How do you solve Course Schedule III on LeetCode?

Course Schedule III (LeetCode #630) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting courses by their last day allows us to make optimal decisions about which courses to take.

What is the time complexity of Course Schedule III?

The optimal solution for Course Schedule III runs in O(n log n) time and uses O(n) space. The sorting step takes O(n log n), and maintaining the heap takes O(log n) for each course, leading to an overall complexity of O(n log n).

What is the brute force solution for Course Schedule III?

The brute force approach for Course Schedule III is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible combination of courses to see which ones can be taken without exceeding their deadlines. This is straightforward but inefficient as it doesn't scale well with larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Course Schedule III?

Course Schedule III uses the following concepts: Array, Greedy, Sorting, Heap (Priority Queue). The recommended patterns to study are: Greedy Algorithms, Heap (Priority Queue), Sorting.

What are the interview tips for Course Schedule III?

Always think about how to optimize your solution after the brute-force approach. Be clear about the data structures you choose and why they are suitable for the problem. Practice explaining your thought process as you code, as this helps clarify your understanding.

What are common mistakes when solving Course Schedule III?

Not considering the order of courses based on their deadlines. Failing to use a data structure like a heap to manage durations efficiently.

#630

Course Schedule III

Hard

ArrayGreedySortingHeap (Priority Queue)Greedy AlgorithmsHeap (Priority Queue)Sorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses a greedy approach combined with a max heap (priority queue) to efficiently manage the courses taken. By sorting the courses by their last day and using a heap to track the longest durations, we can maximize the number of courses while respecting deadlines.

⚙️

Algorithm

5 steps

1Step 1: Sort the courses by their last day.
2Step 2: Initialize a max heap to keep track of the durations of the courses taken.
3Step 3: Iterate through the sorted courses, adding each course's duration to the heap if it can be completed by its last day.
4Step 4: If adding a course exceeds the current total duration, remove the longest course from the heap.
5Step 5: The size of the heap at the end gives the maximum number of courses that can be taken.

solution.py13 lines

1# Full working Python code
2import heapq
3
4def maxCourses(courses):
5    courses.sort(key=lambda x: x[1])
6    max_heap = []
7    total_duration = 0
8    for duration, last_day in courses:
9        total_duration += duration
10        heapq.heappush(max_heap, -duration)
11        if total_duration > last_day:
12            total_duration += heapq.heappop(max_heap)
13    return len(max_heap)

ℹ

Complexity note: The sorting step takes O(n log n), and maintaining the heap takes O(log n) for each course, leading to an overall complexity of O(n log n).

1Sorting courses by their last day allows us to make optimal decisions about which courses to take.
2Using a max heap helps efficiently manage the durations of the courses taken, allowing for quick adjustments.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.