# Course Schedule III LeetCode Solution

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.

# List of all 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 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

## 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();
}
};```Code language: PHP (php)```

## 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;
} else if (pq.size() > 0 && pq.peek() > dur) {
total += dur - pq.poll();
}
}
return pq.size();
}
}```Code language: PHP (php)```

## 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()
};```Code language: JavaScript (javascript)```

## 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)```Code language: HTML, XML (xml)```
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