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0630-course-schedule-iii/README.md

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+<h2><a href="https://leetcode.com/problems/course-schedule-iii">630. Course Schedule III</a></h2><h3>Hard</h3><hr><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>
+
+<p>You will start on the <code>1<sup>st</sup></code> day and you cannot take two or more courses simultaneously.</p>
+
+<p>Return <em>the maximum number of courses that you can take</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> courses = [[100,200],[200,1300],[1000,1250],[2000,3200]]
+<strong>Output:</strong> 3
+Explanation: 
+There are totally 4 courses, but you can take 3 courses at most:
+First, 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.
+Second, 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. 
+Third, take the 2<sup>nd</sup> course, it costs 200 days so you will finish it on the 1300<sup>th</sup> day. 
+The 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.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> courses = [[1,2]]
+<strong>Output:</strong> 1
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> courses = [[3,2],[4,3]]
+<strong>Output:</strong> 0
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= courses.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= duration<sub>i</sub>, lastDay<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
+</ul>