<p>You are given an integer <code>n</code>, which indicates that there are <code>n</code> courses labeled from <code>1</code> to <code>n</code>. You are also given an array <code>relations</code> where <code>relations[i] = [prevCourse<sub>i</sub>, nextCourse<sub>i</sub>]</code>, representing a prerequisite relationship between course <code>prevCourse<sub>i</sub></code> and course <code>nextCourse<sub>i</sub></code>: course <code>prevCourse<sub>i</sub></code> has to be taken before course <code>nextCourse<sub>i</sub></code>. Also, you are given the integer <code>k</code>.</p>

<p>In one semester, you can take <strong>at most</strong> <code>k</code> courses as long as you have taken all the prerequisites in the <strong>previous</strong> semesters for the courses you are taking.</p>

<p>Return <em>the <strong>minimum</strong> number of semesters needed to take all courses</em>. The testcases will be generated such that it is possible to take every course.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/05/22/leetcode_parallel_courses_1.png" style="width: 269px; height: 147px;" />
<pre>
<strong>Input:</strong> n = 4, relations = [[2,1],[3,1],[1,4]], k = 2
<strong>Output:</strong> 3
<strong>Explanation:</strong> The figure above represents the given graph.
In the first semester, you can take courses 2 and 3.
In the second semester, you can take course 1.
In the third semester, you can take course 4.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/05/22/leetcode_parallel_courses_2.png" style="width: 271px; height: 211px;" />
<pre>
<strong>Input:</strong> n = 5, relations = [[2,1],[3,1],[4,1],[1,5]], k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> The figure above represents the given graph.
In the first semester, you can only take courses 2 and 3 since you cannot take more than two per semester.
In the second semester, you can take course 4.
In the third semester, you can take course 1.
In the fourth semester, you can take course 5.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 15</code></li>
	<li><code>1 &lt;= k &lt;= n</code></li>
	<li><code>0 &lt;= relations.length &lt;= n * (n-1) / 2</code></li>
	<li><code>relations[i].length == 2</code></li>
	<li><code>1 &lt;= prevCourse<sub>i</sub>, nextCourse<sub>i</sub> &lt;= n</code></li>
	<li><code>prevCourse<sub>i</sub> != nextCourse<sub>i</sub></code></li>
	<li>All the pairs <code>[prevCourse<sub>i</sub>, nextCourse<sub>i</sub>]</code> are <strong>unique</strong>.</li>
	<li>The given graph is a directed acyclic graph.</li>
</ul>

from collections import deque

class Solution:
    def minNumberOfSemesters(self, numCourses: int, prerequisites: list[list[int]], maxCoursesPerSemester: int) -> int:
        # Initialize the prerequisite bitmask for each course
        prereq_mask = [0] * numCourses
        for pre, course in prerequisites:
            prereq_mask[course - 1] |= 1 << (pre - 1)
        
        # Initialize DP table with infinity
        min_semesters = [float('inf')] * (1 << numCourses)
        min_semesters[0] = 0  # Base case: 0 courses taken requires 0 semesters
        
        # Use BFS to explore all possible states
        queue = deque([0])
        while queue:
            current_mask = queue.popleft()
            available_courses_mask = 0
            for course_index in range(numCourses):
                if (current_mask & prereq_mask[course_index]) == prereq_mask[course_index]:  # Check prerequisites
                    available_courses_mask |= 1 << course_index
            
            available_courses_mask &= ~current_mask  # Remove already taken courses
            
            # Try all subsets of available courses
            subset = available_courses_mask
            while subset > 0:
                if bin(subset).count('1') <= maxCoursesPerSemester:
                    new_mask = current_mask | subset
                    if min_semesters[new_mask] > min_semesters[current_mask] + 1:
                        min_semesters[new_mask] = min_semesters[current_mask] + 1
                        queue.append(new_mask)
                subset = (subset - 1) & available_courses_mask
        
        return min_semesters[(1 << numCourses) - 1]

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