Alice and Bob are traveling to Rome for separate business meetings.
\n\nYou are given 4 strings arriveAlice, leaveAlice, arriveBob, and leaveBob. Alice will be in the city from the dates arriveAlice to leaveAlice (inclusive), while Bob will be in the city from the dates arriveBob to leaveBob (inclusive). Each will be a 5-character string in the format "MM-DD", corresponding to the month and day of the date.
Return the total number of days that Alice and Bob are in Rome together.
\n\nYou can assume that all dates occur in the same calendar year, which is not a leap year. Note that the number of days per month can be represented as: [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31].
\n
Example 1:
\n\n\nInput: arriveAlice = "08-15", leaveAlice = "08-18", arriveBob = "08-16", leaveBob = "08-19"\nOutput: 3\nExplanation: Alice will be in Rome from August 15 to August 18. Bob will be in Rome from August 16 to August 19. They are both in Rome together on August 16th, 17th, and 18th, so the answer is 3.\n\n\n
Example 2:
\n\n\nInput: arriveAlice = "10-01", leaveAlice = "10-31", arriveBob = "11-01", leaveBob = "12-31"\nOutput: 0\nExplanation: There is no day when Alice and Bob are in Rome together, so we return 0.\n\n\n
\n
Constraints:
\n\n"MM-DD".Alice 和 Bob 计划分别去罗马开会。
\n\n给你四个字符串 arriveAlice ，leaveAlice ，arriveBob 和 leaveBob 。Alice 会在日期 arriveAlice 到 leaveAlice 之间在城市里（日期为闭区间），而 Bob 在日期 arriveBob 到 leaveBob 之间在城市里（日期为闭区间）。每个字符串都包含 5 个字符，格式为 \"MM-DD\" ，对应着一个日期的月和日。
请你返回 Alice和 Bob 同时在罗马的天数。
\n\n你可以假设所有日期都在 同一个 自然年，而且 不是 闰年。每个月份的天数分别为：[31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] 。
\n\n
示例 1：
\n\n\n输入：arriveAlice = \"08-15\", leaveAlice = \"08-18\", arriveBob = \"08-16\", leaveBob = \"08-19\"\n输出：3\n解释：Alice 从 8 月 15 号到 8 月 18 号在罗马。Bob 从 8 月 16 号到 8 月 19 号在罗马，他们同时在罗马的日期为 8 月 16、17 和 18 号。所以答案为 3 。\n\n\n
示例 2：
\n\n\n输入：arriveAlice = \"10-01\", leaveAlice = \"10-31\", arriveBob = \"11-01\", leaveBob = \"12-31\"\n输出：0\n解释：Alice 和 Bob 没有同时在罗马的日子，所以我们返回 0 。\n\n\n
\n\n
提示：
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collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\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>\\r\\n\\r\\n \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. 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\ No newline at end of file
diff --git a/leetcode-cn/originData/length-of-the-longest-alphabetical-continuous-substring.json b/leetcode-cn/originData/length-of-the-longest-alphabetical-continuous-substring.json
new file mode 100644
index 00000000..dfef35bc
--- /dev/null
+++ b/leetcode-cn/originData/length-of-the-longest-alphabetical-continuous-substring.json
@@ -0,0 +1,164 @@
+{
+ "data": {
+ "question": {
+ "questionId": "2492",
+ "questionFrontendId": "2414",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 1827108,
+ "title": "Length of the Longest Alphabetical Continuous Substring",
+ "titleSlug": "length-of-the-longest-alphabetical-continuous-substring",
+ "content": " An alphabetical continuous string is a string consisting of consecutive letters in the alphabet. In other words, it is any substring of the string Given a string Example 1: Example 2: Constraints: 字母序连续字符串 是由字母表中连续字母组成的字符串。换句话说，字符串 给你一个仅由小写英文字母组成的字符串 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>, 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\ No newline at end of file
diff --git a/leetcode-cn/originData/maximum-matching-of-players-with-trainers.json b/leetcode-cn/originData/maximum-matching-of-players-with-trainers.json
new file mode 100644
index 00000000..64424cc8
--- /dev/null
+++ b/leetcode-cn/originData/maximum-matching-of-players-with-trainers.json
@@ -0,0 +1,164 @@
+{
+ "data": {
+ "question": {
+ "questionId": "2497",
+ "questionFrontendId": "2410",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 1827365,
+ "title": "Maximum Matching of Players With Trainers",
+ "titleSlug": "maximum-matching-of-players-with-trainers",
+ "content": " You are given a 0-indexed integer array The Return the maximum number of matchings between Example 1: Example 2: Constraints: 给你一个下标从 0 开始的整数数组 如果第 请你返回满足上述要求 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 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\ No newline at end of file
diff --git a/leetcode-cn/originData/minimum-money-required-before-transactions.json b/leetcode-cn/originData/minimum-money-required-before-transactions.json
new file mode 100644
index 00000000..1059dc38
--- /dev/null
+++ b/leetcode-cn/originData/minimum-money-required-before-transactions.json
@@ -0,0 +1,165 @@
+{
+ "data": {
+ "question": {
+ "questionId": "2499",
+ "questionFrontendId": "2412",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 1827367,
+ "title": "Minimum Money Required Before Transactions",
+ "titleSlug": "minimum-money-required-before-transactions",
+ "content": " You are given a 0-indexed 2D integer array The array describes transactions, where each transaction must be completed exactly once in some order. At any given moment, you have a certain amount of Return the minimum amount of Example 1: Example 2: Constraints: 给你一个下标从 0 开始的二维整数数组 数组描述了若干笔交易。其中每笔交易必须以 某种顺序 恰好完成一次。在任意一个时刻，你有一定数目的钱 请你返回 任意一种 交易顺序下，你都能完成所有交易的最少钱数 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>, collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\u7248\\u672c\\uff1aclang 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 1.17<\\/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"abcdefghijklmnopqrstuvwxyz".\n\t
\n\n"abc" is an alphabetical continuous string, while "acb" and "za" are not.s consisting of lowercase letters only, return the length of the longest alphabetical continuous substring.\nInput: s = "abacaba"\nOutput: 2\nExplanation: There are 4 distinct continuous substrings: "a", "b", "c" and "ab".\n"ab" is the longest continuous substring.\n
\n\n\nInput: s = "abcde"\nOutput: 5\nExplanation: "abcde" is the longest continuous substring.\n
\n\n\n\t
\n",
+ "translatedTitle": "最长的字母序连续子字符串的长度",
+ "translatedContent": "1 <= s.length <= 105s consists of only English lowercase letters.\"abcdefghijklmnopqrstuvwxyz\" 的任意子字符串都是 字母序连续字符串 。\n\t
\n\n\"abc\" 是一个字母序连续字符串，而 \"acb\" 和 \"za\" 不是。s ，返回其 最长 的 字母序连续子字符串 的长度。输入：s = \"abacaba\"\n输出：2\n解释：共有 4 个不同的字母序连续子字符串 \"a\"、\"b\"、\"c\" 和 \"ab\" 。\n\"ab\" 是最长的字母序连续子字符串。\n
\n\n输入：s = \"abcde\"\n输出：5\n解释：\"abcde\" 是最长的字母序连续子字符串。\n
\n\n\n\t
\n",
+ "isPaidOnly": false,
+ "difficulty": "Medium",
+ "likes": 4,
+ "dislikes": 0,
+ "isLiked": null,
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+ "contributors": [],
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+ "topicTags": [],
+ "companyTagStats": null,
+ "codeSnippets": [
+ {
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+ "__typename": "CodeSnippetNode"
+ },
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+ "__typename": "CodeSnippetNode"
+ },
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+ "langSlug": "python",
+ "code": "class Solution(object):\n def longestContinuousSubstring(self, s):\n \"\"\"\n :type s: str\n :rtype: int\n \"\"\"",
+ "__typename": "CodeSnippetNode"
+ },
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+ "langSlug": "golang",
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+ "stats": "{\"totalAccepted\": \"7.8K\", \"totalSubmission\": \"12.7K\", \"totalAcceptedRaw\": 7785, \"totalSubmissionRaw\": 12725, \"acRate\": \"61.2%\"}",
+ "hints": [
+ "What is the longest possible continuous substring?",
+ "The size of the longest possible continuous substring is at most 26, so we can just brute force the answer."
+ ],
+ "solution": null,
+ "status": null,
+ "sampleTestCase": "\"abacaba\"",
+ "metaData": "{\n \"name\": \"longestContinuousSubstring\",\n \"params\": [\n {\n \"name\": \"s\",\n \"type\": \"string\"\n }\n ],\n \"return\": {\n \"type\": \"integer\"\n }\n}",
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\n\n\nInput: players = [1,1,1], trainers = [10]\nOutput: 1\nExplanation:\nThe trainer can be matched with any of the 3 players.\nEach player can only be matched with one trainer, so the maximum answer is 1.\n
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+ "translatedTitle": "运动员和训练师的最大匹配数",
+ "translatedContent": "1 <= players.length, trainers.length <= 1051 <= players[i], trainers[j] <= 109players ，其中 players[i] 表示第 i 名运动员的 能力 值，同时给你一个下标从 0 开始的整数数组 trainers ，其中 trainers[j] 表示第 j 名训练师的 训练能力值 。i 名运动员的能力值 小于等于 第 j 名训练师的能力值，那么第 i 名运动员可以 匹配 第 j 名训练师。除此以外，每名运动员至多可以匹配一位训练师，每位训练师最多可以匹配一位运动员。players 和 trainers 的 最大 匹配数。输入：players = [4,7,9], trainers = [8,2,5,8]\n输出：2\n解释：\n得到两个匹配的一种方案是：\n- players[0] 与 trainers[0] 匹配，因为 4 <= 8 。\n- players[1] 与 trainers[3] 匹配，因为 7 <= 8 。\n可以证明 2 是可以形成的最大匹配数。\n
\n\n输入：players = [1,1,1], trainers = [10]\n输出：1\n解释：\n训练师可以匹配所有 3 个运动员\n每个运动员至多只能匹配一个训练师，所以最大答案是 1 。\n
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+ "code": "class Solution {\npublic:\n int matchPlayersAndTrainers(vector1 <= players.length, trainers.length <= 1051 <= players[i], trainers[j] <= 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 1.17<\\/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\\ntransactions, where transactions[i] = [costi, cashbacki].money. In order to complete transaction i, money >= costi must hold true. After performing a transaction, money becomes money - costi + cashbacki.money required before any transaction so that all of the transactions can be completed regardless of the order of the transactions.\nInput: transactions = [[2,1],[5,0],[4,2]]\nOutput: 10\nExplanation:\nStarting with money = 10, the transactions can be performed in any order.\nIt can be shown that starting with money < 10 will fail to complete all transactions in some order.\n
\n\n\nInput: transactions = [[3,0],[0,3]]\nOutput: 3\nExplanation:\n- If transactions are in the order [[3,0],[0,3]], the minimum money required to complete the transactions is 3.\n- If transactions are in the order [[0,3],[3,0]], the minimum money required to complete the transactions is 0.\nThus, starting with money = 3, the transactions can be performed in any order.\n
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+ "translatedTitle": "完成所有交易的初始最少钱数",
+ "translatedContent": "1 <= transactions.length <= 105transactions[i].length == 20 <= costi, cashbacki <= 109transactions，其中transactions[i] = [costi, cashbacki] 。money ，为了完成交易 i ，money >= costi 这个条件必须为真。执行交易后，你的钱数 money 变成 money - costi + cashbacki 。money 是多少。\n输入：transactions = [[2,1],[5,0],[4,2]]\n输出：10\n解释：\n刚开始 money = 10 ，交易可以以任意顺序进行。\n可以证明如果 money < 10 ，那么某些交易无法进行。\n
\n\n\n输入：transactions = [[3,0],[0,3]]\n输出：3\n解释：\n- 如果交易执行的顺序是 [[3,0],[0,3]] ，完成所有交易需要的最少钱数是 3 。\n- 如果交易执行的顺序是 [[0,3],[3,0]] ，完成所有交易需要的最少钱数是 0 。\n所以，刚开始钱数为 3 ，任意顺序下交易都可以全部完成。\n
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+ "code": "class Solution {\npublic:\n long long minimumMoney(vector1 <= transactions.length <= 105transactions[i].length == 20 <= costi, cashbacki <= 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\\n