2050. Parallel Courses III

Problem Statement

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.

Intuition

Approach:

Use Kahn Algorithm to find the start nodes,
Apply that,

Maintain a distance array to store all the distances uptill that point and update at
each step

Links

https://leetcode.com/problems/parallel-courses-iii/description/?envType=daily-question&envId=2023-10-18

Video Links

Approach 1:

C++

class Solution {
public:
    int minimumTime(int n, vector<vector<int>>& relations, vector<int>& time) {
        vector<vector<int>> adj(n+1);
        vector<int> indeg(n+1,0);

        for(auto &it: relations) {
            indeg[it[1]]++;
            adj[it[0]].push_back(it[1]);
        }

        vector<int> dist(n+1, 0);
        queue<int> q;
        for(int i=1; i<=n; i++) {
            if(indeg[i] == 0){
                q.push(i);
                dist[i] = time[i-1];
                cout<<i<<" "<<dist[i]<<endl;
            }
        }

        while(!q.empty()){
            int temp = q.front();
            q.pop();

            for(auto &it: adj[temp]){
                dist[it] = max(dist[it], time[it-1]+dist[temp]);
                if(--indeg[it] == 0)
                    q.push(it);
            }
        }

        return *max_element(dist.begin(), dist.end());
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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