leetcode-problemset/leetcode-cn/originData/minimum-score-after-removals-on-a-tree.json

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            "questionId": "2400",
            "questionFrontendId": "2322",
            "categoryTitle": "Algorithms",
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            "title": "Minimum Score After Removals on a Tree",
            "titleSlug": "minimum-score-after-removals-on-a-tree",
            "content": "<p>There is an undirected connected tree with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code> and <code>n - 1</code> edges.</p>\n\n<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code> where <code>nums[i]</code> represents the value of the <code>i<sup>th</sup></code> node. You are also given a 2D integer array <code>edges</code> of length <code>n - 1</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the tree.</p>\n\n<p>Remove two <strong>distinct</strong> edges of the tree to form three connected components. For a pair of removed edges, the following steps are defined:</p>\n\n<ol>\n\t<li>Get the XOR of all the values of the nodes for <strong>each</strong> of the three components respectively.</li>\n\t<li>The <strong>difference</strong> between the <strong>largest</strong> XOR value and the <strong>smallest</strong> XOR value is the <strong>score</strong> of the pair.</li>\n</ol>\n\n<ul>\n\t<li>For example, say the three components have the node values: <code>[4,5,7]</code>, <code>[1,9]</code>, and <code>[3,3,3]</code>. The three XOR values are <code>4 ^ 5 ^ 7 = <u><strong>6</strong></u></code>, <code>1 ^ 9 = <u><strong>8</strong></u></code>, and <code>3 ^ 3 ^ 3 = <u><strong>3</strong></u></code>. The largest XOR value is <code>8</code> and the smallest XOR value is <code>3</code>. The score is then <code>8 - 3 = 5</code>.</li>\n</ul>\n\n<p>Return <em>the <strong>minimum</strong> score of any possible pair of edge removals on the given tree</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/05/03/ex1drawio.png\" style=\"width: 193px; height: 190px;\" />\n<pre>\n<strong>Input:</strong> nums = [1,5,5,4,11], edges = [[0,1],[1,2],[1,3],[3,4]]\n<strong>Output:</strong> 9\n<strong>Explanation:</strong> The diagram above shows a way to make a pair of removals.\n- The 1<sup>st</sup> component has nodes [1,3,4] with values [5,4,11]. Its XOR value is 5 ^ 4 ^ 11 = 10.\n- The 2<sup>nd</sup> component has node [0] with value [1]. Its XOR value is 1 = 1.\n- The 3<sup>rd</sup> component has node [2] with value [5]. Its XOR value is 5 = 5.\nThe score is the difference between the largest and smallest XOR value which is 10 - 1 = 9.\nIt can be shown that no other pair of removals will obtain a smaller score than 9.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/05/03/ex2drawio.png\" style=\"width: 287px; height: 150px;\" />\n<pre>\n<strong>Input:</strong> nums = [5,5,2,4,4,2], edges = [[0,1],[1,2],[5,2],[4,3],[1,3]]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> The diagram above shows a way to make a pair of removals.\n- The 1<sup>st</sup> component has nodes [3,4] with values [4,4]. Its XOR value is 4 ^ 4 = 0.\n- The 2<sup>nd</sup> component has nodes [1,0] with values [5,5]. Its XOR value is 5 ^ 5 = 0.\n- The 3<sup>rd</sup> component has nodes [2,5] with values [2,2]. Its XOR value is 2 ^ 2 = 0.\nThe score is the difference between the largest and smallest XOR value which is 0 - 0 = 0.\nWe cannot obtain a smaller score than 0.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>3 &lt;= n &lt;= 1000</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>8</sup></code></li>\n\t<li><code>edges.length == n - 1</code></li>\n\t<li><code>edges[i].length == 2</code></li>\n\t<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>\n\t<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>\n\t<li><code>edges</code> represents a valid tree.</li>\n</ul>\n",
            "translatedTitle": "从树中删除边的最小分数",
            "translatedContent": "<p>存在一棵无向连通树，树中有编号从 <code>0</code> 到 <code>n - 1</code> 的 <code>n</code> 个节点， 以及 <code>n - 1</code> 条边。</p>\n\n<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> ，长度为 <code>n</code> ，其中 <code>nums[i]</code> 表示第 <code>i</code> 个节点的值。另给你一个二维整数数组 <code>edges</code> ，长度为 <code>n - 1</code> ，其中 <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> 表示树中存在一条位于节点 <code>a<sub>i</sub></code> 和 <code>b<sub>i</sub></code> 之间的边。</p>\n\n<p>删除树中两条 <strong>不同</strong> 的边以形成三个连通组件。对于一种删除边方案，定义如下步骤以计算其分数：</p>\n\n<ol>\n\t<li>分别获取三个组件 <strong>每个</strong> 组件中所有节点值的异或值。</li>\n\t<li><strong>最大</strong> 异或值和 <strong>最小</strong> 异或值的 <strong>差值</strong> 就是这一种删除边方案的分数。</li>\n</ol>\n\n<ul>\n\t<li>例如，三个组件的节点值分别是：<code>[4,5,7]</code>、<code>[1,9]</code> 和 <code>[3,3,3]</code> 。三个异或值分别是 <code>4 ^ 5 ^ 7 = <em><strong>6</strong></em></code>、<code>1 ^ 9 = <em><strong>8</strong></em></code> 和 <code>3 ^ 3 ^ 3 = <em><strong>3</strong></em></code> 。最大异或值是 <code>8</code> ，最小异或值是 <code>3</code> ，分数是 <code>8 - 3 = 5</code> 。</li>\n</ul>\n\n<p>返回在给定树上执行任意删除边方案可能的 <strong>最小</strong> 分数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/05/03/ex1drawio.png\" style=\"width: 193px; height: 190px;\">\n<pre><strong>输入：</strong>nums = [1,5,5,4,11], edges = [[0,1],[1,2],[1,3],[3,4]]\n<strong>输出：</strong>9\n<strong>解释：</strong>上图展示了一种删除边方案。\n- 第 1 个组件的节点是 [1,3,4] ，值是 [5,4,11] 。异或值是 5 ^ 4 ^ 11 = 10 。\n- 第 2 个组件的节点是 [0] ，值是 [1] 。异或值是 1 = 1 。\n- 第 3 个组件的节点是 [2] ，值是 [5] 。异或值是 5 = 5 。\n分数是最大异或值和最小异或值的差值，10 - 1 = 9 。\n可以证明不存在分数比 9 小的删除边方案。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/05/03/ex2drawio.png\" style=\"width: 287px; height: 150px;\">\n<pre><strong>输入：</strong>nums = [5,5,2,4,4,2], edges = [[0,1],[1,2],[5,2],[4,3],[1,3]]\n<strong>输出：</strong>0\n<strong>解释：</strong>上图展示了一种删除边方案。\n- 第 1 个组件的节点是 [3,4] ，值是 [4,4] 。异或值是 4 ^ 4 = 0 。\n- 第 2 个组件的节点是 [1,0] ，值是 [5,5] 。异或值是 5 ^ 5 = 0 。\n- 第 3 个组件的节点是 [2,5] ，值是 [2,2] 。异或值是 2 ^ 2 = 0 。\n分数是最大异或值和最小异或值的差值，0 - 0 = 0 。\n无法获得比 0 更小的分数 0 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>3 &lt;= n &lt;= 1000</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>8</sup></code></li>\n\t<li><code>edges.length == n - 1</code></li>\n\t<li><code>edges[i].length == 2</code></li>\n\t<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>\n\t<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>\n\t<li><code>edges</code> 表示一棵有效的树</li>\n</ul>\n",
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