<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>

class Solution:
    def minNumberOfSemesters(self, n: int, relations: List[List[int]], k: int) -> int:
        # Step 1: Build the graph
        graph = [[] for _ in range(n)]
        for prev, next in relations:
            graph[next - 1].append(prev - 1)
        
        # Step 2: Topological Sorting
        def topological_sort():
            indegree = [0] * n
            for node in range(n):
                for prev in graph[node]:
                    indegree[node] += 1
            
            queue = [node for node in range(n) if indegree[node] == 0]
            top_order = []
            
            while queue:
                node = queue.pop(0)
                top_order.append(node)
                for next in graph[node]:
                    indegree[next] -= 1
                    if indegree[next] == 0:
                        queue.append(next)
            
            return top_order
        
        top_order = topological_sort()
        
        # Step 3: Dynamic Programming
        dp = [float('inf')] * (1 << n)
        dp[0] = 0  # Base case: no courses taken
        
        for mask in range(1 << n):
            # Find the courses that are not taken yet
            available_courses = [i for i in range(n) if (mask & (1 << i)) == 0]
            
            # Find the courses that have all their prerequisites satisfied
            taken = mask
            can_take = []
            for course in available_courses:
                if all(taken & (1 << prev) for prev in graph[course]):
                    can_take.append(course)
            
            # Try all combinations of taking up to k courses
            from itertools import combinations
            for combo in combinations(can_take, min(k, len(can_take))):
                new_mask = mask
                for course in combo:
                    new_mask |= (1 << course)
                dp[new_mask] = min(dp[new_mask], dp[mask] + 1)
        
        # The answer is the minimum number of semesters to take all courses
        return dp[(1 << n) - 1]

# Example usage:
solution = Solution()
n = 4
relations = [[2,1],[3,1],[1,4]]
k = 2
print(solution.minNumberOfSemesters(n, relations, k))  # Output: 3
16900