## Algorithm

Problem Name: 630. Course Schedule III

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 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
```

Constraints:

• `1 <= courses.length <= 104`
• `1 <= durationi, lastDayi <= 104`

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int scheduleCourse(int[][] courses) {
Arrays.sort(courses, (a, b) -> a[1] - b[1]);
PriorityQueue queue = new PriorityQueue<>((a, b) -> b - a);
int time = 0;
for (int[] course : courses) {
if (time + course[0] <= course[1]) {
time += course[0];
} else if (!queue.isEmpty() && queue.peek() > course[0]) {
time += course[0] - queue.poll();
}
}
return queue.size();
}
}
``````
Copy The Code &

Input

cmd
courses = [[1,2]]

Output

cmd
1

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const scheduleCourse = function (courses) {
const queue = new MaxPriorityQueue({
priority: e => e[0]
})
courses.sort((a, b) => a[1] - b[1])
let time = 0
for(let e of courses) {
time += e[0]
queue.enqueue(e)
if(time > e[1]) {
const tmp = queue.dequeue().element
time -= tmp[0]
}
}
return queue.size()
}
``````
Copy The Code &

Input

cmd
courses = [[1,2]]

Output

cmd
1

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def scheduleCourse(self, courses):
pq = []
start = 0
for t, end in sorted(courses, key = lambda x: x[1]):
start += t
heapq.heappush(pq, -t)
while start > end:
start += heapq.heappop(pq)
return len(pq)
``````
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Input

cmd
courses = [[3,2],[4,3]]

### #4 Code Example with C Programming

```Code - C Programming```

``````
typedef struct {
int *p;
int n;
} heap_t;
int cmp(const void *a, const void *b) {
int x = *(int *)a, y = *(int *)b;
return x < y ? -1 :
x > y ?  1 : 0;
}
int cmp2(const void *a, const void *b) {
int x = (*(int **)a)[1], *y = (*(int **)b)[1];
return x < y ? -1 :
x > y ?  1 : 0;
}
void heap_push(heap_t *heap, int k) {
heap->p[heap->n ++] = k;
qsort(heap->p, heap->n, sizeof(int), cmp);  // optimize here to avoid TLE!!!
}
int heap_pop(heap_t *heap) {
int k = heap->p[-- heap->n];
return k;
}
int scheduleCourse(int** courses, int coursesRowSize, int coursesColSize) {
heap_t heap = { 0 };
int i, days, end, t;

qsort(courses, coursesRowSize, sizeof(int *), cmp2);

heap.p = malloc(coursesRowSize * sizeof(int));
//assert(heap.p);

t = 0;
for (i = 0; i < coursesRowSize; i ++) {
days = courses[i][0];
end  = courses[i][1];
heap_push(&heap, days);
t += days;
if (t > end) {
t -= heap_pop(&heap);
}
}

free(heap.p);

return heap.n;
}
``````
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Input

cmd
courses = [[3,2],[4,3]]