mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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193 lines
18 KiB
JSON
193 lines
18 KiB
JSON
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"title": "Course Schedule III",
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"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> </p>\n<p><strong class=\"example\">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 class=\"example\">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 class=\"example\">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> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= courses.length <= 10<sup>4</sup></code></li>\n\t<li><code>1 <= duration<sub>i</sub>, lastDay<sub>i</sub> <= 10<sup>4</sup></code></li>\n</ul>\n",
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"code": "# @param {Integer[][]} courses\n# @return {Integer}\ndef schedule_course(courses)\n \nend",
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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": "(define/contract (schedule-course courses)\n (-> (listof (listof exact-integer?)) exact-integer?)\n )",
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"code": "-spec schedule_course(Courses :: [[integer()]]) -> integer().\nschedule_course(Courses) ->\n .",
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"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.",
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"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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