Course Schedule IV Leetcode Problem 1462 [Python Solution]

In the world of coding, solving graph problems can be a challenging but rewarding experience; Course Schedule IV Leetcode Problem now!

Today, we’re going to tackle the Course Schedule Iv problem from LeetCode.

This problem falls under the category of graphs and is of medium difficulty.

Our goal is to determine if one course is a prerequisite for another, considering direct and indirect prerequisites.

We will walk you through the problem overview, constraints, and provide you with an efficient Python solution.

So, let’s dive into it!

Problem Overview

Imagine you are a student in a world where courses are labeled from 0 to numCourses - 1.

You are given an array of prerequisites, where each element is a pair [ai, bi].

This pair indicates that you must complete course ai before you can take course bi.

For example, if you have the pair [0, 1], it means you must complete course 0 before you can take course 1. Prerequisites can also be indirect.

If course A is a prerequisite of course B, and course B is a prerequisite of course C, then course A is an indirect prerequisite of course C.

Additionally, you are given an array of queries, each represented as [uj, vj].

For each query, you need to determine whether course uj is a prerequisite of course vj.

Your task is to return a boolean array, where each element in the array answers a query.

Let’s take a look at a few examples to better understand the problem.

Example 1:

Input:

• numCourses = 2
• prerequisites = [[1, 0]]
• queries = [[0, 1], [1, 0]]

Output:

• [False, True]

In this example, the pair [1, 0] indicates that you must take course 1 before course 0. However, course 0 is not a prerequisite for course 1, but the reverse is true.

Example 2:

Input:

• numCourses = 2
• prerequisites = []
• queries = [[1, 0], [0, 1]]

Output:

• [False, False]

In this case, there are no prerequisites, and each course is independent.

Example 3:

Input:

• numCourses = 3
• prerequisites = [[1, 2], [1, 0], [2, 0]]
• queries = [[1, 0], [1, 2]]

Output:

• [True, True]

Here, course 1 is a prerequisite for both course 0 and course 2.

Constraints:

• 2 <= numCourses <= 100
• 0 <= prerequisites.length <= (numCourses * (numCourses – 1) / 2)
• prerequisites[i].length == 2
• 0 <= ai, bi <= n – 1
• ai != bi
• All pairs [ai, bi] are unique.
• The prerequisites graph has no cycles.
• 1 <= queries.length <= 104
• 0 <= ui, vi <= n – 1
• ui != vi

Now that we have a clear understanding of the problem, let’s move on to the solution.

Understanding the Constraints

Before we dive into the solution, let’s understand the constraints provided:

1. The number of courses, numCourses, can vary from 2 to 100. This means we might need an efficient algorithm to handle larger inputs.
2. The prerequisites array prerequisites can have a maximum length of (numCourses * (numCourses - 1) / 2).

This indicates that there could be many prerequisites.

3. The prerequisites pairs [ai, bi] are unique, and there are no cycles in the prerequisites graph.
4. The number of queries can be as high as 10,000.
5. The queries are pairs of courses, and we need to efficiently determine if one is a prerequisite for the other.

Considering these constraints, we need an algorithm that can handle a large number of courses and prerequisites efficiently.

Course Schedule IV LeetCode Problem Solution

1. Bruteforce Approach

Let’s start by discussing a brute-force approach.

We can solve this problem by running a depth-first search (DFS) for each query to check if the first course is a prerequisite for the second course.

We would start from the second course (vj) and traverse the graph to see if we can reach the first course (uj).

The time complexity of this approach would be O(Q * (P + N)), where Q is the number of queries, P is the number of prerequisites, and N is the number of courses.

In the worst case, this could be inefficient, especially for large inputs.

2. An Efficient Approach with a Hash Map

To optimize the solution, we can use a hash map to store the indirect prerequisites for each course.

This way, we only need to perform a single DFS to build the prerequisites for all courses.

Let’s go step by step through the efficient Python solution.

def checkIfPrerequisite(self, numCourses: int, prerequisites: List[List[int]], queries: List[List[int]]) -> List[bool]:
        # Create an adjacency list to represent the prerequisites graph.
        adj = defaultdict(list)
        for prereq, crs in prerequisites:
            adj[crs].append(prereq)

        # Define a depth-first search (DFS) function to find indirect prerequisites.
        def dfs(crs):
            if crs not in prereqMap:
                prereqMap[crs] = set()
                for pre in adj[crs]:
                    prereqMap[crs] |= dfs(pre)
            prereqMap[crs].add(crs)
            return prereqMap[crs]

        # Initialize a dictionary to store the indirect prerequisites.
        prereqMap = {}
        # Build the prerequisites for each course using DFS.
        for crs in range(numCourses):
            dfs(crs)

        # Answer the queries by checking if u is in the prereqMap of v.
        res = []
        for u, v in queries:
            res.append(u in prereqMap[v])
        return res

Let’s break down the code step by step:

• We start by creating an adjacency list adj to represent the prerequisites graph.

This adjacency list helps us efficiently traverse the graph.

• The dfs function is a depth-first search function that finds indirect prerequisites for a given course.

If we haven’t already calculated the prerequisites for a course, we recursively find them.

We also add the course itself to its prerequisites.

This function efficiently handles the depth-first traversal of the graph.

• We initialize a dictionary prereqMap to store the indirect prerequisites.

This dictionary will map each course to its set of indirect prerequisites.

• We then iterate through all courses (0 to numCourses-1) and call the dfs function to populate the prereqMap with indirect prerequisites for each course.
• Next, we answer the queries by checking if u is in the prereqMap of `

v`.

If it is, we append True to the result list; otherwise, we append False.

This efficient approach ensures that we calculate indirect prerequisites for each course only once, reducing the time complexity.

The time and space complexity for this solution is discussed next.

Time and Space Complexity

Time Complexity

The time complexity of this solution is primarily determined by the DFS traversal of the prerequisites graph.

The DFS traversal is performed only once for all courses.

Hence, the time complexity can be represented as O(N + E), where:

• N is the number of courses (nodes), and
• E is the number of prerequisites (edges).

In the worst case, E can be N^2, but it’s still an efficient solution for the given constraints.

The time complexity to answer each query is O(1), as it involves simple dictionary lookups.

Overall, the time complexity is O(N + E) for preprocessing and O(Q) for answering Q queries, where Q is the number of queries.

Space Complexity

The space complexity is determined by the additional data structures used in the solution:

• The adj dictionary representing the prerequisites graph has a space complexity of O(E).
• The prereqMap dictionary stores the indirect prerequisites for each course, resulting in a space complexity of O(N).
• The res list used to store the query results has a space complexity of O(Q).

Therefore, the overall space complexity of the solution is O(N + E + Q).

Reasoning Behind Our Approach

The efficient approach we discussed minimizes redundant work by using a depth-first search to calculate the indirect prerequisites for each course.

By building a dictionary (prereqMap) that maps courses to their indirect prerequisites, we ensure that we only perform this calculation once for all courses.

This optimization reduces the time complexity of answering queries, making the solution efficient for larger inputs.

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Conclusion

In this blog post, we’ve explored the Course Schedule Iv problem on LeetCode, a graph-based problem where we need to determine if one course is a prerequisite for another, considering both direct and indirect prerequisites.

We discussed the problem overview, constraints, and an efficient Python solution using a depth-first search (DFS) approach with a hash map.

The efficient solution optimizes the calculation of indirect prerequisites, ensuring that we perform the DFS traversal only once for all courses.

This results in a time complexity of O(N + E) for preprocessing and O(Q) for answering queries.

We hope this guide has been helpful in understanding the problem and the efficient solution.

If you have any questions, suggestions, or comments, please feel free to share them in the comments section.

Happy coding!

Question Link: Course Schedule IV on LeetCode

Remember, coding is a journey, and learning to solve challenging problems is a great way to improve your skills.

Keep practicing and stay curious!