<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 typing import List

class Solution:
    def minNumberOfSemesters(self, n: int, relations: List[List[int]], k: int) -> int:
        # Prerequisite bitmask for each course
        prereq = [0] * n
        for u, v in relations:
            prereq[v - 1] |= (1 << (u - 1))
        
        # DP table
        dp = {0: 0}
        
        # Iterate over all possible states
        for state in range(1 << n):
            if state not in dp:
                continue
            
            # Determine which courses can be taken next
            can_take = 0
            for i in range(n):
                if (state & (1 << i)) == 0 and (state & prereq[i]) == prereq[i]:
                    can_take |= (1 << i)
            
            # Iterate over all subsets of can_take with at most k courses
            subset = can_take
            while subset:
                if bin(subset).count('1') <= k:
                    new_state = state | subset
                    if new_state not in dp or dp[new_state] > dp[state] + 1:
                        dp[new_state] = dp[state] + 1
                subset = (subset - 1) & can_take
        
        # The result is the minimum number of semesters to take all courses
        return dp[(1 << n) - 1]

# Example usage:
# sol = Solution()
# print(sol.minNumberOfSemesters(4, [[2,1],[3,1],[1,4]], 2))  # Output: 3
# print(sol.minNumberOfSemesters(5, [[2,1],[3,1],[4,1],[1,5]], 2))  # Output: 4
19700