Last updated on October 9th, 2024 at 06:13 pm

Here, We see Course Schedule III LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Course Schedule III LeetCode Solution

Problem Statement

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 1st course, it costs 100 days so you will finish it on the 100th day, and be 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

1. Course Schedule III LeetCode Solution C++

class Solution {
public:
    int scheduleCourse(vector<vector<int>>& courses) {
        sort(courses.begin(), courses.end(), [](vector<int> a, vector<int> b){return a[1] < b[1];});
        priority_queue<int> heap;
        int now = 0;
        for (int i = 0; i < courses.size(); ++ i)
        {
            heap.push(courses[i][0]);
            now += courses[i][0];
            if (now > courses[i][1])
                now -= heap.top(), heap.pop();
        }
        return heap.size();
    }
};

2. Course Schedule III LeetCode Solution Java

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 total = 0;
        for (int[] course : courses) {
            int dur = course[0], end = course[1];
            if (dur + total <= end) {
                total += dur;
                pq.add(dur);
            } else if (pq.size() > 0 && pq.peek() > dur) {
                total += dur - pq.poll();
                pq.add(dur);
            }
        }
        return pq.size();
    }
}

3. Course Schedule III LeetCode Solution JavaScript

var scheduleCourse = function(courses) {
    courses.sort((a,b) => a[1] - b[1])
    let total = 0, pq = new MaxPriorityQueue({priority: x => x})
    for (let [dur, end] of courses)
        if (dur + total <= end)
            total += dur, pq.enqueue(dur)
        else if (pq.front() && pq.front().element > dur)
            total += dur - pq.dequeue().element, pq.enqueue(dur)
    return pq.size()
};

4. Course Schedule III LeetCode Solution Python

class Solution(object):
    def scheduleCourse(self, courses):
        heap, total = [], 0
        for dur, end in sorted(courses, key=lambda el: el[1]):
            if dur + total <= end:
                total += dur
                heappush(heap, -dur)
            elif heap and -heap[0] > dur:
                total += dur + heappop(heap)
                heappush(heap, -dur)
        return len(heap)