2050. Parallel Courses III (H)

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:

Version 1: Topological

class Solution {
    
    public int minimumTime(int n, int[][] relations, int[] time) {
        
        int res = 0;
        List<Integer>[] graph = buildGraph(relations, n);
        int[] indegree = new int[n+1];
        Queue<Node> queue = new LinkedList<Node>();
        int[] cost = new int[n+1];
        
        for(int i = 0; i< relations.length; i++)
        {
            int to = relations[i][1];
            indegree[to]++;
        }
        
        for(int i = 1; i <=n ; i++)
        {
            if(indegree[i] == 0)
            {
                queue.offer(new Node(i, time[i-1]));
                cost[i] = time[i-1];
            }  
        }
        
        while(!queue.isEmpty())
        {
            Node current = queue.poll();
            res = Math.max(res, current.currentCost);
            for(int nextNode: graph[current.courseNum])
            {
                indegree[nextNode]--;
                //这里要在外面更新，你能在if里面。
                cost[nextNode] = Math.max(cost[nextNode], current.currentCost + time[nextNode-1]);
                if(indegree[nextNode] == 0)
                {
                    queue.offer(new Node(nextNode, cost[nextNode]));
                }
            }
        }
        return res;
        
    }

    
    public List<Integer>[] buildGraph(int[][] relations, int n)
    {
        List<Integer>[] graph = new LinkedList[n+1];
        
        for(int i = 0; i<= n; i++)
        {
            graph[i] = new LinkedList<Integer>();
        }
        
        for(int i = 0; i< relations.length; i++)
        {
            int from = relations[i][0];
            int to = relations[i][1];
            graph[from].add(to);
        }
        return graph;
    }
}

class Node
{
    int courseNum;
    int currentCost;
    
    Node(int courseNum, int currentCost)
    {
        this.courseNum = courseNum;
        this.currentCost = currentCost;
    }
}

Version 2: DFS+Memorazition (JAVA --> TLE)

If we have N nodes & E edges or relationships, then: Time Complexity: O(N+E) Space Complexity: O(N+E)

class Solution {
    
    private int[] memo;
    public int minimumTime(int n, int[][] relations, int[] time) 
    {
        int res = 0;
        //when end is n, the max time it took
        memo = new int[n+1];
        List<Integer>[] graph = buildgrah(n, relations);
    
        for(int i = 1; i <=n; i++)
        {
            dfs(i, graph, time);
        }
        
        for(int i = 0; i<=n; i++)
        {
            res = Math.max(res, memo[i]);
        }
        return res;
    }
    
    // maxtime(i) = max of(child)+ times[i-1]
    public int dfs(int current, List<Integer>[] graph, int[] time)
    {
        if(graph[current].size() == 0)
        {
            memo[current] = time[current-1];
            return memo[current];
        }
        
        //memo[current] = max(all of child)+ time[current-1];
        for(int nextNode : graph[current])
        {
            int nextNodeCost = dfs(nextNode, graph, time);
            memo[current] = Math.max(memo[current], nextNodeCost + time[current-1]);
        }
        
        return memo[current];
    }
    
    public List<Integer>[] buildgrah(int n, int[][] relations)
    {
        List<Integer>[] graph = new ArrayList[n+1];
        for(int i = 0; i<= n; i++)
        {
            graph[i] = new ArrayList<Integer>();
        }
        for(int i = 0; i< relations.length; i++)
        {
            int from = relations[i][0];
            int to = relations[i][1];
            graph[from].add(to);
        }
        return graph;
    }
        
       
}