g0601_0700.s0630_course_schedule_iii.Solution Maven / Gradle / Ivy
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Java-based LeetCode algorithm problem solutions, regularly updated
package g0601_0700.s0630_course_schedule_iii;
// #Hard #Array #Greedy #Heap_Priority_Queue
import java.util.Arrays;
import java.util.PriorityQueue;
@SuppressWarnings("java:S135")
public class Solution {
public int scheduleCourse(int[][] courses) {
// Sort the courses based on their deadline date.
Arrays.sort(courses, (a, b) -> 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 int pair.
PriorityQueue pq = new PriorityQueue<>((a, b) -> b - a);
// Total time consumed.
int 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 (int[] course : 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.isEmpty() && 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();
}
}