Course Schedule

Clarify the problem:

• The problem requires us to determine if it is possible to finish all courses given a list of prerequisites.
• We need to check if there is a valid course schedule that satisfies all the prerequisites.
• A course schedule is valid if we can complete all the courses in the given list of prerequisites without violating any dependencies.
• It is assumed that there are no cyclic dependencies in the course prerequisites.

Analyze the problem:

• Input: An integer numCourses representing the total number of courses, and a list of tuples prerequisites representing the prerequisites for each course.
• Output: A boolean indicating whether it is possible to finish all courses.
• Constraints:
  • 2 <= numCourses <= 10^5
  • 0 <= prerequisites.length <= 5000
  • prerequisites[i].length == 2
  • 0 <= prerequisites[i][0], prerequisites[i][1] < numCourses
  • The input prerequisites list does not contain any duplicate pairs.

Design an algorithm:

• We can solve this problem using a graph-based approach.
• First, we need to create a directed graph representation of the courses and their prerequisites.
• We can use a dictionary to represent the graph, where the keys are the course numbers, and the values are lists of course numbers representing the prerequisites.
• After constructing the graph, we can perform a depth-first search (DFS) traversal on the graph to check for any cycles.
• If we encounter a cycle during the DFS traversal, it means there is a circular dependency among the courses, and it is not possible to finish all courses.
• If no cycle is found, it means it is possible to finish all courses.

Explain your approach:

• We will implement a class called CourseSchedule to encapsulate the solution.
• The class will have the following methods:
  • buildGraph(numCourses, prerequisites): Constructs the directed graph representation of the courses and prerequisites.
  • hasCycle(course): Performs a depth-first search (DFS) traversal on the graph to check for cycles, starting from the given course.
  • canFinish(numCourses, prerequisites): Checks if it is possible to finish all courses given the total number of courses and the prerequisites.

Write clean and readable code:

python

class CourseSchedule:
    def __init__(self):
        self.graph = {}

    def buildGraph(self, numCourses, prerequisites):
        self.graph = {course: [] for course in range(numCourses)}
        for course, prereq in prerequisites:
            self.graph[course].append(prereq)

    def hasCycle(self, course):
        if course in self.visiting:
            return True
        if course in self.visited:
            return False

        self.visiting.add(course)
        for prereq in self.graph[course]:
            if self.hasCycle(prereq):
                return True
        self.visiting.remove(course)
        self.visited.add(course)
        return False

    def canFinish(self, numCourses, prerequisites):
        self.buildGraph(numCourses, prerequisites)
        self.visiting = set()
        self.visited = set()

        for course in range(numCourses):
            if self.hasCycle(course):
                return False

        return True

Test your code:

python

schedule = CourseSchedule()
print(schedule.canFinish(2, [[1, 0]]))             # Output: True
print(schedule.canFinish(2, [[1, 0], [0, 1]]))     # Output: False
print(schedule.canFinish(3, [[0, 1], [0, 2]]))     # Output: True
print(schedule.canFinish(3, [[0, 1], [1, 2], [2, 0]]))  # Output: False

Explanation:

• We test the solution with different scenarios, including cases with valid course schedules and cases with cyclic dependencies.
• In the first test case, there is only one prerequisite relationship (1 -> 0), and it is possible to finish all courses.
• In the second test case, there is a cyclic dependency between courses 0 and 1, making it impossible to finish all courses.
• In the third test case, there is a valid course schedule without any cyclic dependencies.
• In the fourth test case, there is a cyclic dependency among courses 0, 1, and 2, making it impossible to finish all courses.

Optimize if necessary:

• The solution utilizes a graph-based approach with a depth-first search (DFS) traversal to detect cycles.
• It is a commonly used and efficient approach to solve this type of problem, and further optimization is not necessary.

Handle error cases:

• The code assumes that the input follows the problem constraints. It does not explicitly handle invalid input cases.

Discuss complexity analysis:

• Let N be the total number of courses, and E be the total number of prerequisites.
• Building the graph takes O(E) time since we iterate through all the prerequisites.
• The DFS traversal to check for cycles has a time complexity of O(N + E) in the worst case, as we visit each course and each prerequisite once.
• Therefore, the overall time complexity of the canFinish function is O(N + E).
• The space complexity is O(N + E) as well since we use additional sets and dictionaries to store the graph and visited nodes.