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2050. Parallel Courses III

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2050. 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.

Similar Questions:

Solution 1. Topological Sort (BFS)

A node's distance is its time plus the maximum distance of all predecessor nodes. We can calculate the distances via a topological sort. The answer is the maximum distance.

// OJ: https://leetcode.com/problems/parallel-courses-iii/
// Author: github.com/lzl124631x
// Time: O(N + E)
// Space: O(N + E)
class Solution {
public:
    int minimumTime(int n, vector<vector<int>>& E, vector<int>& T) {
        vector<vector<int>> G(n);
        vector<int> indegree(n), dist(n);
        for (auto &e : E) { // build graph and count indegrees
            G[e[0] - 1].push_back(e[1] - 1);
            indegree[e[1] - 1]++;
        }
        queue<int> q;
        for (int i = 0; i < n; ++i) {
            if (indegree[i] == 0) { // enqueue nodes with 0 indegree.
                q.push(i);
                dist[i] = T[i]; // source nodes' distance is their corresponding time
            } 
        }
        while (q.size()) {
            int u = q.front();
            q.pop();
            for (int v : G[u]) {
                dist[v] = max(dist[u] + T[v], dist[v]); // update the distance of node `v` using the maximum distance of predecessor nodes.
                if (--indegree[v] == 0) q.push(v); // enqueue node `v` when its indegree drops to 0
            }
        }
        return *max_element(begin(dist), end(dist)); // the answer is the maximum distance.
    }
};

Solution 2. Topological Sort (DFS)

DFS version topological sort is Post-order Traversal + Memo.

// OJ: https://leetcode.com/problems/parallel-courses-iii/
// Author: github.com/lzl124631x
// Time: O(N + E)
// Space: O(N + E)
class Solution {
public:
    int minimumTime(int n, vector<vector<int>>& E, vector<int>& T) {
        vector<vector<int>> G(n);
        vector<int> dist(n);
        for (auto &e : E) G[e[1] - 1].push_back(e[0] - 1);
        function<int(int)> dfs = [&](int u) {
            if (dist[u]) return dist[u];
            int mx = 0;
            for (int v : G[u]) mx = max(mx, dfs(v));
            return dist[u] = mx + T[u];
        };
        for (int i = 0; i < n; ++i) dfs(i);
        return *max_element(begin(dist), end(dist));
    }
};

Discuss

https://leetcode.com/problems/parallel-courses-iii/discuss/1537501/C%2B%2B-Topological-Sort