How do you solve Parallel Courses II on LeetCode?

Parallel Courses II (LeetCode #1494) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(2^n) space complexity. Key insight: Understanding bit manipulation is crucial for optimizing state representation.

What is the brute force solution for Parallel Courses II?

The brute force approach for Parallel Courses II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves exploring all possible combinations of courses that can be taken each semester, while respecting the prerequisite constraints. This method is straightforward but inefficient for larger inputs due to its combinatorial nature. The optimal Optimal Solution approach improves this to O(n * 2^n).

What algorithm or data structure is used to solve Parallel Courses II?

Parallel Courses II uses the following concepts: Dynamic Programming, Bit Manipulation, Graph Theory, Bitmask. The recommended patterns to study are: Dynamic Programming, Bit Manipulation.

What are the interview tips for Parallel Courses II?

Break down the problem into smaller parts and think about how to represent states. Practice bit manipulation and dynamic programming problems beforehand. Explain your thought process clearly during the interview.

What are common mistakes when solving Parallel Courses II?

Failing to correctly track prerequisites when updating states. Not considering all combinations of courses that can be taken each semester.

#1494

Parallel Courses II

Hard

Dynamic ProgrammingBit ManipulationGraph TheoryBitmaskDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 2^n)
Space	O(1)	O(2^n)

💡

Intuition

Time O(n * 2^n)Space O(2^n)

The optimal solution leverages dynamic programming and bit manipulation to efficiently track which courses have been taken and which can be taken next, minimizing the number of semesters needed. This approach is more efficient than brute force because it avoids redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Create a graph representation of courses and their prerequisites.
2Step 2: Use a bitmask to represent the set of courses taken.
3Step 3: Use dynamic programming to calculate the minimum semesters needed by exploring valid combinations of courses that can be taken each semester.

solution.py30 lines

1def minNumberOfSemesters(n, relations, k):
2    from collections import defaultdict
3    graph = defaultdict(list)
4    indegree = [0] * (n + 1)
5    for prev, next in relations:
6        graph[prev].append(next)
7        indegree[next] += 1
8
9    dp = [float('inf')] * (1 << n)
10    dp[0] = 0
11
12    for mask in range(1 << n):
13        available = []
14        for i in range(1, n + 1):
15            if (mask & (1 << (i - 1))) == 0 and indegree[i] == 0:
16                available.append(i)
17        for i in range(1 << len(available)):
18            if bin(i).count('1') <= k:
19                next_mask = mask
20                for j in range(len(available)):
21                    if i & (1 << j):
22                        next_mask |= (1 << (available[j] - 1))
23                        for neighbor in graph[available[j]]:
24                            indegree[neighbor] -= 1
25                dp[next_mask] = min(dp[next_mask], dp[mask] + 1)
26                for j in range(len(available)):
27                    if i & (1 << j):
28                        for neighbor in graph[available[j]]:
29                            indegree[neighbor] += 1
30    return dp[(1 << n) - 1]

ℹ

Complexity note: The time complexity is O(n * 2^n) because we are iterating through all subsets of courses (2^n) and for each subset, we check the available courses (up to n). The space complexity is O(2^n) for storing the dp array.

1Understanding bit manipulation is crucial for optimizing state representation.
2Dynamic programming can significantly reduce redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.