630. Course Schedule III
ndifferent online courses numbered from
n. Each course has some duration(course length)
tand closed on
dthday. A course should be taken continuously for
tdays and must be finished before or on the
dthday. You will start at the
nonline courses represented by pairs
(t,d), your task is to find the maximal number of courses that can be taken.
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.
- The integer 1 <= d, t, n <= 10,000.
- You can't take two courses simultaneously.
sort course by end time, remove course by descending course length
Rigorous proof here
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)