Parallel Courses

class Solution {
public:
    int minSemesterCount(int count, vector<vector<int>>& prerequisites) {
        vector<vector<int>> directedGraph(count + 1);
        for (auto& pre : prerequisites) {
            directedGraph[pre[0]].push_back(pre[1]);
        }
    
        vector<int> visited(count + 1, 0);
        int maximumDepth = 1;
        for (int vertex = 1; vertex <= count; vertex++) {
            int depth = explore(vertex, directedGraph, visited);
            // Cycle detection
            if (depth == -1) {
                return -1;
            }
            maximumDepth = max(depth, maximumDepth);
        }
        return maximumDepth;
    }
    
private:
    int explore(int vertex, vector<vector<int>>& directedGraph, vector<int>& visited) {
        // finds the depth of dependency
        if (visited[vertex] != 0) {
            return visited[vertex];
        } else {
            // visiting
            visited[vertex] = -1;
        }
        int maxDepth = 1;
        for (auto& nextVertex : directedGraph[vertex]) {
            int currentDepth = explore(nextVertex, directedGraph, visited);
            // Checking for a cycle
            if (currentDepth == -1) {
                return -1;
            }
            maxDepth = max(currentDepth + 1, maxDepth);
        }
        // Visited and processed
        visited[vertex] = maxDepth;
        return maxDepth;
    }
};

The given C++ code provides a solution to find the minimum number of semesters required to complete all courses given their prerequisites. The problem is approached using depth-first search (DFS) on a directed graph representation of course dependencies.

Key Components of the Solution:

• Graph Construction:

  • A directed graph is formed where nodes represent courses and edges represent prerequisites.
  • Each course points to the courses that are prerequisites for it.

• Depth-first Search (DFS) and Cycle Detection:

  • The code carries out a DFS search starting from each course to determine the maximum depth or the longest path starting from that course.
  • A visited vector is used to keep track of visited nodes to detect cycles. It marks unvisited nodes as 0, visited nodes as -1 (cycle detection step), and fully processed nodes (all children visited) with their maximum depth.

• Cycle Handling:

  • While exploring each vertex, if at any step a node referring to itself or among its dependencies the same course is encountered, this indicates a cycle (a situation where a course is a prerequisite of itself directly or indirectly).
  • If a cycle is detected (when explore function returns -1), the function immediately returns -1, indicating it's not possible to complete all courses.

• Result Calculation:

  • The solution iterates through each node and calculates the maximum depth of dependent courses.
  • The final result, which represents the minimum number of semesters needed, is the maximum depth found among all courses.

Overall, this solution explores graph-based depth-first searches to effectively measure course dependency lengths and handle complex prerequisite scenarios, including cycles, efficiently determining the number of semesters needed or detecting an impossible situation to complete the courses.

class Solution {
    public int minSemesters(int totalCourses, int[][] dependencies) {
        List<List<Integer>> adjacencyList = new ArrayList<>(totalCourses + 1);
        for (int i = 0; i < totalCourses + 1; i++) {
            adjacencyList.add(new ArrayList<Integer>());
        }
        for (int[] dependency : dependencies) {
            adjacencyList.get(dependency[0]).add(dependency[1]);
        }
        int[] courseStatus = new int[totalCourses + 1];
    
        int longestPath = 1;
        for (int course = 1; course < totalCourses + 1; course++) {
            int pathLength = findPath(course, adjacencyList, courseStatus);
            // Cycle detection
            if (pathLength == -1) {
                return -1;
            }
            longestPath = Math.max(pathLength, longestPath);
        }
        return longestPath;
    }
    
    private int findPath(int course, List<List<Integer>> adjacencyList, int[] courseStatus) {
        // Check if the course has been calculated or visited
        if (courseStatus[course] != 0) {
            return courseStatus[course];
        } else {
            // Mark as currently visiting
            courseStatus[course] = -1;
        }
        int maxPathLength = 1;
        for (int successor : adjacencyList.get(course)) {
            int pathLength = findPath(successor, adjacencyList, courseStatus);
            // Cycle detection
            if (pathLength == -1) {
                return -1;
            }
            maxPathLength = Math.max(pathLength + 1, maxPathLength);
        }
        // Mark as fully visited
        courseStatus[course] = maxPathLength;
        return maxPathLength;
    }
}

The Java solution to the "Parallel Courses" problem involves determining the minimum number of semesters required to complete a specified number of courses given the dependencies between them. This problem is approached using topological sorting with depth-first search (DFS) to find the longest path in a directed acyclic graph (DAG), representing the courses and their dependencies.

The solution contains the following main components:

• Adjacency List Creation:

  • An adjacency list is initialized to represent the directed graph of course dependencies.
  • Dependencies are added to the adjacency list as specified by the input array.

• DFS Implementation for Path Finding:

  • A findPath function is implemented using DFS to calculate the maximum path length from each course. The function returns the length of the longest path starting from a given course and also handles cycle detection.
  • If a cycle is detected (i.e., a course is revisited before the current search path terminates), the function returns -1, indicating that the course schedule cannot be completed due to circular dependencies.

• Course Status Array:

  • An array courseStatus is utilized to store the status of each course: unvisited (0), visiting (-1), and fully processed with a known path length (path length value).

• Loop through Courses:

  • The main function loops through all courses to invoke findPath starting from each one.
  • If findPath returns -1 for any course, immediate termination occurs with a return of -1 to indicate that not all courses can be completed due to a cycle.
  • Otherwise, the maximum path length found among all initial courses determines the minimal semesters required as each path length represents the depth of dependency chains.

Overall, the implementation efficiently uses graph theory concepts to determine course scheduling feasibility and calculates the minimum number of semesters required considering the dependencies. The approach is optimized to handle cyclic dependencies by aborting the process and returning an error condition.

class Solution:
    def calculateMinimumSemesters(self, totalCourses: int, prerequisites: List[List[int]]) -> int:
        courseGraph = {i: [] for i in range(1, totalCourses + 1)}
        for pre, course in prerequisites:
            courseGraph[pre].append(course)
    
        courseStatus = {}
    
        def explore(course: int) -> int:
            if course in courseStatus:
                return courseStatus[course]
    
            courseStatus[course] = -1
                
            longestPath = 1
            for nextCourse in courseGraph[course]:
                result = explore(nextCourse)
                if result == -1:
                    return -1
                longestPath = max(result + 1, longestPath)
    
            courseStatus[course] = longestPath
            return longestPath
    
        semesterCount = -1
        for course in courseGraph.keys():
            pathLength = explore(course)
            if pathLength == -1:
                return -1
            semesterCount = max(pathLength, semesterCount)
        return semesterCount

The provided Python solution addresses the problem of finding the minimum number of semesters required to complete all given courses considering the prerequisites. The solution uses Depth-First Search (DFS) to explore course dependencies and determine the longest path in the course graph.

• Start by building a graph where each course points to its list of dependent courses.
• Use a dictionary to track the status of each course exploration:
  • A status of -1 indicates that the course is currently being processed (used for detecting cycles).
  • A positive integer value indicates the maximum duration from the current course to the end of its dependency chain.
• Implement a helper function explore that uses DFS to calculate the longest path for any given course.
  • If encountering a course marked with -1, a cycle is detected, making course completion impossible, hence return -1.
  • Compute the maximum path length recursively, ensuring it's incremented at each level of exploration.
• Iterate through all courses and for each initiate a DFS using the explore function.
  • Keep track of the highest number from these results since it represents the minimum semesters required if no cycle is present.
  • If any course exploration returns -1, it implies a cycle exists in the prerequisites, making it impossible to finish, and the function should return -1.

The solution ensures that every course is explored to calculate the maximum depth of its dependency chain, thus computing the overall minimum number of semesters needed. The algorithm's efficiency and correctness hinge upon accurately building the course graph and adeptly managing the course exploration statuses to avoid revisiting or cyclic computation errors.