630. Course Schedule III

There arendifferent online courses numbered from1ton. Each course has some duration(course length)tand closed ondthday. A course should be taken continuously fortdays and must be finished before or on thedthday. You will start at the1stday.

Givennonline courses represented by pairs(t,d), your task is to find the maximal number of courses that can be taken.

Example:

Input:
 [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]]

Output:
 3

Explanation:

There're 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.

Note:

1. The integer 1 <= d, t, n <= 10,000.

2. You can't take two courses simultaneously.

Thoughts:

Greedy:

sort course by end time, remove course by descending course length

Rigorous proof here

Code (Python):

class Solution(object):
    def scheduleCourse(self, courses):
        """
        :type courses: List[List[int]]
        :rtype: int
        """
        pq = [] 
        total = 0
        for t, end in sorted(courses, key = lambda(t,end):end):
            total += t
            heapq.heappush(pq, -t) #-t here since heapq is a minheap by default
            if total > end:
                total += heapq.heappop(pq) # add negative number
        return len(pq)