</code> ，或</li>
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<li>逆时针方向的顶点 <code>(i - 1 + n) % n</code> 。</li>
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</ul>
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<p>如果移动后至少有两个猴子位于同一顶点，则会发生 <strong>碰撞</strong> 。</p>
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<p>返回猴子至少发生 <strong>一次碰撞 </strong>的移动方法数。由于答案可能非常大，请返回对 <code>10<sup>9</sup>+7</code> 取余后的结果。</p>
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<p><strong>注意</strong>，每只猴子只能移动一次。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre><strong>输入：</strong>n = 3
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<strong>输出：</strong>6
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<strong>解释：</strong>共计 8 种移动方式。
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下面列出两种会发生碰撞的方式：
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- 猴子 1 顺时针移动；猴子 2 逆时针移动；猴子 3 顺时针移动。猴子 1 和猴子 2 碰撞。
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- 猴子 1 逆时针移动；猴子 2 逆时针移动；猴子 3 顺时针移动。猴子 1 和猴子 3 碰撞。
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可以证明，有 6 种让猴子碰撞的方法。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><strong>输入：</strong>n = 4
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<strong>输出：</strong>14
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<strong>解释：</strong>可以证明，有 14 种让猴子碰撞的方法。</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>3 <= n <= 10<sup>9</sup></code></li>
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</ul>
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<p>给你一个长度为 <code>n</code> 下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> ，它包含 <code>1</code> 到 <code>n</code> 的所有数字，请你返回上升四元组的数目。</p>
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<p>如果一个四元组 <code>(i, j, k, l)</code> 满足以下条件，我们称它是上升的：</p>
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<ul>
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<li><code>0 <= i < j < k < l < n</code> 且</li>
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<li><code>nums[i] < nums[k] < nums[j] < nums[l]</code> 。</li>
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</ul>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>nums = [1,3,2,4,5]
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<b>输出：</b>2
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<b>解释：</b>
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- 当 i = 0 ，j = 1 ，k = 2 且 l = 3 时，有 nums[i] < nums[k] < nums[j] < nums[l] 。
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- 当 i = 0 ，j = 1 ，k = 2 且 l = 4 时，有 nums[i] < nums[k] < nums[j] < nums[l] 。
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没有其他的四元组，所以我们返回 2 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>nums = [1,2,3,4]
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<b>输出：</b>0
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<b>解释：</b>只存在一个四元组 i = 0 ，j = 1 ，k = 2 ，l = 3 ，但是 nums[j] < nums[k] ，所以我们返回 0 。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>4 <= nums.length <= 4000</code></li>
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<li><code>1 <= nums[i] <= nums.length</code></li>
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<li><code>nums</code> 中所有数字 <strong>互不相同</strong> ，<code>nums</code> 是一个排列。</li>
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</ul>
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<p>给你一个正整数 <code>n</code> ，开始时，它放在桌面上。在 <code>10<sup>9</sup></code> 天内，每天都要执行下述步骤：</p>
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<ul>
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<li>对于出现在桌面上的每个数字 <code>x</code> ，找出符合 <code>1 <= i <= n</code> 且满足 <code>x % i == 1</code> 的所有数字 <code>i</code> 。</li>
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<li>然后，将这些数字放在桌面上。</li>
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</ul>
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<p>返回在 <code>10<sup>9</sup></code> 天之后，出现在桌面上的 <strong>不同</strong> 整数的数目。</p>
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<p><strong>注意：</strong></p>
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<ul>
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<li>一旦数字放在桌面上，则会一直保留直到结束。</li>
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<li><code>%</code> 表示取余运算。例如，<code>14 % 3</code> 等于 <code>2</code> 。</li>
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</ul>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<strong>输入：</strong>n = 5
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<strong>输出：</strong>4
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<strong>解释：</strong>最开始，5 在桌面上。
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第二天，2 和 4 也出现在桌面上，因为 5 % 2 == 1 且 5 % 4 == 1 。
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再过一天 3 也出现在桌面上，因为 4 % 3 == 1 。
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在十亿天结束时，桌面上的不同数字有 2 、3 、4 、5 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>n = 3
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<strong>输出：</strong>2
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<strong>解释：</strong>
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因为 3 % 2 == 1 ，2 也出现在桌面上。
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在十亿天结束时，桌面上的不同数字只有两个：2 和 3 。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 <= n <= 100</code></li>
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</ul>
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<p>You have <code>k</code> bags. You are given a <strong>0-indexed</strong> integer array <code>weights</code> where <code>weights[i]</code> is the weight of the <code>i<sup>th</sup></code> marble. You are also given the integer <code>k.</code></p>
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<p>Divide the marbles into the <code>k</code> bags according to the following rules:</p>
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<ul>
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<li>No bag is empty.</li>
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<li>If the <code>i<sup>th</sup></code> marble and <code>j<sup>th</sup></code> marble are in a bag, then all marbles with an index between the <code>i<sup>th</sup></code> and <code>j<sup>th</sup></code> indices should also be in that same bag.</li>
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<li>If a bag consists of all the marbles with an index from <code>i</code> to <code>j</code> inclusively, then the cost of the bag is <code>weights[i] + weights[j]</code>.</li>
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</ul>
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<p>The <strong>score</strong> after distributing the marbles is the sum of the costs of all the <code>k</code> bags.</p>
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<p>Return <em>the <strong>difference</strong> between the <strong>maximum</strong> and <strong>minimum</strong> scores among marble distributions</em>.</p>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> weights = [1,3,5,1], k = 2
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<strong>Output:</strong> 4
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<strong>Explanation:</strong>
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The distribution [1],[3,5,1] results in the minimal score of (1+1) + (3+1) = 6.
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The distribution [1,3],[5,1], results in the maximal score of (1+3) + (5+1) = 10.
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Thus, we return their difference 10 - 6 = 4.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> weights = [1, 3], k = 2
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> The only distribution possible is [1],[3].
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Since both the maximal and minimal score are the same, we return 0.
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 <= k <= weights.length <= 10<sup>5</sup></code></li>
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<li><code>1 <= weights[i] <= 10<sup>9</sup></code></li>
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</ul>
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<p>There is a regular convex polygon with <code>n</code> vertices. The vertices are labeled from <code>0</code> to <code>n - 1</code> in a clockwise direction, and each vertex has <strong>exactly one monkey</strong>. The following figure shows a convex polygon of <code>6</code> vertices.</p>
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<img alt="" src="https://assets.leetcode.com/uploads/2023/01/22/hexagon.jpg" style="width: 300px; height: 293px;" />
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<p>Each monkey moves simultaneously to a neighboring vertex. A neighboring vertex for a vertex <code>i</code> can be:</p>
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<ul>
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<li>the vertex <code>(i + 1) % n</code> in the clockwise direction, or</li>
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<li>the vertex <code>(i - 1 + n) % n</code> in the counter-clockwise direction.</li>
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</ul>
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<p>A <strong>collision</strong> happens if at least two monkeys reside on the same vertex after the movement.</p>
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<p>Return <em>the number of ways the monkeys can move so that at least <strong>one collision</strong></em> <em> happens</em>. Since the answer may be very large, return it modulo <code>10<sup>9 </sup>+ 7</code>.</p>
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<p><strong>Note</strong> that each monkey can only move once.</p>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> n = 3
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<strong>Output:</strong> 6
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<strong>Explanation:</strong> There are 8 total possible movements.
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Two ways such that they collide at some point are:
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- Monkey 1 moves in a clockwise direction; monkey 2 moves in an anticlockwise direction; monkey 3 moves in a clockwise direction. Monkeys 1 and 2 collide.
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- Monkey 1 moves in an anticlockwise direction; monkey 2 moves in an anticlockwise direction; monkey 3 moves in a clockwise direction. Monkeys 1 and 3 collide.
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It can be shown 6 total movements result in a collision.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> n = 4
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<strong>Output:</strong> 14
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<strong>Explanation:</strong> It can be shown that there are 14 ways for the monkeys to collide.
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>3 <= n <= 10<sup>9</sup></code></li>
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</ul>
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<p>Given a <strong>0-indexed</strong> integer array <code>nums</code> of size <code>n</code> containing all numbers from <code>1</code> to <code>n</code>, return <em>the number of increasing quadruplets</em>.</p>
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<p>A quadruplet <code>(i, j, k, l)</code> is increasing if:</p>
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<ul>
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<li><code>0 <= i < j < k < l < n</code>, and</li>
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<li><code>nums[i] < nums[k] < nums[j] < nums[l]</code>.</li>
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</ul>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,3,2,4,5]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong>
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- When i = 0, j = 1, k = 2, and l = 3, nums[i] < nums[k] < nums[j] < nums[l].
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- When i = 0, j = 1, k = 2, and l = 4, nums[i] < nums[k] < nums[j] < nums[l].
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There are no other quadruplets, so we return 2.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,3,4]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> There exists only one quadruplet with i = 0, j = 1, k = 2, l = 3, but since nums[j] < nums[k], we return 0.
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>4 <= nums.length <= 4000</code></li>
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<li><code>1 <= nums[i] <= nums.length</code></li>
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<li>All the integers of <code>nums</code> are <strong>unique</strong>. <code>nums</code> is a permutation.</li>
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</ul>
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<p>You are given a positive integer <code>n</code>, that is initially placed on a board. Every day, for <code>10<sup>9</sup></code> days, you perform the following procedure:</p>
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<ul>
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<li>For each number <code>x</code> present on the board, find all numbers <code>1 <= i <= n</code> such that <code>x % i == 1</code>.</li>
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<li>Then, place those numbers on the board.</li>
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</ul>
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<p>Return<em> the number of <strong>distinct</strong> integers present on the board after</em> <code>10<sup>9</sup></code> <em>days have elapsed</em>.</p>
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<p><strong>Note:</strong></p>
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<ul>
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<li>Once a number is placed on the board, it will remain on it until the end.</li>
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<li><code>%</code> stands for the modulo operation. For example, <code>14 % 3</code> is <code>2</code>.</li>
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</ul>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> n = 5
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> Initially, 5 is present on the board.
|
||||
The next day, 2 and 4 will be added since 5 % 2 == 1 and 5 % 4 == 1.
|
||||
After that day, 3 will be added to the board because 4 % 3 == 1.
|
||||
At the end of a billion days, the distinct numbers on the board will be 2, 3, 4, and 5.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 3
|
||||
<strong>Output:</strong> 2
|
||||
<strong>Explanation:</strong>
|
||||
Since 3 % 2 == 1, 2 will be added to the board.
|
||||
After a billion days, the only two distinct numbers on the board are 2 and 3.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n <= 100</code></li>
|
||||
</ul>
|
||||
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187
leetcode/originData/count-distinct-numbers-on-board.json
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187
leetcode/originData/count-distinct-numbers-on-board.json
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193
leetcode/originData/count-increasing-quadruplets.json
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193
leetcode/originData/count-increasing-quadruplets.json
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187
leetcode/originData/put-marbles-in-bags.json
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187
leetcode/originData/put-marbles-in-bags.json
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|
||||
<p>There is a regular convex polygon with <code>n</code> vertices. The vertices are labeled from <code>0</code> to <code>n - 1</code> in a clockwise direction, and each vertex has <strong>exactly one monkey</strong>. The following figure shows a convex polygon of <code>6</code> vertices.</p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/22/hexagon.jpg" style="width: 300px; height: 293px;" />
|
||||
<p>Each monkey moves simultaneously to a neighboring vertex. A neighboring vertex for a vertex <code>i</code> can be:</p>
|
||||
|
||||
<ul>
|
||||
<li>the vertex <code>(i + 1) % n</code> in the clockwise direction, or</li>
|
||||
<li>the vertex <code>(i - 1 + n) % n</code> in the counter-clockwise direction.</li>
|
||||
</ul>
|
||||
|
||||
<p>A <strong>collision</strong> happens if at least two monkeys reside on the same vertex after the movement.</p>
|
||||
|
||||
<p>Return <em>the number of ways the monkeys can move so that at least <strong>one collision</strong></em> <em> happens</em>. Since the answer may be very large, return it modulo <code>10<sup>9 </sup>+ 7</code>.</p>
|
||||
|
||||
<p><strong>Note</strong> that each monkey can only move once.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 3
|
||||
<strong>Output:</strong> 6
|
||||
<strong>Explanation:</strong> There are 8 total possible movements.
|
||||
Two ways such that they collide at some point are:
|
||||
- Monkey 1 moves in a clockwise direction; monkey 2 moves in an anticlockwise direction; monkey 3 moves in a clockwise direction. Monkeys 1 and 2 collide.
|
||||
- Monkey 1 moves in an anticlockwise direction; monkey 2 moves in an anticlockwise direction; monkey 3 moves in a clockwise direction. Monkeys 1 and 3 collide.
|
||||
It can be shown 6 total movements result in a collision.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 4
|
||||
<strong>Output:</strong> 14
|
||||
<strong>Explanation:</strong> It can be shown that there are 14 ways for the monkeys to collide.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>3 <= n <= 10<sup>9</sup></code></li>
|
||||
</ul>
|
||||
44
leetcode/problem/count-distinct-numbers-on-board.html
Normal file
44
leetcode/problem/count-distinct-numbers-on-board.html
Normal file
@ -0,0 +1,44 @@
|
||||
<p>You are given a positive integer <code>n</code>, that is initially placed on a board. Every day, for <code>10<sup>9</sup></code> days, you perform the following procedure:</p>
|
||||
|
||||
<ul>
|
||||
<li>For each number <code>x</code> present on the board, find all numbers <code>1 <= i <= n</code> such that <code>x % i == 1</code>.</li>
|
||||
<li>Then, place those numbers on the board.</li>
|
||||
</ul>
|
||||
|
||||
<p>Return<em> the number of <strong>distinct</strong> integers present on the board after</em> <code>10<sup>9</sup></code> <em>days have elapsed</em>.</p>
|
||||
|
||||
<p><strong>Note:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li>Once a number is placed on the board, it will remain on it until the end.</li>
|
||||
<li><code>%</code> stands for the modulo operation. For example, <code>14 % 3</code> is <code>2</code>.</li>
|
||||
</ul>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 5
|
||||
<strong>Output:</strong> 4
|
||||
<strong>Explanation:</strong> Initially, 5 is present on the board.
|
||||
The next day, 2 and 4 will be added since 5 % 2 == 1 and 5 % 4 == 1.
|
||||
After that day, 3 will be added to the board because 4 % 3 == 1.
|
||||
At the end of a billion days, the distinct numbers on the board will be 2, 3, 4, and 5.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 3
|
||||
<strong>Output:</strong> 2
|
||||
<strong>Explanation:</strong>
|
||||
Since 3 % 2 == 1, 2 will be added to the board.
|
||||
After a billion days, the only two distinct numbers on the board are 2 and 3.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n <= 100</code></li>
|
||||
</ul>
|
||||
37
leetcode/problem/count-increasing-quadruplets.html
Normal file
37
leetcode/problem/count-increasing-quadruplets.html
Normal file
@ -0,0 +1,37 @@
|
||||
<p>Given a <strong>0-indexed</strong> integer array <code>nums</code> of size <code>n</code> containing all numbers from <code>1</code> to <code>n</code>, return <em>the number of increasing quadruplets</em>.</p>
|
||||
|
||||
<p>A quadruplet <code>(i, j, k, l)</code> is increasing if:</p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= i < j < k < l < n</code>, and</li>
|
||||
<li><code>nums[i] < nums[k] < nums[j] < nums[l]</code>.</li>
|
||||
</ul>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,3,2,4,5]
|
||||
<strong>Output:</strong> 2
|
||||
<strong>Explanation:</strong>
|
||||
- When i = 0, j = 1, k = 2, and l = 3, nums[i] < nums[k] < nums[j] < nums[l].
|
||||
- When i = 0, j = 1, k = 2, and l = 4, nums[i] < nums[k] < nums[j] < nums[l].
|
||||
There are no other quadruplets, so we return 2.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,2,3,4]
|
||||
<strong>Output:</strong> 0
|
||||
<strong>Explanation:</strong> There exists only one quadruplet with i = 0, j = 1, k = 2, l = 3, but since nums[j] < nums[k], we return 0.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>4 <= nums.length <= 4000</code></li>
|
||||
<li><code>1 <= nums[i] <= nums.length</code></li>
|
||||
<li>All the integers of <code>nums</code> are <strong>unique</strong>. <code>nums</code> is a permutation.</li>
|
||||
</ul>
|
||||
42
leetcode/problem/put-marbles-in-bags.html
Normal file
42
leetcode/problem/put-marbles-in-bags.html
Normal file
@ -0,0 +1,42 @@
|
||||
<p>You have <code>k</code> bags. You are given a <strong>0-indexed</strong> integer array <code>weights</code> where <code>weights[i]</code> is the weight of the <code>i<sup>th</sup></code> marble. You are also given the integer <code>k.</code></p>
|
||||
|
||||
<p>Divide the marbles into the <code>k</code> bags according to the following rules:</p>
|
||||
|
||||
<ul>
|
||||
<li>No bag is empty.</li>
|
||||
<li>If the <code>i<sup>th</sup></code> marble and <code>j<sup>th</sup></code> marble are in a bag, then all marbles with an index between the <code>i<sup>th</sup></code> and <code>j<sup>th</sup></code> indices should also be in that same bag.</li>
|
||||
<li>If a bag consists of all the marbles with an index from <code>i</code> to <code>j</code> inclusively, then the cost of the bag is <code>weights[i] + weights[j]</code>.</li>
|
||||
</ul>
|
||||
|
||||
<p>The <strong>score</strong> after distributing the marbles is the sum of the costs of all the <code>k</code> bags.</p>
|
||||
|
||||
<p>Return <em>the <strong>difference</strong> between the <strong>maximum</strong> and <strong>minimum</strong> scores among marble distributions</em>.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> weights = [1,3,5,1], k = 2
|
||||
<strong>Output:</strong> 4
|
||||
<strong>Explanation:</strong>
|
||||
The distribution [1],[3,5,1] results in the minimal score of (1+1) + (3+1) = 6.
|
||||
The distribution [1,3],[5,1], results in the maximal score of (1+3) + (5+1) = 10.
|
||||
Thus, we return their difference 10 - 6 = 4.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> weights = [1, 3], k = 2
|
||||
<strong>Output:</strong> 0
|
||||
<strong>Explanation:</strong> The only distribution possible is [1],[3].
|
||||
Since both the maximal and minimal score are the same, we return 0.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= k <= weights.length <= 10<sup>5</sup></code></li>
|
||||
<li><code>1 <= weights[i] <= 10<sup>9</sup></code></li>
|
||||
</ul>
|
||||
Loading…
Reference in New Issue
Block a user