leetcode-problemset/leetcode-cn/originData/course-schedule-iii.json

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    "data": {
        "question": {
            "questionId": "630",
            "questionFrontendId": "630",
            "categoryTitle": "Algorithms",
            "boundTopicId": 1700,
            "title": "Course Schedule III",
            "titleSlug": "course-schedule-iii",
            "content": "<p>There are <code>n</code> different online courses numbered from <code>1</code> to <code>n</code>. You are given an array <code>courses</code> where <code>courses[i] = [duration<sub>i</sub>, lastDay<sub>i</sub>]</code> indicate that the <code>i<sup>th</sup></code> course should be taken <b>continuously</b> for <code>duration<sub>i</sub></code> days and must be finished before or on <code>lastDay<sub>i</sub></code>.</p>\n\n<p>You will start on the <code>1<sup>st</sup></code> day and you cannot take two or more courses simultaneously.</p>\n\n<p>Return <em>the maximum number of courses that you can take</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> courses = [[100,200],[200,1300],[1000,1250],[2000,3200]]\n<strong>Output:</strong> 3\nExplanation: \nThere are totally 4 courses, but you can take 3 courses at most:\nFirst, take the 1<sup>st</sup> course, it costs 100 days so you will finish it on the 100<sup>th</sup> day, and ready to take the next course on the 101<sup>st</sup> day.\nSecond, take the 3<sup>rd</sup> course, it costs 1000 days so you will finish it on the 1100<sup>th</sup> day, and ready to take the next course on the 1101<sup>st</sup> day. \nThird, take the 2<sup>nd</sup> course, it costs 200 days so you will finish it on the 1300<sup>th</sup> day. \nThe 4<sup>th</sup> course cannot be taken now, since you will finish it on the 3300<sup>th</sup> day, which exceeds the closed date.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> courses = [[1,2]]\n<strong>Output:</strong> 1\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> courses = [[3,2],[4,3]]\n<strong>Output:</strong> 0\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= courses.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= duration<sub>i</sub>, lastDay<sub>i</sub> &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "课程表 III",
            "translatedContent": "<p>这里有 <code>n</code> 门不同的在线课程，按从 <code>1</code> 到 <code>n</code>&nbsp;编号。给你一个数组 <code>courses</code> ，其中 <code>courses[i] = [duration<sub>i</sub>, lastDay<sub>i</sub>]</code> 表示第 <code>i</code> 门课将会 <strong>持续</strong> 上 <code>duration<sub>i</sub></code> 天课，并且必须在不晚于 <code>lastDay<sub>i</sub></code> 的时候完成。</p>\n\n<p>你的学期从第 <code>1</code> 天开始。且不能同时修读两门及两门以上的课程。</p>\n\n<p>返回你最多可以修读的课程数目。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>courses = [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]]\n<strong>输出：</strong>3\n<strong>解释：</strong>\n这里一共有 4 门课程，但是你最多可以修 3 门：\n首先，修第 1 门课，耗费 100 天，在第 100 天完成，在第 101 天开始下门课。\n第二，修第 3 门课，耗费 1000 天，在第 1100 天完成，在第 1101 天开始下门课程。\n第三，修第 2 门课，耗时 200 天，在第 1300 天完成。\n第 4 门课现在不能修，因为将会在第 3300 天完成它，这已经超出了关闭日期。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>courses = [[1,2]]\n<strong>输出：</strong>1\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>courses = [[3,2],[4,3]]\n<strong>输出：</strong>0\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= courses.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= duration<sub>i</sub>, lastDay<sub>i</sub> &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    int scheduleCourse(vector<vector<int>>& courses) {\n\n    }\n};",
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                    "code": "class Solution(object):\n    def scheduleCourse(self, courses):\n        \"\"\"\n        :type courses: List[List[int]]\n        :rtype: int\n        \"\"\"",
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                    "code": "\n\nint scheduleCourse(int** courses, int coursesSize, int* coursesColSize){\n\n}",
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                    "code": "public class Solution {\n    public int ScheduleCourse(int[][] courses) {\n\n    }\n}",
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                    "code": "/**\n * @param {number[][]} courses\n * @return {number}\n */\nvar scheduleCourse = function(courses) {\n\n};",
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                    "code": "# @param {Integer[][]} courses\n# @return {Integer}\ndef schedule_course(courses)\n\nend",
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                    "code": "class Solution {\n    fun scheduleCourse(courses: Array<IntArray>): Int {\n\n    }\n}",
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                    "code": "impl Solution {\n    pub fn schedule_course(courses: Vec<Vec<i32>>) -> i32 {\n\n    }\n}",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer[][] $courses\n     * @return Integer\n     */\n    function scheduleCourse($courses) {\n\n    }\n}",
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            "hints": [
                "During iteration, say I want to add the current course, currentTotalTime being total time of all courses taken till now, but adding the current course might exceed my deadline or it doesn’t.</br></br>\r\n\r\n1. If it doesn’t, then I have added one new course. Increment the currentTotalTime with duration of current course.",
                "2. If it exceeds deadline, I can swap current course with current courses that has biggest duration.</br>\r\n* No harm done and I might have just reduced the currentTotalTime, right? </br>\r\n* What preprocessing do I need to do on my course processing order so that this swap is always legal?"
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