<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 itertools import combinations
from functools import lru_cache
from collections import defaultdict
from typing import List

class Solution:
    @lru_cache(None)
    def recurse(self, mask, in_degrees):
        # Base case: if all courses are taken, return 0 semesters needed
        if not mask:
            return 0
        
        # Find all courses that can be taken in the current semester
        nodes = [i for i in range(self.n) if mask & (1 << i) and in_degrees[i] == 0]
        
        ans = float('inf')
        # Try all combinations of courses that can be taken this semester
        for k_nodes in combinations(nodes, min(self.k, len(nodes))):
            new_mask, new_in_degrees = mask, list(in_degrees)
            
            # Update the mask and in-degrees for the selected courses
            for node in k_nodes:
                new_mask ^= 1 << node  # Mark the course as taken
                for child in self.graph[node]:
                    new_in_degrees[child] -= 1  # Reduce in-degree of child courses
            
            # Recurse with the updated state and add 1 semester
            ans = min(ans, 1 + self.recurse(new_mask, tuple(new_in_degrees)))
        
        return ans
    
    def minNumberOfSemesters(self, n: int, relations: List[List[int]], k: int) -> int:
        self.n = n
        self.k = k
        in_degrees = [0] * self.n
        self.graph = defaultdict(list)
        
        # Build the graph and in-degree array
        for prev_course, next_course in relations:
            in_degrees[next_course - 1] += 1
            self.graph[prev_course - 1].append(next_course - 1)
        
        # Start with all courses needing to be taken
        return self.recurse((1 << self.n) - 1, tuple(in_degrees))
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