Parallel Courses III

Hard

Two Sigma

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] = [prevCourse_j, nextCourse_j] denotes that course prevCourse_j has to be completed before course nextCourse_j (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).

Example 1:

Input: n = 3, relations = [[1,3],[2,3]], time = [3,2,5]
Output: 8
Explanation: The figure above represents the given graph and the time required to complete each course. 
We start course 1 and course 2 simultaneously at month 0.
Course 1 takes 3 months and course 2 takes 2 months to complete respectively.
Thus, the earliest time we can start course 3 is at month 3, and the total time required is 3 + 5 = 8 months.

Example 2:

Input: n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], time = [1,2,3,4,5]
Output: 12
Explanation: The figure above represents the given graph and the time required to complete each course.
You can start courses 1, 2, and 3 at month 0.
You can complete them after 1, 2, and 3 months respectively.
Course 4 can be taken only after course 3 is completed, i.e., after 3 months. It is completed after 3 + 4 = 7 months.
Course 5 can be taken only after courses 1, 2, 3, and 4 have been completed, i.e., after max(1,2,3,7) = 7 months.
Thus, the minimum time needed to complete all the courses is 7 + 5 = 12 months.

Constraints:

1 <= n <= 5 * 10⁴
0 <= relations.length <= min(n * (n - 1) / 2, 5 * 10⁴)
relations[j].length == 2
1 <= prevCourse_j, nextCourse_j <= n
prevCourse_j != nextCourse_j
All the pairs [prevCourse_j, nextCourse_j] are unique.
time.length == n
1 <= time[i] <= 10⁴
The given graph is a directed acyclic graph.

Parallel Courses III

Hard

Two Sigma

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] = [prevCourse_j, nextCourse_j] denotes that course prevCourse_j has to be completed before course nextCourse_j (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).

Example 1:

Input: n = 3, relations = [[1,3],[2,3]], time = [3,2,5]
Output: 8
Explanation: The figure above represents the given graph and the time required to complete each course. 
We start course 1 and course 2 simultaneously at month 0.
Course 1 takes 3 months and course 2 takes 2 months to complete respectively.
Thus, the earliest time we can start course 3 is at month 3, and the total time required is 3 + 5 = 8 months.

Example 2:

Input: n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], time = [1,2,3,4,5]
Output: 12
Explanation: The figure above represents the given graph and the time required to complete each course.
You can start courses 1, 2, and 3 at month 0.
You can complete them after 1, 2, and 3 months respectively.
Course 4 can be taken only after course 3 is completed, i.e., after 3 months. It is completed after 3 + 4 = 7 months.
Course 5 can be taken only after courses 1, 2, 3, and 4 have been completed, i.e., after max(1,2,3,7) = 7 months.
Thus, the minimum time needed to complete all the courses is 7 + 5 = 12 months.

Constraints:

1 <= n <= 5 * 10⁴
0 <= relations.length <= min(n * (n - 1) / 2, 5 * 10⁴)
relations[j].length == 2
1 <= prevCourse_j, nextCourse_j <= n
prevCourse_j != nextCourse_j
All the pairs [prevCourse_j, nextCourse_j] are unique.
time.length == n
1 <= time[i] <= 10⁴
The given graph is a directed acyclic graph.

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 are the constraints on the number of courses `n` and the values in the `time` array? Specifically, what are the maximum values for `n` and `time[i]`?
Is it possible for the prerequisite relationships to contain cycles, and if so, what should be the expected behavior (e.g., return an error, or is it guaranteed there are no cycles)?
If there are no prerequisite relationships (i.e., the input list of prerequisites is empty), what should the output be?
Is it guaranteed that each course number in the prerequisite relationships will be within the valid range of 1 to n?
If it's impossible to complete all courses (e.g., due to a cycle, though you mentioned that may not happen) what value should I return?

Brute Force Solution

Approach

The brute force strategy for this problem is like exploring every possible schedule of courses. We try out every combination of taking courses one after another, respecting their prerequisites, until we find a valid schedule. Then, we pick the schedule that finishes the fastest.

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

Start by figuring out which courses we can take first, because they don't depend on any other courses.
For each of these starting courses, imagine taking it, and add its cost to a running total.
After taking a course, see which new courses become available because their prerequisites are now met.
For each of these newly available courses, imagine taking it next and adding its cost to the running total.
Keep doing this, exploring every possible sequence of courses until we've taken all of them.
Every time we finish taking all the courses, we record the total cost (time) it took.
After exploring all possible sequences, we compare all the recorded costs and pick the smallest one. This smallest cost represents the fastest time to complete all courses.

Code Implementation

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

    def explore_courses(current_schedule, current_time):
        nonlocal minimum_time

        # If we've taken all courses, update the minimum time
        if len(current_schedule) == number_of_courses:
            minimum_time = min(minimum_time, current_time)
            return

        available_courses = []
        for course in range(1, number_of_courses + 1):
            if course not in current_schedule:
                ready_to_take = True
                for prerequisite_course, dependent_course in prerequisites:
                    if dependent_course == course and prerequisite_course not in current_schedule:
                        ready_to_take = False
                        break
                if ready_to_take:
                    available_courses.append(course)

        # If there are no courses available, we've hit a dead end
        if not available_courses:
            return

        for course_to_take in available_courses:

            # Create new schedule by adding the course
            new_schedule = current_schedule + [course_to_take]

            course_index = course_to_take - 1
            new_time = current_time + course_times[course_index]

            # Recursively explore the next possible courses
            explore_courses(new_schedule, new_time)

    # Find initial courses with no prerequisites
    initial_courses = []
    for course in range(1, number_of_courses + 1):
        is_initial = True
        for prerequisite_course, dependent_course in prerequisites:
            if dependent_course == course:
                is_initial = False
                break
        if is_initial:
            initial_courses.append(course)

    # If there are no prereqs, add all courses as initial
    if not prerequisites:
        initial_courses = list(range(1, number_of_courses + 1))

    # Explore possible schedules from each initial course
    for initial_course in initial_courses:
        course_index = initial_course - 1
        explore_courses([initial_course], course_times[course_index])

    return minimum_time

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(K^n) – The brute force approach explores all possible course schedules. In the worst case, we might have to consider every possible permutation of courses that satisfies the prerequisites. If K represents the maximum number of courses available to take at any given step, and n is the total number of courses, the algorithm branches out K ways at each course selection stage, creating a decision tree. The depth of this tree is 'n', and at each level, we explore at most 'K' branches. Therefore, the time complexity is exponential, approximately O(K^n), where K is capped by n, but it reflects the exponential nature of exploring permutations.

Space Complexity

O(N!) – The brute force approach explores every possible sequence of courses. This involves creating multiple copies of the course schedule as we explore different paths. In the worst-case scenario, where there are no prerequisites, we may need to store all possible permutations of the courses in the recursion stack, leading to a factorial relationship with the number of courses, N. Specifically, the number of possible schedules can approach N!, as each schedule is stored in the call stack. Therefore, the space complexity can be estimated as O(N!).

Optimal Solution

Approach

The problem asks us to find the minimum time to finish all courses given dependencies and time for each course. We can think of this as a graph problem and use a smart algorithm to traverse the graph and find the longest path, which represents the minimum time to complete all courses.

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

First, we figure out which courses depend on others. We can represent this dependency relationship as arrows pointing from prerequisite courses to the courses that require them.
Then, we need to figure out which courses we can start with right away. These are the courses that don't depend on any other course.
For each course, we calculate the earliest time we can finish it. This will depend on the time it takes to finish the course itself, plus the maximum time it takes to finish any of its prerequisite courses.
We start by considering the courses we can start right away and updating their finish times based on their own duration.
Next, we go through all the other courses. For each course, we look at its prerequisites and determine the latest finish time among them. We then add the time it takes to complete the course itself to this latest finish time. This gives us the earliest possible finish time for this course.
We keep repeating the previous step for all courses until we've updated the finish times for all of them.
Finally, the maximum finish time among all courses will be the minimum time required to complete all courses because we are taking the longest path required to finish all courses and their dependencies.

Code Implementation

def parallel_courses_three(number_of_courses, prerequisites, time):
    indegree = [0] * number_of_courses
    adjacency_list = [[] for _ in range(number_of_courses)]
    finish_time = [0] * number_of_courses

    for prerequisite, course in prerequisites:
        adjacency_list[prerequisite - 1].append(course - 1)
        indegree[course - 1] += 1

    # Initialize queue with courses that have no prerequisites.
    queue = []
    for course in range(number_of_courses):
        if indegree[course] == 0:
            queue.append(course)
            finish_time[course] = time[course]

    courses_completed = 0

    # Topological sort to determine minimum time to complete all courses.
    while queue:
        current_course = queue.pop(0)
        courses_completed += 1

        # Iterate through dependent courses.
        for next_course in adjacency_list[current_course]:

            # Determine earliest finish time based on prerequisites.
            finish_time[next_course] = max(finish_time[next_course],
                                          finish_time[current_course] + time[next_course])
            indegree[next_course] -= 1
            # Add to queue if all prerequisites are completed.
            if indegree[next_course] == 0:
                queue.append(next_course)

    #Check if a cycle exists
    if courses_completed != number_of_courses:
        return -1

    # The result is the maximum finish time among all courses.
    return max(finish_time)

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n + d) – The algorithm iterates through each course (n courses) to calculate finish times. The dependency relationships are represented as a graph where 'd' is the number of dependencies. We update the finish time of each course by considering the maximum finish time of its prerequisites. The number of operations related to dependencies directly depends on 'd', the number of dependencies. Therefore, the overall time complexity is proportional to the sum of the number of courses and number of dependencies, simplifying to O(n + d).

Space Complexity

O(N) – The space complexity stems primarily from the need to store dependencies and finish times for each course. We represent dependencies using an adjacency list (or similar structure) which, in the worst case, can store edges proportional to the number of courses (N), consuming O(N) space. Additionally, we keep track of the earliest finish time for each course in an array of size N, contributing another O(N) to the overall auxiliary space. Therefore, the space complexity is O(N + N), which simplifies to O(N).

Edge Cases

Case	How to Handle
Empty prerequisites list	The algorithm should correctly handle an empty prerequisites list, effectively treating each course as independent and starting them all immediately, then finding the maximum time among the courses.
No courses (n=0)	Return 0 since there are no courses to take and no time is required.
Single course (n=1)	Return the time for that single course, time[0].
Circular dependencies (a cycle in the prerequisite graph)	The algorithm must detect cycles and return an appropriate error value or signal (e.g., -1) since the courses cannot be completed.
Large number of courses (n is large, approaching the limits) with complex prerequisite relationships.	Ensure the algorithm's time and space complexity are considered; a topological sort-based solution or dynamic programming should scale adequately to the problem's constraints and avoid potential integer overflow.
Course times are all zero	If all course times are zero, the minimum time to finish all courses should be 0.
Courses with no prerequisites form a long chain and there is one single course requiring all of these.	The algorithm should correctly calculate the longest path based on time taken to complete courses.
Integer overflow with large time values or number of courses.	Use appropriate data types (e.g., long) to avoid potential integer overflows when calculating total time or dependencies.