# 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] = [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.

## Solutions

• class Solution {
public int minimumTime(int n, int[][] relations, int[] time) {
List<Integer>[] g = new List[n];
Arrays.setAll(g, k -> new ArrayList<>());
int[] indeg = new int[n];
for (int[] e : relations) {
int a = e[0] - 1, b = e[1] - 1;
++indeg[b];
}
Deque<Integer> q = new ArrayDeque<>();
int[] f = new int[n];
int ans = 0;
for (int i = 0; i < n; ++i) {
int v = indeg[i], t = time[i];
if (v == 0) {
q.offer(i);
f[i] = t;
ans = Math.max(ans, t);
}
}
while (!q.isEmpty()) {
int i = q.pollFirst();
for (int j : g[i]) {
f[j] = Math.max(f[j], f[i] + time[j]);
ans = Math.max(ans, f[j]);
if (--indeg[j] == 0) {
q.offer(j);
}
}
}
return ans;
}
}

• class Solution {
public:
int minimumTime(int n, vector<vector<int>>& relations, vector<int>& time) {
vector<vector<int>> g(n);
vector<int> indeg(n);
for (auto& e : relations) {
int a = e[0] - 1, b = e[1] - 1;
g[a].push_back(b);
++indeg[b];
}
queue<int> q;
vector<int> f(n);
int ans = 0;
for (int i = 0; i < n; ++i) {
int v = indeg[i], t = time[i];
if (v == 0) {
q.push(i);
f[i] = t;
ans = max(ans, t);
}
}
while (!q.empty()) {
int i = q.front();
q.pop();
for (int j : g[i]) {
if (--indeg[j] == 0) {
q.push(j);
}
f[j] = max(f[j], f[i] + time[j]);
ans = max(ans, f[j]);
}
}
return ans;
}
};

• class Solution:
def minimumTime(self, n: int, relations: List[List[int]], time: List[int]) -> int:
g = defaultdict(list)
indeg = [0] * n
for a, b in relations:
g[a - 1].append(b - 1)
indeg[b - 1] += 1
q = deque()
f = [0] * n
ans = 0
for i, (v, t) in enumerate(zip(indeg, time)):
if v == 0:
q.append(i)
f[i] = t
ans = max(ans, t)
while q:
i = q.popleft()
for j in g[i]:
f[j] = max(f[j], f[i] + time[j])
ans = max(ans, f[j])
indeg[j] -= 1
if indeg[j] == 0:
q.append(j)
return ans


• func minimumTime(n int, relations [][]int, time []int) int {
g := make([][]int, n)
indeg := make([]int, n)
for _, e := range relations {
a, b := e[0]-1, e[1]-1
g[a] = append(g[a], b)
indeg[b]++
}
f := make([]int, n)
q := []int{}
ans := 0
for i, v := range indeg {
if v == 0 {
q = append(q, i)
f[i] = time[i]
ans = max(ans, time[i])
}
}
for len(q) > 0 {
i := q[0]
q = q[1:]
for _, j := range g[i] {
indeg[j]--
if indeg[j] == 0 {
q = append(q, j)
}
f[j] = max(f[j], f[i]+time[j])
ans = max(ans, f[j])
}
}
return ans
}

• function minimumTime(n: number, relations: number[][], time: number[]): number {
const g: number[][] = Array(n)
.fill(0)
.map(() => []);
const indeg: number[] = Array(n).fill(0);
for (const [a, b] of relations) {
g[a - 1].push(b - 1);
++indeg[b - 1];
}
const q: number[] = [];
const f: number[] = Array(n).fill(0);
let ans: number = 0;
for (let i = 0; i < n; ++i) {
if (indeg[i] === 0) {
q.push(i);
f[i] = time[i];
ans = Math.max(ans, f[i]);
}
}
while (q.length > 0) {
const i = q.shift()!;
for (const j of g[i]) {
f[j] = Math.max(f[j], f[i] + time[j]);
ans = Math.max(ans, f[j]);
if (--indeg[j] === 0) {
q.push(j);
}
}
}
return ans;
}