# Course Schedule III

This page explains Java solution to problem `Course Schedule III` using `Priority Queue` data structure.

## Problem Statement

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 1:

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.

## Solution

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``````
package com.vc.hard;

import java.util.*;

class CourseScheduleIii {
public int scheduleCourse(int[][] courses) {
//Sort courses by, closed on day in ascending order
Arrays.sort(courses, new Comparator<int[]>(){
public int compare(int[] x, int[] y) {
return Integer.compare(x[1], y[1]);
}
});

//Priority Queue which returns course with greater duration first
PriorityQueue<Integer> durationPQ = new PriorityQueue<Integer>(new Comparator<Integer>(){
public int compare(Integer x, Integer y) {
return Integer.compare(y, x);
}
});

int time = 0;
for(int[] course: courses) {
int duration = course[0];
int endBefore = course[1];

if(time + duration <= endBefore) {
durationPQ.offer(duration);
time += duration;
}
else if(!durationPQ.isEmpty() && durationPQ.peek() > duration) {
time -= durationPQ.poll();
time += duration;
durationPQ.offer(duration);
}
}

return durationPQ.size();
}
}
``````

## Time Complexity

O(N * log N) Where
N is total number of course in an input array.

## Space Complexity

O(N) Where
N is total number of course in an input array.