course-schedule-iii.py

# Time:  O(nlogn)
# Space: O(k), k is the number of courses you can take

# There are n different online courses numbered from 1 to n.
# Each course has some duration(course length) t and closed on dth day.
# A course should be taken continuously for t days and must be finished before or on the dth day.
# You will start at the 1st day.
#
# Given n online 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:
# The integer 1 <= d, t, n <= 10,000.
# You can't take two courses simultaneously.

class Solution(object):
    def scheduleCourse(self, courses):
        """
        :type courses: List[List[int]]
        :rtype: int
        """
        courses.sort(key=lambda(t, end): end)
        max_heap = []
        now = 0
        for t, end in courses:
            now += t
            heapq.heappush(max_heap, -t)
            if now > end:
                now += heapq.heappop(max_heap)
        return len(max_heap)