Problem Description

In this problem, you're essentially a student planning out your semesters in advance to ensure you can take all the courses you need to graduate. You've got n different courses numbered from 1 to n, and a list of prerequisite relationships such that you have to complete some courses before starting others. These prerequisites are indicated by pairs [prevCourse, nextCourse] where prevCourse is a course you must take before nextCourse.

You have one more constraint: in any given semester, you can only take up to k courses. Given these rules, your challenge is to determine the minimum number of semesters required to complete all n courses. The problem guarantees that it's possible to take all courses by eventually meeting all prerequisites.

Intuition

The core of the solution lies in understanding that this is a scheduling problem that can be represented as a directed graph, where each node is a course, and the edges represent prerequisite relationships. To minimize the number of semesters, we want to take as many courses as we can each semester, taking care not to exceed the limit k and ensuring we've completed all prerequisites needed.

To approach this algorithmically, we represent our course schedule using bit manipulation, where a bit set to 1 indicates a course is completed. We initialize with 0, representing that no courses have been taken. Each bit position corresponds to a specific course, so an integer can represent any combination of completed courses.

We iterate over all courses, checking prerequisites using bitwise AND operations. If a course's prerequisites are satisfied, indicated by a 1 in all prerequisite course's positions in our bitmask, we can consider it for selection in the current semester.

The challenge involves choosing the right courses when facing the k course limit. We use depth-first search to traverse the different combinations of courses we can take each semester, branching out into different scenarios and making sure we don't revisit the same state (combination of courses) again.

The algorithm uses a queue to process these combinations in a breadth-first manner. For each state, we branch into new states that represent taking a different subset of k or fewer courses, marking states as visited to avoid redundant work. We repeat this until we reach a state where all courses are complete, and since we're using breadth-first search, the first time we reach this state will be the minimum number of semesters required.

The bit_count method (in Python, you might see this as bin(x).count('1') or a similar alternative) is used to quickly count the number of bits set to 1 to ensure we don't exceed k courses in a semester. When we have a selection of courses that meets the prerequisites and doesn't exceed the k limit, we add that state to our queue and continue the search until we've found the optimal number of semesters.

Learn more about Graph, Dynamic Programming and Bitmask patterns.

Solution Approach

The solution employs bit manipulation, breadth-first search, and a queue data structure to efficiently explore different possibilities for course schedules while tracking progress and prerequisites.

Here's a step-by-step breakdown of the implementation:

• Each combination of completed courses is represented as a bit sequence, with 1s indicating completed courses. A deque (double-ended queue) from Python's collections is initiated with the starting point (0, 0), which represents no courses taken (0) in 0 semesters.

• A vis (visited) set is used to record states we have already considered, preventing us from repeating the same combinations, and thus reducing unnecessary computations.

• The core of the solution is a loop that continues as long as there are states in our queue to process. On each iteration, we:

  • Pop the first state cur from the deque, which represents the current combination of courses taken up to the present, and t, the number of semesters taken so far.

  • If cur matches our goal state (1 << (n + 1)) - 2, which means all courses except the course numbered 0 (which doesn't exist) are completed, we return t as the minimum number of semesters needed.

  • Calculate which new courses (nxt) can be taken given the current state. This involve iterating over each course i and checking if its prerequisites (d[i]) are all in cur. If yes, this course can be added (nxt |= 1 << i).

  • Remove courses from nxt that are already in cur, leaving a bitmask of courses that are potential candidates for the current semester.

  • If the number of potential courses is less than or equal to k, we can take all of them in the current semester. The new state (nxt | cur) is added to the queue if it's not already visited.

  • If there are more than k courses available, we have to consider all subsets of nxt of size k and put each into the queue if not visited. This is done through a combination of bit manipulation steps that count the bits, check the subset size, and ensure all subsets are considered.

Overall, the solution uses a breadth-first search to traverse through a graph representation of course schedules. The graph is implicit, with each node (state) representing a set of completed courses, and edges determined by possible courses to be taken each semester, following prerequisites and the limit k. Bit manipulation allows efficient representation and manipulation of course combinations.

Example Walkthrough

Assume we have n = 4 different courses with corresponding prerequisite pairs [[1, 2], [1, 3], [3, 4]], indicating that you must complete course 1 before taking courses 2 and 3, and course 3 before taking course 4. We also have a limit k = 2, meaning you can take up to two courses in any given semester.

Let's walk through an application of the solution approach:

1. Initialization: We represent our course schedule with bit manipulation:

   • No courses are taken initially, so our starting bit sequence is 0000.
   • We begin with a deque initialized with this starting state (0, 0), meaning no courses have been completed and 0 semesters have passed.
   • We also initialize a vis (visited) set to keep track of states we've already examined.

2. Process Queue:

   • We start processing the states in our queue.
   • Our goal state is (1 << (n + 1)) - 2 = 1110, representing all courses except the non-existent course 0 being completed. (The bit sequence uses n+1 because we are using 1-indexing for our courses.)

3. Selecting Courses:

   • At the start, with the current state being 0000, we identify which courses have their prerequisites fulfilled and can be taken. Course 1 has no prerequisites, so it can be taken immediately, resulting in the next state being 0001.
   • We then add (0001, 1) to our queue to represent that course 1 is completed after 1 semester.

4. Branching for the Next Semester:

   • We dequeue the state (0001, 1) and identify potential new courses that can be taken. Given the prerequisites, we can now take courses 2 and 3, resulting in the bitmask 0110.
   • We now have three potential states to consider for the next semester: 0101, 0011, and 0110. Each of these represents taking either course 2 or 3 along with the already completed course 1, fulfilling the prerequisite conditions.

5. Considering Course Limit k:

   • The limit of k=2 means we can only consider combinations of up to 2 courses per semester. Since we do not exceed this limit by taking either course 2 or 3, we can add both states 0101 and 0011 to our queue.

6. Reaching the Goal:

   • Proceeding with either of the new states (0101 or 0011), we follow similar steps and eventually can take course 4, leading us to our goal state of 1110 after a minimum of 2 semesters, since by the second semester all prerequisites for the remaining courses would have been satisfied, allowing us to take the final course and complete the schedule.

In conclusion, using this example with n = 4, prerequisites [[1, 2], [1, 3], [3, 4]], and course limit k = 2, we can see that the minimum number of semesters required to complete all courses is 2. This approach, combining bit manipulation, a queue, and breadth-first search, allows us to efficiently arrive at this solution.

Solution Implementation

Time and Space Complexity

The given Python code aims to find the minimum number of semesters to take all courses given pre-requisite relations and a limit on the number of courses that can be taken in one semester. The code employs a breadth-first search (BFS) strategy.

Time Complexity

Let n be the number of courses and k be the maximum number of courses that can be taken in a semester. The time complexity primarily depends on the operations inside the while-loop.

• The BFS explores all possible combinations of courses one can take each semester.
• For each state in the queue, the code first identifies all courses whose prerequisites are satisfied (the for-loop over n courses), which takes O(n) time.
• The main complexity comes from generating the next set of courses to take. In the worst case, it iterates through all combinations of courses that can be taken in the next step. Since there are n courses, there can be up to 2^n combinations. However, because we only care about combinations of size k, the number of combinations checked effectively becomes C(n,k) (combinatorial), which is upper-bounded by O(n^k).
• Each level of BFS, therefore, has a maximum time complexity of O(n * n^k) since we multiply the number of operations to identify the next set of courses O(n^k) by the operations required to find the courses that can be taken O(n).

Hence, the time complexity is O((2^n) * (n * n^k)). Since C(n, k) is smaller than 2^n for small values of k, a tighter bound could be O((2^n) * n * C(n, k)).

Space Complexity

• The space complexity involves storing the queue and the visited states.
• Each state is a bitmask of length n, and there are 2^n possible states. However, due to the pruning by checking visited states, not all states will be stored in the vis set simultaneously.
• The space complexity becomes O(2^n) for storing visited states, which is the dominant factor compared to the queue.

Overall, the space complexity is O(2^n). Note that this analysis assumes that standard operations like bit-counting take O(1) time, which is typically the case in modern architectures.

Learn more about how to find time and space complexity quickly using problem constraints.