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Java-based LeetCode algorithm problem solutions, regularly updated
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package g2001_2100.s2050_parallel_courses_iii;

// #Hard #Dynamic_Programming #Graph #Topological_Sort
// #2022_05_26_Time_33_ms_(86.04%)_Space_127.1_MB_(28.87%)

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
import java.util.Queue;

/**
 * 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] = [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:**
 *
 * **![](https://assets.leetcode.com/uploads/2021/10/07/ex1.png)**
 *
 * **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:**
 *
 * **![](https://assets.leetcode.com/uploads/2021/10/07/ex2.png)**
 *
 * **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 <= prevCourse_j, nextCourse_j <= n
 * *   prevCourse_j != nextCourse_j
 * *   All the pairs [prevCourse_j, nextCourse_j] are **unique**.
 * *   `time.length == n`
 * *   1 <= time[i] <= 10⁴
 * *   The given graph is a directed acyclic graph.
**/
public class Solution {
    public int minimumTime(int n, int[][] relations, int[] time) {
        int v = time.length;
        List> adj = new ArrayList<>();
        for (int i = 0; i < v; i++) {
            adj.add(new ArrayList<>());
        }
        int[] indegree = new int[v];
        int[] requiredTime = new int[v];
        for (int[] relation : relations) {
            List vertices = adj.get(relation[0] - 1);
            vertices.add(relation[1] - 1);
            indegree[relation[1] - 1]++;
        }

        Queue q = new ArrayDeque<>();
        for (int i = 0; i < v; i++) {
            if (indegree[i] == 0) {
                q.add(i);
                requiredTime[i] = time[i];
            }
        }

        while (!q.isEmpty()) {
            int vertex = q.poll();
            List edges = adj.get(vertex);
            for (Integer e : edges) {
                indegree[e]--;
                if (indegree[e] == 0) {
                    q.add(e);
                }
                int totalTime = time[e] + requiredTime[vertex];
                if (requiredTime[e] < totalTime) {
                    requiredTime[e] = totalTime;
                }
            }
        }
        int maxMonth = 0;
        for (int i = 0; i < n; i++) {
            maxMonth = Math.max(maxMonth, requiredTime[i]);
        }
        return maxMonth;
    }
}