Solution Explanation:

This problem can be solved using bit manipulation and breadth-first search (BFS). The core idea is to represent the courses taken in each semester using a bitmask. Each bit in the bitmask corresponds to a course; if the bit is set, the course has been taken.

1. Preprocessing:

• We create a d array (or equivalent data structure in other languages) to store the prerequisites for each course. d[i] represents a bitmask where each set bit indicates a prerequisite course for course i. This is done by iterating through the relations array.

2. BFS:

• We start with an initial state where no courses have been taken (bitmask 0).
• The BFS queue q stores pairs (current_state, semesters_taken).
• In each iteration, we:
  • Check if all courses have been taken (current_state == (1 << (n + 1)) - 2). If so, we have found the solution.
  • Calculate nxt, a bitmask representing the courses that can be taken in the current semester. A course can be taken if all its prerequisites are already satisfied (checked using bitwise AND with d[i]). nxt is then XORed with cur to get only the courses that haven't been taken.
  • Course Selection: If the number of courses in nxt is less than or equal to k, we can take all of them in this semester. Otherwise, we need to choose a subset of k courses from nxt. This is where the bit manipulation gets slightly more complex. We iterate through all possible subsets of nxt using bit manipulation (subtracting 1 and bitwise ANDing). For each subset, we check if its size is k and if it hasn't been visited before.
  • Add the new state (updated bitmask) and increased semester count to the queue if it hasn't been visited before.

3. Bit Manipulation Subset Generation:

The key part of the code is the generation of subsets of size k when nxt has more courses than k. It's done using a clever bit manipulation technique:

x = nxt
while nxt:
    if nxt.bit_count() == k and (nxt | cur) not in vis:
        q.append((nxt | cur, t + 1))
    nxt = (nxt - 1) & x

This loop iterates through all possible subsets of x. nxt = (nxt - 1) & x is the crucial part. Let's break down how it works:

• nxt - 1: Decrements the bitmask. This flips the least significant set bit to 0 and sets all bits to the right to 1.
• & x: The bitwise AND with the original x keeps only the bits that were originally set in x. Effectively, this systematically removes one bit at a time from the original x and generates all possible subsets.

Time Complexity Analysis:

The time complexity is dominated by the BFS. In the worst case, we could explore all possible states, which is O(2n). The subset generation loop within each BFS step can also add to the complexity but remains within the same order because the number of subsets is also bounded by O(2n).

Space Complexity Analysis:

The space complexity is dominated by the vis set which stores visited states and the queue q. In the worst case, both can store up to O(2n) states.

Therefore, the overall time and space complexity are both O(2n) in the worst case. However, because n is limited to 15 in the problem constraints, this algorithm remains efficient enough to solve the problem.