![\"\"](\"https://assets.leetcode.com/uploads/2023/03/01/graph-2.png\")
There exists an undirected and unrooted tree with n nodes indexed from 0 to n - 1. You are given an integer n and a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. You are also given an array coins of size n where coins[i] can be either 0 or 1, where 1 indicates the presence of a coin in the vertex i.
Initially, you choose to start at any vertex in the tree. Then, you can perform the following operations any number of times:
\n\n2 from the current vertex, orFind the minimum number of edges you need to go through to collect all the coins and go back to the initial vertex.
\n\nNote that if you pass an edge several times, you need to count it into the answer several times.
\n\n\n
Example 1:
\n\n\nInput: coins = [1,0,0,0,0,1], edges = [[0,1],[1,2],[2,3],[3,4],[4,5]]\nOutput: 2\nExplanation: Start at vertex 2, collect the coin at vertex 0, move to vertex 3, collect the coin at vertex 5 then move back to vertex 2.\n\n\n
Example 2:
\n![\"\"](\"https://assets.leetcode.com/uploads/2023/03/02/graph-4.png\")
\n\nInput: coins = [0,0,0,1,1,0,0,1], edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[5,6],[5,7]]\nOutput: 2\nExplanation: Start at vertex 0, collect the coins at vertices 4 and 3, move to vertex 2, collect the coin at vertex 7, then move back to vertex 0.\n\n\n
\n
Constraints:
\n\nn == coins.length1 <= n <= 3 * 1040 <= coins[i] <= 1edges.length == n - 1edges[i].length == 20 <= ai, bi < nai != biedges represents a valid tree.给你一个 n 个节点的无向无根树，节点编号从 0 到 n - 1 。给你整数 n 和一个长度为 n - 1 的二维整数数组 edges ，其中 edges[i] = [ai, bi] 表示树中节点 ai 和 bi 之间有一条边。再给你一个长度为 n 的数组 coins ，其中 coins[i] 可能为 0 也可能为 1 ，1 表示节点 i 处有一个金币。
一开始，你需要选择树中任意一个节点出发。你可以执行下述操作任意次：
\n\n2 以内的所有金币，或者你需要收集树中所有的金币，并且回到出发节点，请你返回最少经过的边数。
\n\n如果你多次经过一条边，每一次经过都会给答案加一。
\n\n\n\n
示例 1：
\n\n
输入：coins = [1,0,0,0,0,1], edges = [[0,1],[1,2],[2,3],[3,4],[4,5]]\n输出：2\n解释：从节点 2 出发，收集节点 0 处的金币，移动到节点 3 ，收集节点 5 处的金币，然后移动回节点 2 。\n\n\n
示例 2：
\n\n
输入：coins = [0,0,0,1,1,0,0,1], edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[5,6],[5,7]]\n输出：2\n解释：从节点 0 出发，收集节点 4 和 3 处的金币，移动到节点 2 处，收集节点 7 处的金币，移动回节点 0 。\n\n\n
\n\n
提示：
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diff --git a/leetcode-cn/originData/k-items-with-the-maximum-sum.json b/leetcode-cn/originData/k-items-with-the-maximum-sum.json
new file mode 100644
index 00000000..84afc2ff
--- /dev/null
+++ b/leetcode-cn/originData/k-items-with-the-maximum-sum.json
@@ -0,0 +1,165 @@
+{
+ "data": {
+ "question": {
+ "questionId": "2715",
+ "questionFrontendId": "6354",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2187837,
+ "title": "K Items With the Maximum Sum",
+ "titleSlug": "k-items-with-the-maximum-sum",
+ "content": " There is a bag that consists of items, each item has a number You are given four non-negative integers The bag initially contains: We want to pick exactly Example 1: Example 2: Constraints: 袋子中装有一些物品，每个物品上都标记着数字 给你四个非负整数 袋子最初包含： 现计划从这些物品中恰好选出 示例 1： 示例 2： 提示： \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a 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diff --git a/leetcode-cn/originData/minimum-operations-to-make-all-array-elements-equal.json b/leetcode-cn/originData/minimum-operations-to-make-all-array-elements-equal.json
new file mode 100644
index 00000000..6c7dd90d
--- /dev/null
+++ b/leetcode-cn/originData/minimum-operations-to-make-all-array-elements-equal.json
@@ -0,0 +1,164 @@
+{
+ "data": {
+ "question": {
+ "questionId": "2718",
+ "questionFrontendId": "6357",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2187608,
+ "title": "Minimum Operations to Make All Array Elements Equal",
+ "titleSlug": "minimum-operations-to-make-all-array-elements-equal",
+ "content": " You are given an array You are also given an integer array Return an array Note that after each query the array is reset to its original state. Example 1: Example 2: Constraints: 给你一个正整数数组 同时给你一个长度为 请你返回一个长度为 注意，每次查询后，数组变回最开始的值。 示例 1： 示例 2： 提示： \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, 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8.1<\\/code>.<\\/p>\\r\\n\\r\\n1, 0, or -1 written on it.numOnes, numZeros, numNegOnes, and k.\n\t
\n\nnumOnes items with 1s written on them.numZeroes items with 0s written on them.numNegOnes items with -1s written on them.k items among the available items. Return the maximum possible sum of numbers written on the items.\nInput: numOnes = 3, numZeros = 2, numNegOnes = 0, k = 2\nOutput: 2\nExplanation: We have a bag of items with numbers written on them {1, 1, 1, 0, 0}. We take 2 items with 1 written on them and get a sum in a total of 2.\nIt can be proven that 2 is the maximum possible sum.\n\n\n\nInput: numOnes = 3, numZeros = 2, numNegOnes = 0, k = 4\nOutput: 3\nExplanation: We have a bag of items with numbers written on them {1, 1, 1, 0, 0}. We take 3 items with 1 written on them, and 1 item with 0 written on it, and get a sum in a total of 3.\nIt can be proven that 3 is the maximum possible sum.\n\n\n\n\t
\n",
+ "translatedTitle": "K 件物品的最大和",
+ "translatedContent": "0 <= numOnes, numZeros, numNegOnes <= 500 <= k <= numOnes + numZeros + numNegOnes1 、0 或 -1 。numOnes 、numZeros 、numNegOnes 和 k 。\n\t
\n\nnumOnes 件标记为 1 的物品。numZeroes 件标记为 0 的物品。numNegOnes 件标记为 -1 的物品。k 件物品。返回所有可行方案中，物品上所标记数字之和的最大值。输入：numOnes = 3, numZeros = 2, numNegOnes = 0, k = 2\n输出：2\n解释：袋子中的物品分别标记为 {1, 1, 1, 0, 0} 。取 2 件标记为 1 的物品，得到的数字之和为 2 。\n可以证明 2 是所有可行方案中的最大值。\n\n输入：numOnes = 3, numZeros = 2, numNegOnes = 0, k = 4\n输出：3\n解释：袋子中的物品分别标记为 {1, 1, 1, 0, 0} 。取 3 件标记为 1 的物品，1 件标记为 0 的物品，得到的数字之和为 3 。\n可以证明 3 是所有可行方案中的最大值。\n\n\n\n\t
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+ "code": "/**\n * @param {number} numOnes\n * @param {number} numZeros\n * @param {number} numNegOnes\n * @param {number} k\n * @return {number}\n */\nvar kItemsWithMaximumSum = function(numOnes, numZeros, numNegOnes, k) {\n\n};",
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+ "code": "impl Solution {\n pub fn k_items_with_maximum_sum(num_ones: i32, num_zeros: i32, num_neg_ones: i32, k: i32) -> i32 {\n\n }\n}",
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+ "hints": [
+ "It is always optimal to take items with the number 1 written on them as much as possible.",
+ "If k > numOnes, after taking all items with the number 1, it is always optimal to take items with the number 0 written on them as much as possible.",
+ "If k > numOnes + numZeroes we are forced to take k - numOnes - numZeroes -1s."
+ ],
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+ "status": null,
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\n\n1.answer of size m where answer[i] is the minimum number of operations to make all elements of nums equal to queries[i].\nInput: nums = [3,1,6,8], queries = [1,5]\nOutput: [14,10]\nExplanation: For the first query we can do the following operations:\n- Decrease nums[0] 2 times, so that nums = [1,1,6,8].\n- Decrease nums[2] 5 times, so that nums = [1,1,1,8].\n- Decrease nums[3] 7 times, so that nums = [1,1,1,1].\nSo the total number of operations for the first query is 2 + 5 + 7 = 14.\nFor the second query we can do the following operations:\n- Increase nums[0] 2 times, so that nums = [5,1,6,8].\n- Increase nums[1] 4 times, so that nums = [5,5,6,8].\n- Decrease nums[2] 1 time, so that nums = [5,5,5,8].\n- Decrease nums[3] 3 times, so that nums = [5,5,5,5].\nSo the total number of operations for the second query is 2 + 4 + 1 + 3 = 10.\n
\n\n\nInput: nums = [2,9,6,3], queries = [10]\nOutput: [20]\nExplanation: We can increase each value in the array to 10. The total number of operations will be 8 + 1 + 4 + 7 = 20.\n
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+ "translatedTitle": "使数组元素全部相等的最少操作次数",
+ "translatedContent": "n == nums.lengthm == queries.length1 <= n, m <= 1051 <= nums[i], queries[i] <= 109nums 。m 的整数数组 queries 。第 i 个查询中，你需要将 nums 中所有元素变成 queries[i] 。你可以执行以下操作 任意 次：\n\t
\n\n1 。m 的数组 answer ，其中 answer[i]是将 nums 中所有元素变成 queries[i] 的 最少 操作次数。输入：nums = [3,1,6,8], queries = [1,5]\n输出：[14,10]\n解释：第一个查询，我们可以执行以下操作：\n- 将 nums[0] 减小 2 次，nums = [1,1,6,8] 。\n- 将 nums[2] 减小 5 次，nums = [1,1,1,8] 。\n- 将 nums[3] 减小 7 次，nums = [1,1,1,1] 。\n第一个查询的总操作次数为 2 + 5 + 7 = 14 。\n第二个查询，我们可以执行以下操作：\n- 将 nums[0] 增大 2 次，nums = [5,1,6,8] 。\n- 将 nums[1] 增大 4 次，nums = [5,5,6,8] 。\n- 将 nums[2] 减小 1 次，nums = [5,5,5,8] 。\n- 将 nums[3] 减小 3 次，nums = [5,5,5,5] 。\n第二个查询的总操作次数为 2 + 4 + 1 + 3 = 10 。\n
\n\n输入：nums = [2,9,6,3], queries = [10]\n输出：[20]\n解释：我们可以将数组中所有元素都增大到 10 ，总操作次数为 8 + 1 + 4 + 7 = 20 。\n
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+ "code": "class Solution {\npublic:\n vectorn == nums.lengthm == queries.length1 <= n, m <= 1051 <= nums[i], queries[i] <= 109clang 11<\\/code> \\u91c7\\u7528\\u6700\\u65b0C++ 17\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O2<\\/code>\\u7ea7\\u4f18\\u5316\\u3002AddressSanitizer<\\/a> \\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\nOpenJDK 17<\\/code>\\u3002\\u53ef\\u4ee5\\u4f7f\\u7528Java 8\\u7684\\u7279\\u6027\\u4f8b\\u5982\\uff0clambda expressions \\u548c stream API\\u3002<\\/p>\\r\\n\\r\\nPython 2.7.12<\\/code><\\/p>\\r\\n\\r\\nGCC 8.2<\\/code>\\uff0c\\u91c7\\u7528GNU11\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O1<\\/code>\\u7ea7\\u4f18\\u5316\\u3002 AddressSanitizer<\\/a>\\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n\\r\\nstruct hash_entry {\\r\\n int id; \\/* we'll use this field as the key *\\/\\r\\n char name[10];\\r\\n UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry *s) {\\r\\n HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nstruct hash_entry *find_user(int user_id) {\\r\\n struct hash_entry *s;\\r\\n HASH_FIND_INT(users, &user_id, s);\\r\\n return s;\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nvoid delete_user(struct hash_entry *user) {\\r\\n HASH_DEL(users, user); \\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\"],\"csharp\":[\"C#\",\"Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f \\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\nRuby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\nSwift 5.5.2<\\/code><\\/p>\\r\\n\\r\\nGo 1.18<\\/code><\\/p>\\r\\n\\r\\nPython 3.10<\\/code><\\/p>\\r\\n\\r\\nScala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"Kotlin 1.3.10<\\/code><\\/p>\"],\"rust\":[\"Rust\",\"rust 1.58.1<\\/code><\\/p>\\r\\n\\r\\nPHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n