Skip to content

630. Course Schedule III

Description

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.

Β 

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

Β 

Constraints:

  • 1 <= courses.length <= 104
  • 1 <= durationi, lastDayi <= 104

Solutions

Solution 1: Greedy + Priority Queue (Max-Heap)

We can sort the courses in ascending order by their end time, and each time select the course with the earliest deadline to take.

If the total time \(s\) of the selected courses exceeds the end time \(last\) of the current course, we remove the course with the longest duration from the previously selected courses, until the constraint of the current course's end time is satisfied. Here we use a priority queue (max-heap) \(pq\) to maintain the durations of the currently selected courses, and each time we pop the course with the longest duration from the priority queue to remove it.

Finally, the number of elements in the priority queue is the maximum number of courses we can take.

The time complexity is \(O(n \times \log n)\) and the space complexity is \(O(n)\), where \(n\) is the number of courses.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution:
    def scheduleCourse(self, courses: List[List[int]]) -> int:
        courses.sort(key=lambda x: x[1])
        pq = []
        s = 0
        for duration, last in courses:
            heappush(pq, -duration)
            s += duration
            while s > last:
                s += heappop(pq)
        return len(pq)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
class Solution {
    public int scheduleCourse(int[][] courses) {
        Arrays.sort(courses, (a, b) -> a[1] - b[1]);
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
        int s = 0;
        for (var e : courses) {
            int duration = e[0], last = e[1];
            pq.offer(duration);
            s += duration;
            while (s > last) {
                s -= pq.poll();
            }
        }
        return pq.size();
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
class Solution {
public:
    int scheduleCourse(vector<vector<int>>& courses) {
        sort(courses.begin(), courses.end(), [](const vector<int>& a, const vector<int>& b) {
            return a[1] < b[1];
        });
        priority_queue<int> pq;
        int s = 0;
        for (auto& e : courses) {
            int duration = e[0], last = e[1];
            pq.push(duration);
            s += duration;
            while (s > last) {
                s -= pq.top();
                pq.pop();
            }
        }
        return pq.size();
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
func scheduleCourse(courses [][]int) int {
    sort.Slice(courses, func(i, j int) bool { return courses[i][1] < courses[j][1] })
    pq := &hp{}
    s := 0
    for _, e := range courses {
        duration, last := e[0], e[1]
        s += duration
        pq.push(duration)
        for s > last {
            s -= pq.pop()
        }
    }
    return pq.Len()
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
func (h *hp) Push(v any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
    a := h.IntSlice
    v := a[len(a)-1]
    h.IntSlice = a[:len(a)-1]
    return v
}
func (h *hp) push(v int) { heap.Push(h, v) }
func (h *hp) pop() int   { return heap.Pop(h).(int) }

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
function scheduleCourse(courses: number[][]): number {
    courses.sort((a, b) => a[1] - b[1]);
    const pq = new MaxPriorityQueue<number>();
    let s = 0;
    for (const [duration, last] of courses) {
        pq.enqueue(duration);
        s += duration;
        while (s > last) {
            s -= pq.dequeue();
        }
    }
    return pq.size();
}

Comments