How do you solve Parallel Courses III on LeetCode?

Parallel Courses III (LeetCode #2050) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n) space complexity. Key insight: Courses can be taken simultaneously if prerequisites are met.

What is the brute force solution for Parallel Courses III?

The brute force approach for Parallel Courses III is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of course completions, which can be quite inefficient. We would simulate the completion of courses and track the time taken, but this leads to an exponential number of possibilities. The optimal Optimal Solution approach improves this to O(n + m).

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

Parallel Courses III uses the following concepts: Array, Dynamic Programming, Graph Theory, Topological Sort. The recommended patterns to study are: Graph Theory, Topological Sort, Dynamic Programming.

What are the interview tips for Parallel Courses III?

Understand the graph representation of the problem. Practice topological sorting and dynamic programming concepts. Think about edge cases, such as courses with no prerequisites.

What are common mistakes when solving Parallel Courses III?

Not considering simultaneous course completions. Failing to handle the graph structure correctly.

#2050

Parallel Courses III

Hard

ArrayDynamic ProgrammingGraph TheoryTopological SortGraph TheoryTopological SortDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + m)
Space	O(1)	O(n)

💡

Intuition

Time O(n + m)Space O(n)

The optimal solution uses dynamic programming and topological sorting to efficiently calculate the minimum time required to complete all courses. By processing courses in order of their prerequisites, we can ensure that we always know when a course can be started.

⚙️

Algorithm

3 steps

1Step 1: Build a graph from the relations and an indegree array to track prerequisites.
2Step 2: Use a queue to perform a topological sort, processing courses with no prerequisites first.
3Step 3: For each course processed, update the minimum time required for its dependent courses.

solution.py28 lines

1# Full working Python code
2from collections import deque
3
4class Solution:
5    def minimumTime(self, n, relations, time):
6        graph = [[] for _ in range(n)]
7        indegree = [0] * n
8        for prev, next in relations:
9            graph[prev - 1].append(next - 1)
10            indegree[next - 1] += 1
11
12        min_time = [0] * n
13        queue = deque()
14
15        for i in range(n):
16            if indegree[i] == 0:
17                queue.append(i)
18                min_time[i] = time[i]
19
20        while queue:
21            course = queue.popleft()
22            for next_course in graph[course]:
23                indegree[next_course] -= 1
24                min_time[next_course] = max(min_time[next_course], min_time[course] + time[next_course])
25                if indegree[next_course] == 0:
26                    queue.append(next_course)
27
28        return max(min_time)

ℹ

Complexity note: The complexity is O(n + m) where n is the number of courses and m is the number of relations, as we process each course and each relation once.

1Courses can be taken simultaneously if prerequisites are met.
2Topological sorting helps in determining the order of course completion.

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