Problem

There are n different online courses numbered from 1 to n. You are given an array courses where courses[i] = [durationi, lastDayi] indicate that the ith course should be taken continuously for durationi days and must be finished before or on lastDayi.

You will start on the 1st day and you cannot take two or more courses simultaneously.

Return the maximum number of courses that you can take.

Examples

Example 1:

Input: courses =[[100,200],[200,1300],[1000,1250],[2000,3200]]
Output: 3
Explanation: 
There are totally 4 courses, but you can take 3 courses at most:
First, take the 1st course, it costs 100 days so you will finish it on the 100th day, and ready to take the next course on the 101st day.
Second, take the 3rd course, it costs 1000 days so you will finish it on the 1100th day, and ready to take the next course on the 1101st day. 
Third, take the 2nd course, it costs 200 days so you will finish it on the 1300th day. 
The 4th course cannot be taken now, since you will finish it on the 3300th day, which exceeds the closed date.

Example 2:

Input: courses =[[1,2]]
Output: 1

Example 3:

Input: courses =[[3,2],[4,3]]
Output: 0

Solution

This problem is not close to other course schedule problems. But close to Interval Concepts and Problems Index.

Method 1 - Sort by Duration and Use maxHeap on Time

Algo

  1. Sort the courses by their deadlines (Greedy! We have to deal with courses with early deadlines first)
  2. Now iterate over each course and add current course duration. Now we can ignore this course
    1. ignore this course OR
    2. We can replace this course with the longest course we added earlier.

Code

Java
public int scheduleCourse(int[][] courses) {
	Arrays.sort(courses, (a, b) -> a[1] - b[1]);
	Queue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
	int time = 0;
	for (int[] course: courses) {
		int duration = course[0];
		int lastDay = course[1];
		
		time += duration;// time elapses
		maxHeap.add(duration);
		// if time is more than lastDay
		// drop the previous course which costs the most time.
		if (time > course[1]) {
			time -= maxHeap.poll();
		}
	}
	return maxHeap.size();
}