2050. Parallel Courses III

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2050. Parallel Courses III

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<sub>j</sub>, nextCourse<sub>j</sub>] denotes that course prevCourse<sub>j</sub> has to be completed before course nextCourse<sub>j</sub> (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:

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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:

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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 <= prevCoursej, nextCoursej <= n
  • prevCoursej != nextCoursej
  • All the pairs [prevCoursej, nextCoursej] are unique .
  • time.length == n
  • 1 <= time[i] <= 10^4
  • The given graph is a directed acyclic graph.

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    int minimumTime(int n, vector<vector<int>>& relations, vector<int>& time) {
        // 1. build the prev -> next graph
        // 2. we need two vector of vectors: from and two
        // 3. the algorithm:
        //      1. enqueue all nodes with no pre-requiste
        //      2. from each of them, their value should be time[idx]
        //      3. check each of their to point, if the point doesn't have a pre-requisite, then enque int
        vector<vector<int>> from(n);
        vector<int> inDegree(n, 0);
        for (auto r : relations) {
            int u = r[0] - 1, v = r[1] - 1;
            from[u].push_back(v);
            inDegree[v]++;
        }
        // vector<int> minTime(n, );
        queue<int> courseProcessing;
        vector<int> maxTime(n, 0);
        for (int i = 0; i < n; i++) {
            if (inDegree[i] == 0) {
                courseProcessing.push(i);
                maxTime[i] = time[i];
            }
        }
        int res = 0, curTime = 0;
        while (!courseProcessing.empty()) {
            int curCourse = courseProcessing.front();
            courseProcessing.pop();
            for (int v : from[curCourse]) {
                inDegree[v]--;
                maxTime[v] = max(maxTime[v], maxTime[curCourse] + time[v]);
                if (inDegree[v] == 0) {
                    courseProcessing.push(v);
                }
            }
        }
        return ranges::max(maxTime);

    }
};