Alice and Bob are playing a turn-based game on a circular field surrounded by flowers. The circle represents the field, and there are x flowers in the clockwise direction between Alice and Bob, and y flowers in the anti-clockwise direction between them.
The game proceeds as follows:
\n\nGiven two integers, n and m, the task is to compute the number of possible pairs (x, y) that satisfy the conditions:
x in the clockwise direction must be in the range [1,n].y in the anti-clockwise direction must be in the range [1,m].Return the number of possible pairs (x, y) that satisfy the conditions mentioned in the statement.
\n
Example 1:
\n\n\nInput: n = 3, m = 2\nOutput: 3\nExplanation: The following pairs satisfy conditions described in the statement: (1,2), (3,2), (2,1).\n\n\n
Example 2:
\n\n\nInput: n = 1, m = 1\nOutput: 0\nExplanation: No pairs satisfy the conditions described in the statement.\n\n\n
\n
Constraints:
\n\n1 <= n, m <= 105Alice 和 Bob 在一个长满鲜花的环形草地玩一个回合制游戏。环形的草地上有一些鲜花，Alice 到 Bob 之间顺时针有 x 朵鲜花，逆时针有 y 朵鲜花。
游戏过程如下：
\n\n给你两个整数 n 和 m ，你的任务是求出满足以下条件的所有 (x, y) 对：
x 必须在区间 [1,n] 之间。y 必须在区间 [1,m] 之间。请你返回满足题目描述的数对 (x, y) 的数目。
\n\n
示例 1：
\n\n\n输入：n = 3, m = 2\n输出：3\n解释：以下数对满足题目要求：(1,2) ，(3,2) ，(2,1) 。\n\n\n
示例 2：
\n\n\n输入：n = 1, m = 1\n输出：0\n解释：没有数对满足题目要求。\n\n\n
\n\n
提示：
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\ No newline at end of file
diff --git a/leetcode-cn/originData/find-the-maximum-number-of-elements-in-subset.json b/leetcode-cn/originData/find-the-maximum-number-of-elements-in-subset.json
new file mode 100644
index 00000000..e3c29f8a
--- /dev/null
+++ b/leetcode-cn/originData/find-the-maximum-number-of-elements-in-subset.json
@@ -0,0 +1,185 @@
+{
+ "data": {
+ "question": {
+ "questionId": "3299",
+ "questionFrontendId": "3020",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2619046,
+ "title": "Find the Maximum Number of Elements in Subset",
+ "titleSlug": "find-the-maximum-number-of-elements-in-subset",
+ "content": " You are given an array of positive integers You need to select a subset of Return the maximum number of elements in a subset that satisfies these conditions. Example 1: Example 2: Constraints: 给你一个 正整数 数组 你需要从数组中选出一个满足下述条件的子集： 返回满足这些条件的子集中，元素数量的 最大值 。 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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+ }
+}
\ No newline at end of file
diff --git a/leetcode-cn/originData/minimize-or-of-remaining-elements-using-operations.json b/leetcode-cn/originData/minimize-or-of-remaining-elements-using-operations.json
new file mode 100644
index 00000000..942a224b
--- /dev/null
+++ b/leetcode-cn/originData/minimize-or-of-remaining-elements-using-operations.json
@@ -0,0 +1,188 @@
+{
+ "data": {
+ "question": {
+ "questionId": "3261",
+ "questionFrontendId": "3022",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2619049,
+ "title": "Minimize OR of Remaining Elements Using Operations",
+ "titleSlug": "minimize-or-of-remaining-elements-using-operations",
+ "content": " You are given a 0-indexed integer array In one operation, you can pick any index Return the minimum possible value of the bitwise Example 1: Example 2: Example 3: Constraints: 给你一个下标从 0 开始的整数数组 一次操作中，你可以选择 请你返回 至多 示例 1： 示例 2： 示例 3： 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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\ No newline at end of file
diff --git a/leetcode-cn/originData/number-of-changing-keys.json b/leetcode-cn/originData/number-of-changing-keys.json
new file mode 100644
index 00000000..7f8de8ab
--- /dev/null
+++ b/leetcode-cn/originData/number-of-changing-keys.json
@@ -0,0 +1,170 @@
+{
+ "data": {
+ "question": {
+ "questionId": "3312",
+ "questionFrontendId": "3019",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2619045,
+ "title": "Number of Changing Keys",
+ "titleSlug": "number-of-changing-keys",
+ "content": " You are given a 0-indexed string Return the number of times the user had to change the key. Note: Modifiers like Example 1: Example 2: Constraints: 给你一个下标从 0 开始的字符串 返回用户输入过程中按键变更的次数。 注意： 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 clang 17<\\/code> \\u91c7\\u7528\\u6700\\u65b0C++ 20\\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 21<\\/code>\\u3002\\u4f7f\\u7528\\u7f16\\u8bd1\\u53c2\\u6570 --enable-preview --release 21<\\/code><\\/p>\\r\\n\\r\\nPython 2.7.12<\\/code><\\/p>\\r\\n\\r\\nGCC 11<\\/code>\\uff0c\\u91c7\\u7528GNU11\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O2<\\/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 20.10.0<\\/code><\\/p>\\r\\n\\r\\n--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f \\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\nRuby 3.2<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\nSwift 5.9<\\/code><\\/p>\\r\\n\\r\\nGo 1.21<\\/code><\\/p>\\r\\n\\r\\nPython 3.11<\\/code><\\/p>\\r\\n\\r\\nScala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"Kotlin 1.9.0<\\/code><\\/p>\\r\\n\\r\\nrust 1.74.1<\\/code><\\/p>\\r\\n\\r\\nPHP 8.2<\\/code>.<\\/p>\\r\\n\\r\\nnums.nums which satisfies the following condition:\n\t
\n\n[x, x2, x4, ..., xk/2, xk, xk/2, ..., x4, x2, x] (Note that k can be be any non-negative power of 2). For example, [2, 4, 16, 4, 2] and [3, 9, 3] follow the pattern while [2, 4, 8, 4, 2] does not.\nInput: nums = [5,4,1,2,2]\nOutput: 3\nExplanation: We can select the subset {4,2,2}, which can be placed in the array as [2,4,2] which follows the pattern and 22 == 4. Hence the answer is 3.\n\n\n\nInput: nums = [1,3,2,4]\nOutput: 1\nExplanation: We can select the subset {1}, which can be placed in the array as [1] which follows the pattern. Hence the answer is 1. Note that we could have also selected the subsets {2}, {4}, or {3}, there may be multiple subsets which provide the same answer. \n\n\n\n\t
\n",
+ "translatedTitle": "子集中元素的最大数量",
+ "translatedContent": "2 <= nums.length <= 1051 <= nums[i] <= 109nums 。\n\t
\n\n[x, x2, x4, ..., xk/2, xk, xk/2, ..., x4, x2, x]（注意，k 可以是任何 非负 的 2 的幂）。例如，[2, 4, 16, 4, 2] 和 [3, 9, 3] 都符合这一模式，而 [2, 4, 8, 4, 2] 则不符合。\n输入：nums = [5,4,1,2,2]\n输出：3\n解释：选择子集 {4,2,2} ，将其放在数组 [2,4,2] 中，它遵循该模式，且 22 == 4 。因此答案是 3 。\n\n\n\n输入：nums = [1,3,2,4]\n输出：1\n解释：选择子集 {1}，将其放在数组 [1] 中，它遵循该模式。因此答案是 1 。注意我们也可以选择子集 {2} 、{4} 或 {3} ，可能存在多个子集都能得到相同的答案。\n\n\n\n\t
\n",
+ "isPaidOnly": false,
+ "difficulty": "Medium",
+ "likes": 6,
+ "dislikes": 0,
+ "isLiked": null,
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+ "contributors": [],
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+ "topicTags": [
+ {
+ "name": "Array",
+ "slug": "array",
+ "translatedName": "数组",
+ "__typename": "TopicTagNode"
+ },
+ {
+ "name": "Hash Table",
+ "slug": "hash-table",
+ "translatedName": "哈希表",
+ "__typename": "TopicTagNode"
+ },
+ {
+ "name": "Enumeration",
+ "slug": "enumeration",
+ "translatedName": "枚举",
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+ }
+ ],
+ "companyTagStats": null,
+ "codeSnippets": [
+ {
+ "lang": "C++",
+ "langSlug": "cpp",
+ "code": "class Solution {\npublic:\n int maximumLength(vector2 <= nums.length <= 1051 <= nums[i] <= 1091’s.",
+ "Put all the values into a HashSet. We can start from each x > 1 as the smallest chosen value and we can find the longest subset by checking the new values (which are the square of the previous value) in the set by brute force.",
+ "Note when x > 1, x2, x4, x8, … increases very fast, the longest subset with smallest value x cannot be very long. (The length is O(log(log(109))).",
+ "Hence we can directly check all lengths less than 10 for all values of x."
+ ],
+ "solution": null,
+ "status": null,
+ "sampleTestCase": "[5,4,1,2,2]",
+ "metaData": "{\n \"name\": \"maximumLength\",\n \"params\": [\n {\n \"name\": \"nums\",\n \"type\": \"integer[]\"\n }\n ],\n \"return\": {\n \"type\": \"integer\"\n }\n}",
+ "judgerAvailable": true,
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\n\n\nInput: nums = [7,3,15,14,2,8], k = 4\nOutput: 2\nExplanation: Let's do the following operations:\n1. Replace nums[0] and nums[1] with (nums[0] & nums[1]) so that nums becomes equal to [3,15,14,2,8]. \n2. Replace nums[0] and nums[1] with (nums[0] & nums[1]) so that nums becomes equal to [3,14,2,8].\n3. Replace nums[0] and nums[1] with (nums[0] & nums[1]) so that nums becomes equal to [2,2,8].\n4. Replace nums[1] and nums[2] with (nums[1] & nums[2]) so that nums becomes equal to [2,0].\nThe bitwise-or of the final array is 2.\nIt can be shown that 2 is the minimum possible value of the bitwise OR of the remaining elements of nums after applying at most k operations.\n
\n\n\nInput: nums = [10,7,10,3,9,14,9,4], k = 1\nOutput: 15\nExplanation: Without applying any operations, the bitwise-or of nums is 15.\nIt can be shown that 15 is the minimum possible value of the bitwise OR of the remaining elements of nums after applying at most k operations.\n
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+ "translatedTitle": "给定操作次数内使剩余元素的或值最小",
+ "translatedContent": "1 <= nums.length <= 1050 <= nums[i] < 2300 <= k < nums.lengthnums 和一个整数 k 。nums 中满足 0 <= i < nums.length - 1 的一个下标 i ，并将 nums[i] 和 nums[i + 1] 替换为数字 nums[i] & nums[i + 1] ，其中 & 表示按位 AND 操作。k 次操作以内，使 nums 中所有剩余元素按位 OR 结果的 最小值 。\n输入：nums = [3,5,3,2,7], k = 2\n输出：3\n解释：执行以下操作：\n1. 将 nums[0] 和 nums[1] 替换为 (nums[0] & nums[1]) ，得到 nums 为 [1,3,2,7] 。\n2. 将 nums[2] 和 nums[3] 替换为 (nums[2] & nums[3]) ，得到 nums 为 [1,3,2] 。\n最终数组的按位或值为 3 。\n3 是 k 次操作以内，可以得到的剩余元素的最小按位或值。
\n\n\n输入：nums = [7,3,15,14,2,8], k = 4\n输出：2\n解释：执行以下操作：\n1. 将 nums[0] 和 nums[1] 替换为 (nums[0] & nums[1]) ，得到 nums 为 [3,15,14,2,8] 。\n2. 将 nums[0] 和 nums[1] 替换为 (nums[0] & nums[1]) ，得到 nums 为 [3,14,2,8] 。\n3. 将 nums[0] 和 nums[1] 替换为 (nums[0] & nums[1]) ，得到 nums 为 [2,2,8] 。\n4. 将 nums[1] 和 nums[2] 替换为 (nums[1] & nums[2]) ，得到 nums 为 [2,0] 。\n最终数组的按位或值为 2 。\n2 是 k 次操作以内，可以得到的剩余元素的最小按位或值。\n
\n\n\n输入：nums = [10,7,10,3,9,14,9,4], k = 1\n输出：15\n解释：不执行任何操作，nums 的按位或值为 15 。\n15 是 k 次操作以内，可以得到的剩余元素的最小按位或值。\n
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+ "code": "class Solution {\npublic:\n int minOrAfterOperations(vector1 <= nums.length <= 1050 <= nums[i] < 2300 <= k < nums.lengthmask.",
+ "For a fixed bit, add it to mask then check if there exists some sequence of k operations such that mask & answer == 0 where answer is the bitwise-or of the remaining elements of nums. If there is no such sequence of operations, remove the current bit from mask. How can we perform this check?",
+ "Let x be the bitwise-and of all elements of nums. If x AND mask != 0, there is no sequence of operations that satisfies the condition in the previous hint. This is because even if we perform this operation n - 1 times on the array, we will end up with x as the final element.",
+ "Otherwise, there exists at least one such sequence. It is sufficient to check if the number of operations in such a sequence is less than k. Let’s calculate the minimum number of operations in such a sequence.",
+ "Iterate over the array from left to right, if nums[i] & mask != 0, apply the operation on index i.",
+ "After iterating over all elements, let x be the bitwise-and of all elements of nums. If x == 0, then we have found the minimum number of operations. Otherwise, It can be proven that we need exactly one more operation so that x == 0.",
+ "The condition in the second hint is satisfied if and only if the minimum number of operations is less than or equal to k."
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\n\n\nInput: s = "AaAaAaaA"\nOutput: 0\nExplanation: There is no change of key since only the letters 'a' and 'A' are pressed which does not require change of key.\n
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+ "translatedTitle": "按键变更的次数",
+ "translatedContent": "1 <= s.length <= 100s consists of only upper case and lower case English letters.s ，该字符串由用户输入。按键变更的定义是：使用与上次使用的按键不同的键。例如 s = \"ab\" 表示按键变更一次，而 s = \"bBBb\" 不存在按键变更。shift 或 caps lock 等修饰键不计入按键变更，也就是说，如果用户先输入字母 'a' 然后输入字母 'A' ，不算作按键变更。\n输入：s = \"aAbBcC\"\n输出：2\n解释： \n从 s[0] = 'a' 到 s[1] = 'A'，不存在按键变更，因为不计入 caps lock 或 shift 。\n从 s[1] = 'A' 到 s[2] = 'b'，按键变更。\n从 s[2] = 'b' 到 s[3] = 'B'，不存在按键变更，因为不计入 caps lock 或 shift 。\n从 s[3] = 'B' 到 s[4] = 'c'，按键变更。\n从 s[4] = 'c' 到 s[5] = 'C'，不存在按键变更，因为不计入 caps lock 或 shift 。\n
\n\n\n输入：s = \"AaAaAaaA\"\n输出：0\n解释： 不存在按键变更，因为这个过程中只按下字母 'a' 和 'A' ，不需要进行按键变更。\n
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