Formatted question description: https://leetcode.ca/all/630.html

# 630. Course Schedule III

Hard

## Description

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:

1. The integer 1 <= d, t, n <= 10,000.
2. You can’t take two courses simultaneously.

## Solution

Core logic: finish shorter duration courses first.

Sort the courses according to the close days. It is always optimal to take a course that has an earlier end day.

Use priority queue, where the maximum element is polled from the priority queue first. For each course, obtain its duration and close day.

• If the current course can be finished on the close day of the course (previous courses are also considered, so consider the total days used for taking the courses so far), then o
• ffer the course’s duration to the priority queue,
• add the duration to the total days,
• and add the counter of courses by 1.
• If the current course can’t be finished on the close day of the course, then check whether the maximum duration in the priority queue is greater than the current course's duration.
• If so, replace the maximum duration with the current course’s duration, and update the number of days used to take the courses so far.

Finally, return the number of courses that can be taken.

• 
public class Course_Schedule_III {

// https://leetcode.com/articles/course-schedule-iii/
public class Solution {
public int scheduleCourse(int[][] courses) {

Arrays.sort(courses, (a, b) -> a[1] - b[1]);
PriorityQueue<Integer> queue = new PriorityQueue<>((a, b) -> b - a); // duration time放入heap

int time = 0;
for (int[] c: courses) {
if (time + c[0] <= c[1]) {
queue.offer(c[0]);
time += c[0];
} else if (!queue.isEmpty() && queue.peek() > c[0]) {
time += c[0] - queue.poll();
queue.offer(c[0]);
}
}

return queue.size();
}
}
}

• // OJ: https://leetcode.com/problems/course-schedule-iii/
// Time: O(NT)
// Space: O(NT)
class Solution {
vector<vector<int>> memo;
int dp(vector<vector<int>> &A, int i, int time) {
if (i == A.size()) return 0;
if (memo[i][time] != -1) return memo[i][time];
int pick = 0;
if (time + A[i][0] <= A[i][1]) pick = 1 + dp(A, i + 1, time + A[i][0]);
int skip = dp(A, i + 1, time);
return memo[i][time] = max(pick, skip);
}
public:
int scheduleCourse(vector<vector<int>>& A) {
sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
memo.assign(A.size(), vector<int>(A.back()[1] + 1, -1));
return dp(A, 0, 0);
}
};

• # idea:
# sort all courses by deadline
# iterate all sorted courses
# if current course is able to be taken, then take it
# if not, check if we can remove some courses from courses we already taken
# if the one has the maximal duration is greater than current course's duration
# then replace it by current course
# (why? because sorted by deadline, current deadline must be later than all deadlines of taken courses,
# so it must be valid)
# moreover, we have more available time for taking more courses

class Solution(object):
def scheduleCourse(self, courses):
"""
:type courses: List[List[int]]
:rtype: int
"""
now = 0
heap = []
for t, d in sorted(courses, key=lambda x: x[1]):
if now + t <= d:
now += t
heapq.heappush(heap, -t)
elif heap and -heap[0] > t:
now += t + heapq.heappop(heap)
heapq.heappush(heap, -t)
return len(heap)