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2176. Parallel Courses Iii

Difficulty: Hard

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2176. Parallel Courses III

Hard

You are given an integer n, which indicates that there are n courses labeled from 1 to n. You are also given a 2D integer array relations where relations[j] = [prevCoursej, nextCoursej] denotes that course prevCoursej has to be completed before course nextCoursej (prerequisite relationship). Furthermore, you are given a 0-indexed integer array time where time[i] denotes how many months it takes to complete the (i+1)th course.

You must find the minimum number of months needed to complete all the courses following these rules:

  • You may start taking a course at any time if the prerequisites are met.
  • Any number of courses can be taken at the same time.

Return the minimum number of months needed to complete all the courses.

Note: The test cases are generated such that it is possible to complete every course (i.e., the graph is a directed acyclic graph).

 

Example 1:

Input: n = 3, relations = [[1,3],[2,3]], time = [3,2,5]
Output: 8
Explanation: The figure above represents the given graph and the time required to complete each course. 
We start course 1 and course 2 simultaneously at month 0.
Course 1 takes 3 months and course 2 takes 2 months to complete respectively.
Thus, the earliest time we can start course 3 is at month 3, and the total time required is 3 + 5 = 8 months.

Example 2:

Input: n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], time = [1,2,3,4,5]
Output: 12
Explanation: The figure above represents the given graph and the time required to complete each course.
You can start courses 1, 2, and 3 at month 0.
You can complete them after 1, 2, and 3 months respectively.
Course 4 can be taken only after course 3 is completed, i.e., after 3 months. It is completed after 3 + 4 = 7 months.
Course 5 can be taken only after courses 1, 2, 3, and 4 have been completed, i.e., after max(1,2,3,7) = 7 months.
Thus, the minimum time needed to complete all the courses is 7 + 5 = 12 months.

 

Constraints:

  • 1 <= n <= 5 * 104
  • 0 <= relations.length <= min(n * (n - 1) / 2, 5 * 104)
  • relations[j].length == 2
  • 1 <= prevCoursej, nextCoursej <= n
  • prevCoursej != nextCoursej
  • All the pairs [prevCoursej, nextCoursej] are unique.
  • time.length == n
  • 1 <= time[i] <= 104
  • The given graph is a directed acyclic graph.

Solution

class Solution {
    public int minimumTime(int n, int[][] relations, int[] time) {
        ArrayList<ArrayList<Integer>> adj = new ArrayList<>();
        for (int i = 0; i <= n; i++)
            adj.add(new ArrayList<>());

        int[] indegree = new int[n + 1];
        for (int[] edge : relations) {
            int u = edge[0], v = edge[1];
            adj.get(u).add(v);
            indegree[v]++;
        }

        Queue<Integer> q = new LinkedList<>();
        int[] finishTime = new int[n + 1]; 

        for (int i = 1; i <= n; i++) {
            if (indegree[i] == 0) {
                q.offer(i);
                finishTime[i] = time[i - 1]; 
            }
        }

        int res = 0;
        while (!q.isEmpty()) {
            int u = q.poll();
            res = Math.max(res, finishTime[u]);
            for (int v : adj.get(u)) {
                finishTime[v] = Math.max(finishTime[v], finishTime[u] + time[v - 1]);
                if (--indegree[v] == 0) {
                    q.offer(v);
                }
            }
        }
        return res;
    }
}

Complexity Analysis

  • Time Complexity: O(?)
  • Space Complexity: O(?)

Approach

Detailed explanation of the approach will be added here