Parallel Courses III

Hard

Amazon

1 view

Topics:

GraphsDynamic Programming

You are given an integer n, which indicates that there are n courses labeled from 1 to n. You are also given a 2D integer array relations where relations[j] = [prevCoursej, nextCoursej] denotes that course prevCoursej has to be completed before course nextCoursej (prerequisite relationship). Furthermore, you are given a 0-indexed integer array time where time[i] denotes how many months it takes to complete the (i+1)th course.

You must find the minimum number of months needed to complete all the courses following these rules:

You may start taking a course at any time if the prerequisites are met.
Any number of courses can be taken at the same time.

Return the minimum number of months needed to complete all the courses.

Note: The test cases are generated such that it is possible to complete every course (i.e., the graph is a directed acyclic graph).

For example:

If n = 3, relations = [[1,3],[2,3]], and time = [3,2,5], the output should be 8. Course 1 takes 3 months, course 2 takes 2 months, and course 3 takes 5 months. Courses 1 and 2 can be started simultaneously. Course 3 can only be started after courses 1 and 2 are complete, which happens at month 3 (due to course 1). Thus the total time is 3 + 5 = 8 months.

As another example, If n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], and time = [1,2,3,4,5], the output should be 12. Courses 1, 2, and 3 can be started at month 0 and completed in 1, 2, and 3 months, respectively. Course 4 can only be started after course 3, so it completes at month 7 (3+4). Course 5 can only be started after courses 1, 2, 3, and 4 are complete, so it starts at month 7 and completes at month 12 (7+5).

Parallel Courses III

Hard

Amazon

1 view

Topics:

GraphsDynamic Programming

You are given an integer n, which indicates that there are n courses labeled from 1 to n. You are also given a 2D integer array relations where relations[j] = [prevCoursej, nextCoursej] denotes that course prevCoursej has to be completed before course nextCoursej (prerequisite relationship). Furthermore, you are given a 0-indexed integer array time where time[i] denotes how many months it takes to complete the (i+1)th course.

You must find the minimum number of months needed to complete all the courses following these rules:

You may start taking a course at any time if the prerequisites are met.
Any number of courses can be taken at the same time.

Return the minimum number of months needed to complete all the courses.

Note: The test cases are generated such that it is possible to complete every course (i.e., the graph is a directed acyclic graph).

For example:

If n = 3, relations = [[1,3],[2,3]], and time = [3,2,5], the output should be 8. Course 1 takes 3 months, course 2 takes 2 months, and course 3 takes 5 months. Courses 1 and 2 can be started simultaneously. Course 3 can only be started after courses 1 and 2 are complete, which happens at month 3 (due to course 1). Thus the total time is 3 + 5 = 8 months.

As another example, If n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], and time = [1,2,3,4,5], the output should be 12. Courses 1, 2, and 3 can be started at month 0 and completed in 1, 2, and 3 months, respectively. Course 4 can only be started after course 3, so it completes at month 7 (3+4). Course 5 can only be started after courses 1, 2, 3, and 4 are complete, so it starts at month 7 and completes at month 12 (7+5).

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum number of courses (n) and what is the range for the time it takes to complete each course?
Is it possible to have cycles in the prerequisites (i.e., is it guaranteed to be a DAG), and if so, what should I return if a cycle exists? Should I throw an error or return a specific value such as -1?
If there are multiple courses that can be taken at the same time, can I assume that there are enough 'slots' available to take them simultaneously?
Is the `prerequisites` list guaranteed to only contain valid course numbers (i.e., indices within the range [1, n])?
If it's impossible to complete all courses, what should I return? Is it valid to return -1 in this case, or is there another specific value I should return?

Brute Force Solution

Approach

The brute force approach for this problem involves exploring every possible order of taking the courses. We will try every combination of courses we can take at each time step and find the order that minimizes the total time taken.

Here's how the algorithm would work step-by-step:

Start with all possible courses we can take first (courses with no prerequisites).
For each of those starting courses, determine the time it takes to complete that course.
After taking that course, determine which new courses are now available (courses whose prerequisites have been met).
For each of these new possible courses, determine the time to complete them, adding this time to the time we already calculated.
Continue this process, exploring all possible sequences of courses, until all courses are taken.
Keep track of the total time for each possible sequence of courses.
Compare the total times for all the sequences and choose the smallest one. This smallest time is the minimum time needed to complete all courses.

Code Implementation

def parallel_courses_iii_brute_force(number_of_courses, prerequisites, time_to_complete):
    minimum_time = float('inf')

    def solve(current_course_taken, time_elapsed, completed_courses):
        nonlocal minimum_time

        # If all courses are completed, update the minimum time.
        if len(completed_courses) == number_of_courses:
            minimum_time = min(minimum_time, time_elapsed)
            return

        available_courses = []
        for course_index in range(number_of_courses):
            if course_index not in completed_courses:
                prerequisites_met = True
                for prerequisite_course, course in prerequisites:
                    if course - 1 == course_index and prerequisite_course - 1 not in completed_courses:
                        prerequisites_met = False
                        break
                if prerequisites_met:
                    available_courses.append(course_index)

        # If there are no available courses, return.
        if not available_courses:
            return

        # Iterate through all available courses and recursively solve for the next course.
        for course_to_take in available_courses:
            solve(
                course_to_take,
                time_elapsed + time_to_complete[course_to_take],
                completed_courses | {course_to_take},
            )

    # Determine the initial available courses.
    initial_available_courses = []
    for course_index in range(number_of_courses):
        is_available = True
        for prerequisite_course, course in prerequisites:
            if course - 1 == course_index:
                is_available = False
                break
        if is_available:
            initial_available_courses.append(course_index)

    # Start the recursive solving process for each initial available course.
    for starting_course in initial_available_courses:
        solve(
            starting_course,
            time_to_complete[starting_course],
            {starting_course},
        )

    # Handle the case where the graph has cycles and no courses can be taken.
    if minimum_time == float('inf'):
        return -1

    return minimum_time

Big(O) Analysis

Time Complexity

O(n!) – The described brute force approach explores every possible order of taking the courses. In the worst-case scenario, where there are few or no prerequisites, the algorithm essentially generates all permutations of the 'n' courses. Generating all permutations of 'n' items takes O(n!) time because there are n! possible arrangements. For each permutation, we calculate the total time, which takes O(n) at most. Therefore the total time complexity is O(n * n!), dominated by the permutation generation. Thus, the overall time complexity is O(n!).

Space Complexity

O(N!) – The brute force approach explores all possible orderings of courses, which can lead to a significant amount of space usage. Specifically, at each step, we are effectively creating a new branch in our exploration for each possible course we can take, leading to a tree-like structure. The maximum space used will be proportional to the number of possible orderings of courses, which is on the order of N! where N is the number of courses. This space is used to store the call stack for each branch of exploration and temporary storage related to the current course sequence being evaluated.

Optimal Solution

Approach

The core idea is to figure out the earliest time we can finish each course, considering its prerequisites and duration. We use a strategy similar to planning a project where some tasks must be done before others, focusing on minimizing the overall completion time.

Here's how the algorithm would work step-by-step:

Think of each course as a task, and the course dependencies as a chain of tasks that need to be completed in a specific order.
Calculate how many courses depend on each course - in other words, how many courses list it as a prerequisite. This tells us which courses can start immediately since they don't have any prerequisites.
Start with courses that have no prerequisites. The earliest time you can finish them is simply their duration.
For courses that *do* have prerequisites, find the course with the longest completion time among all its prerequisites. The earliest you can start such a course is the completion time of that longest prerequisite.
Add the duration of the course itself to this start time to determine its earliest completion time.
Keep doing this for all the courses, updating the earliest completion times as you go. If a course has multiple prerequisites, you need to consider all of them to determine the latest completion time among them before starting the course.
The answer is the largest (maximum) of all earliest completion times calculated because we want to know the earliest time we can finish *all* courses.

Code Implementation

def parallel_courses_iii(number_of_courses, relations, duration):
    incoming_edges = [0] * number_of_courses
    outgoing_courses = [[] for _ in range(number_of_courses)]
    completion_time = [0] * number_of_courses

    for previous_course, next_course in relations:
        outgoing_courses[previous_course - 1].append(next_course - 1)
        incoming_edges[next_course - 1] += 1

    # Courses with no incoming edges can start immediately.
    queue = []
    for course in range(number_of_courses):
        if incoming_edges[course] == 0:
            queue.append(course)
            completion_time[course] = duration[course]

    max_completion_time = 0
    while queue:
        current_course = queue.pop(0)
        max_completion_time = max(max_completion_time, completion_time[current_course])

        # Iterate through the courses dependent on this course.
        for next_course in outgoing_courses[current_course]:
            incoming_edges[next_course] -= 1

            # The course can start once its prerequisites are met.
            completion_time[next_course] = max(completion_time[next_course],
                                                completion_time[current_course] + duration[next_course])
            if incoming_edges[next_course] == 0:
                queue.append(next_course)

    return max_completion_time

Big(O) Analysis

Time Complexity

O(n + m) – The algorithm first calculates the in-degrees of each course, which takes O(n) time where n is the number of courses. Then, it uses a topological sort-like approach. We iterate through each course and its prerequisites, updating completion times. The loop iterates through all n courses, and for each course, it iterates through its dependencies. The total number of dependency relationships is represented by 'm', the number of edges. Therefore, the dependency update part takes O(m) time. Finally, finding the maximum completion time across all courses takes O(n) time, so O(n + m + n) which simplifies to O(n + m).

Space Complexity

O(N) – The algorithm uses auxiliary space for data structures like an indegree array (to store the number of prerequisites for each course), a queue (to hold courses with no prerequisites), and an array to store the earliest finish time for each course. All these data structures store information for each of the N courses. Therefore, the space complexity is O(N), where N is the number of courses.

Edge Cases

Case	How to Handle
Empty input (n = 0, courses = [], prereqs = [])	Return 0 immediately as there are no courses to take.
Single course (n = 1, courses = [1], prereqs = [])	Return the duration of the single course courses[0].
All courses have zero duration (courses = [0, 0, 0, ...])	The algorithm should still correctly calculate the minimum time based on dependencies, resulting in 0 if there are no cycles.
Cyclic dependencies exist in prereqs	Detect cycles using topological sort or DFS and return -1 or an appropriate error value, as a valid schedule is impossible.
Large number of courses (n approaching maximum allowed value)	Ensure the graph representation and topological sort algorithm scale efficiently (e.g., using adjacency lists instead of matrices).
All courses are independent (prereqs = [])	Return the maximum value in the duration array.
Course durations are very large, leading to potential integer overflow in intermediate calculations	Use long data types for storing accumulated times to prevent overflow.
Course durations are negative, representing some impossible scenario.	Return an error value (e.g., -1) or throw an exception since durations cannot be negative.