Parallel Courses III

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Description

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] = [prevCourse_j, nextCourse_j] denotes that course prevCourse_j has to be completed before course nextCourse_j (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 * 10^4
• 0 <= relations.length <= min(n * (n - 1) / 2, 5 * 10^4)
• relations[j].length == 2
• 1 <= prevCourse_j, nextCourse_j <= n
• prevCourse_j != nextCourse_j
• All the pairs [prevCourse_j, nextCourse_j] are unique.
• time.length == n
• 1 <= time[i] <= 10^4
• The given graph is a directed acyclic graph.

Solution

class Solution {
public:
    int minimumTime(int n, vector<vector<int>>& relations, vector<int>& time) {
        vector<vector<int>> edges(n + 1);
        vector<int> inDegree(n + 1);

        for (const auto& info : relations) {
            edges[info[0]].push_back(info[1]);
            inDegree[info[1]]++;
        }

        queue<int> que;
        for (int i = 1; i <= n; i++) {
            if (inDegree[i] == 0)
                que.push(i);
        }

        vector<int> cost(n + 1);
        int res = 0;
        while (!que.empty()) {
            int current = que.front();
            que.pop();

            cost[current] += time[current - 1];
            res = max(res, cost[current]);
            for (int v : edges[current]) {
                cost[v] = max(cost[v], cost[current]);
                if (--inDegree[v] == 0)
                    que.push(v);
            }
        }

        return res;
    }
};

• Time complexity: \(O(V+E)\);
• Space complexity: \(O(V+E)\).