LeetCode in Kotlin

630. Course Schedule III

Hard

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.

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 1^st course, it costs 100 days so you will finish it on the 100^th day, and ready to take the next course on the 101^st day. Second, take the 3^rd course, it costs 1000 days so you will finish it on the 1100^th day, and ready to take the next course on the 1101^st day. Third, take the 2^nd course, it costs 200 days so you will finish it on the 1300^th day. The 4^th course cannot be taken now, since you will finish it on the 3300^th day, which exceeds the closed date.

Example 2:

Input: courses = [[1,2]]

Output: 1

Example 3:

Input: courses = [[3,2],[4,3]]

Output: 0

Constraints:

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

Solution

import java.util.PriorityQueue

class Solution {
    fun scheduleCourse(courses: Array<IntArray>): Int {
        // Sort the courses based on their deadline date.
        courses.sortWith { a: IntArray, b: IntArray ->
            a[1] - b[1]
        }
        // Only the duration is stored. We don't care which course
        // is the longest, we only care about the total courses can
        // be taken.
        // If the question wants the course ids to be returned.
        // Consider use a Pair<Duration, CourseId> int pair.
        val pq = PriorityQueue { a: Int, b: Int -> b - a }
        // Total time consumed.
        var time = 0
        // At the given time `course`, the overall "time limit" is
        // course[1]. All courses in pq is already 'valid'. But
        // adding this course[0] might exceed the course[1] limit.
        for (course in courses) {
            // If adding this course doesn't exceed. Let's add it
            // for now. (Greedy algo). We might remove it later if
            // we have a "better" solution at that time.
            if (time + course[0] <= course[1]) {
                time += course[0]
                pq.offer(course[0])
            } else {
                // If adding this ecxeeds the limit. We can still add it
                // if-and-only-if there are courses longer than current
                // one. If so, by removing a longer course, current shorter
                // course can fit in for sure. Although the total course
                // count is the same, the overall time consumed is shorter.
                // Which gives us more room for future courses.
                // Remove any course that is longer than current course
                // will work, but we remove the longest one with the help
                // of heap (pq).
                if (pq.isNotEmpty() && pq.peek() > course[0]) {
                    time -= pq.poll()
                    time += course[0]
                    pq.offer(course[0])
                }
                // If no course in consider (pq) is shorter than the
                // current course. It is safe to discard it.
            }
        }
        return pq.size
    }
}

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