Course Schedule III

                                            
const scheduleCourse = (courses) =>{

    courses.sort((a,b)=>a[1]-b[1])
    const maxHeap = new MaxPriorityQueue(); 
    let maxTime = 0;

    for(const [time, endTime] of courses){
        if(time + maxTime <= endTime){
            maxHeap.enqueue(time);
            maxTime += time;
        }else if(maxHeap.front() && maxHeap.front().element > time){
            maxTime += time - maxHeap.dequeue().element
            maxHeap.enqueue(time);
        }
    }

    return maxHeap.size(); 
};

To solve this coding challenge, you can use a greedy algorithm approach combined with the use of a max heap (priority queue). The goal is to maximize the number of courses one can take given the constraints of course duration and latest finish day. Here's how the solution works and the corresponding pseudo code:

Solution Explanation:

1. Sort Courses:

First, the courses are sorted based on their lastDay. This is done so that courses with earlier end dates are considered first, aligning with the strategy of minimizing the risk of exceeding course deadlines.

2. Use Max Heap (Priority Queue):

A max heap is used to keep track of the courses taken. In this context, the heap stores the durations of the courses. The max heap is helpful because it allows easy access to the course with the longest duration taken so far, which is the first candidate to be dropped if necessary.

3. Iterate and Decide:

Iterate through each course. For each course, there are two main decisions:

If the course can be taken within its deadline considering the total duration of courses taken so far (maxTime), add it to the heap and update maxTime.
If the course cannot be taken by its deadline, check if there's a course that has been taken with a longer duration than the current one (the top of the max heap). If so, replace that course with the current one by removing the longest course from the heap and adding the current course. This is done because replacing a longer course with a shorter one may allow for more courses to be taken.

4. Return Result:

The size of the heap at the end of the iteration represents the maximum number of courses that can be taken, as it contains all the courses that have been successfully scheduled.

Pseudo Code:

                                        
function scheduleCourse(courses):
    # Sort courses by their end time
    sort courses by lastDay

    # Initialize a max heap and a variable for tracking total time
    maxHeap = new MaxHeap()
    maxTime = 0

    # Iterate through each course
    for each course in courses:
        time, endTime = course

        # If course can be taken within its deadline
        if time + maxTime <= endTime:
            add time to maxHeap
            maxTime += time
        # If not within deadline, but a longer course exists in the heap
        else if maxHeap is not empty and maxHeap.top() > time:
            maxTime += time - maxHeap.pop()
            add time to maxHeap

    # Return the number of courses in the heap
    return size of maxHeap

This approach ensures that you always prioritize taking courses that have the earliest deadlines, while also being flexible enough to replace longer-duration courses with shorter ones if it allows for a more optimal schedule.