LeetCode 2050: Parallel Courses III Solution

Master LeetCode problem 2050 (Parallel Courses III), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2050. Parallel Courses III

Hard

Array Dynamic Programming Graph Topological Sort

Problem Explanation

Explanation:

To solve this problem, we can use a topological sorting approach along with dynamic programming. We will create a directed acyclic graph (DAG) based on the given relations and then calculate the minimum time needed to complete all courses by considering the dependencies and the time required for each course. We will keep track of the earliest possible time to start each course and update it based on the completion time of its prerequisites.

Create a directed graph from the given relations.
Perform a topological sort to find the order in which courses can be taken.
Traverse the sorted order and calculate the minimum time needed to complete each course.
Update the earliest possible start time for each course based on the completion time of its prerequisites.
Finally, return the maximum time needed to complete all courses.

Time Complexity: O(n + m) where n is the number of courses and m is the number of relations. Space Complexity: O(n) for storing the time needed for each course and the earliest start time for each course.

Solution Code

class Solution {
    public int minNumberOfSemesters(int n, int[][] relations, int[] time) {
        List<List<Integer>> graph = new ArrayList<>();
        int[] indegree = new int[n+1];
        int[] dp = new int[1 << n];
        
        for (int i = 0; i <= n; i++) {
            graph.add(new ArrayList<>());
        }
        
        for (int[] rel : relations) {
            graph.get(rel[0]).add(rel[1]);
            indegree[rel[1]]++;
        }
        
        for (int mask = 1; mask < (1 << n); mask++) {
            int bits = 0;
            for (int i = 0; i < n; i++) {
                if ((mask & (1 << i)) != 0 && indegree[i + 1] == 0) {
                    bits |= 1 << i;
                }
            }
            dp[mask] = Integer.MAX_VALUE / 2;
            for (int subset = mask; subset > 0; subset = (subset - 1) & mask) {
                if ((subset & bits) == subset) {
                    int tmp = 0;
                    for (int i = 0; i < n; i++) {
                        if ((subset & (1 << i)) != 0) {
                            tmp += time[i];
                        }
                    }
                    dp[mask] = Math.min(dp[mask], dp[mask ^ subset] + tmp);
                }
            }
        }
        
        return dp[(1 << n) - 1];
    }
}

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