Parallel Courses III solution and explanation
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The problem
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 * 1040 <= relations.length <= min(n * (n - 1) / 2, 5 * 104)relations[j].length == 21 <= prevCoursej, nextCoursej <= nprevCoursej != nextCoursej- All the pairs
[prevCoursej, nextCoursej]are unique. time.length == n1 <= time[i] <= 104- The given graph is a directed acyclic graph.
Approach overview
Problem Overview: You are given n courses where each course takes a specific amount of time to complete. Some courses depend on others through prerequisite relations. Multiple courses can run in parallel as long as their prerequisites are finished. The task is to compute the minimum time required to complete all courses.
Approach 1: Topological Sort with Dynamic Programming (O(n + m) time, O(n + m) space)
This problem naturally maps to a directed acyclic graph (DAG). Each course is a node, and each prerequisite relation is a directed edge. The key idea is that a course can only start after all prerequisite courses finish, so the completion time becomes the maximum finish time among its prerequisites plus its own duration.
Build an adjacency list and track the in-degree of each node. Run Kahn’s algorithm for topological sort using a queue. Maintain a DP array where dp[i] stores the earliest time course i can finish. Initialize it with the course duration. When processing an edge u → v, update dp[v] = max(dp[v], dp[u] + time[v]). This ensures the longest prerequisite chain determines the start time.
Once all nodes are processed, the answer is the maximum value in the DP array. This approach works because a topological order guarantees that all prerequisites of a course are processed before the course itself. The algorithm runs in linear time relative to nodes and edges, making it ideal for large DAGs commonly seen in scheduling problems on graph structures.
Approach 2: DFS with Memoization (O(n + m) time, O(n + m) space)
An alternative approach computes the longest path in the DAG using depth-first search. Treat each course as a node and recursively calculate the total time required to complete it. The DFS explores all prerequisite chains and returns the maximum time among them plus the current course duration.
Memoization avoids recomputing results. Maintain a cache where memo[i] stores the time required to finish course i including all prerequisites. During DFS, iterate through neighbors in the adjacency list and compute time[i] + max(dfs(next)). If a value already exists in the memo table, reuse it immediately.
This effectively finds the longest path in a DAG, a common pattern in dynamic programming on graphs. The recursion depth corresponds to prerequisite chains, while memoization guarantees each node is evaluated once.
Recommended for interviews: Topological Sort with Dynamic Programming is the approach most interviewers expect. It clearly models dependency resolution and scheduling while staying iterative and efficient. Implementing DFS with memoization also works and demonstrates understanding of longest-path problems in DAGs, but the topological approach tends to be easier to reason about and debug under interview constraints.
Approach 1: Topological Sort with Dynamic Programming
This approach involves topologically sorting the courses based on their prerequisites and then calculating the minimal time to complete each course using dynamic programming.
- Create a directed graph where each course is a node, and a prerequisite relationship
(a,b)is a directed edge fromatob. - Perform a topological sort on the courses. This step ensures that all prerequisites of a course are considered before the course itself.
- Use dynamic programming to calculate the minimum time required to complete each course by iteratively adding its prerequisite's completion time.
```python
def minTime(n, relations, time):
from collections import deque, defaultdict
adj = defaultdict(list)
indegree = [0] * (n + 1)
dp = [0] * (n + 1)
for u, v in relations:
adj[u].append(v)
indegree[v] += 1
queue = deque()
for i in range(1, n + 1):
if indegree[i] == 0:
queue.append(i)
dp[i] = time[i - 1]
while queue:
u = queue.popleft()
for v in adj[u]:
dp[v] = max(dp[v], dp[u] + time[v - 1])
indegree[v] -= 1
if indegree[v] == 0:
queue.append(v)
return max(dp)
# Example usage:
n = 5
relations = [[1, 5], [2, 5], [3, 5], [3, 4], [4, 5]]
time = [1, 2, 3, 4, 5]
print(minTime(n, relations, time)) # Output: 12
``````java
import java.util.*;
public class ParallelCourses {
public static int minTime(int n, int[][] relations, int[] time) {
List<Integer>[] adj = new ArrayList[n+1];
int[] indegree = new int[n+1];
int[] dp = new int[n+1];
for (int i = 0; i <= n; i++) {
adj[i] = new ArrayList<>();
}
for (int[] relation : relations) {
adj[relation[0]].add(relation[1]);
indegree[relation[1]]++;
}
Queue<Integer> q = new LinkedList<>();
for (int i = 1; i <= n; i++) {
if (indegree[i] == 0) {
q.offer(i);
dp[i] = time[i-1];
}
}
while (!q.isEmpty()) {
int u = q.poll();
for (int v : adj[u]) {
dp[v] = Math.max(dp[v], dp[u] + time[v-1]);
if (--indegree[v] == 0) {
q.offer(v);
}
}
}
int maxTime = 0;
for (int t : dp) {
maxTime = Math.max(maxTime, t);
}
return maxTime;
}
public static void main(String[] args) {
int n = 5;
int[][] relations = {{1,5},{2,5},{3,5},{3,4},{4,5}};
int[] time = {1, 2, 3, 4, 5};
System.out.println(minTime(n, relations, time));
}
}
``````cpp
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
int minTime(int n, vector<vector<int>>& relations, vector<int>& time) {
vector<int> adj[n+1];
vector<int> indegree(n+1, 0);
vector<int> dp(n+1, 0);
for (const auto& rel : relations) {
adj[rel[0]].push_back(rel[1]);
indegree[rel[1]]++;
}
queue<int> q;
for (int i = 1; i <= n; ++i) {
if (indegree[i] == 0) {
q.push(i);
dp[i] = time[i-1];
}
}
while (!q.empty()) {
int u = q.front(); q.pop();
for (int v : adj[u]) {
dp[v] = max(dp[v], dp[u] + time[v-1]);
if (--indegree[v] == 0) {
q.push(v);
}
}
}
return *max_element(dp.begin(), dp.end());
}
int main() {
int n = 5;
vector<vector<int>> relations = {{1,5},{2,5},{3,5},{3,4},{4,5}};
vector<int> time = {1, 2, 3, 4, 5};
cout << minTime(n, relations, time) << endl;
return 0;
}
``````javascript
function minTime(n, relations, time) {
const indegree = new Array(n + 1).fill(0);
const adj = Array.from({ length: n + 1 }, () => []);
const dp = new Array(n + 1).fill(0);
for (const [u, v] of relations) {
adj[u].push(v);
indegree[v]++;
}
const queue = [];
for (let i = 1; i <= n; i++) {
if (indegree[i] === 0) {
queue.push(i);
dp[i] = time[i - 1];
}
}
while (queue.length > 0) {
const u = queue.shift();
for (const v of adj[u]) {
dp[v] = Math.max(dp[v], dp[u] + time[v - 1]);
indegree[v]--;
if (indegree[v] === 0) {
queue.push(v);
}
}
}
return Math.max(...dp);
}
// Example usage:
const n = 5;
const relations = [[1, 5], [2, 5], [3, 5], [3, 4], [4, 5]];
const time = [1, 2, 3, 4, 5];
console.log(minTime(n, relations, time)); // Output: 12
``````c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 50001
int minMonths(int n, int relationsSize, int** relations, int* time){
int indegree[n+1];
memset(indegree, 0, sizeof(indegree));
int* adj[n+1];
int adjSize[n+1];
memset(adjSize, 0, sizeof(adjSize));
for(int i = 0; i < n; i++)
adj[i + 1] = (int*)malloc((n + 1) * sizeof(int));
for(int i = 0; i < relationsSize; i++){
int u = relations[i][0];
int v = relations[i][1];
adj[u][adjSize[u]++] = v;
indegree[v]++;
}
int dp[n+1];
memcpy(dp, time, n * sizeof(int));
int queue[n];
int front = 0, rear = -1;
for(int i = 1; i <= n; i++)
if(indegree[i] == 0) queue[++rear] = i;
while(front <= rear){
int u = queue[front++];
for(int i = 0; i < adjSize[u]; i++){
int v = adj[u][i];
dp[v] = (dp[v] > dp[u] + time[v-1]) ? dp[v] : dp[u] + time[v-1];
if(--indegree[v] == 0) queue[++rear] = v;
}
}
int maxTime = 0;
for(int i = 1; i <= n; i++)
if(dp[i] > maxTime) maxTime = dp[i];
for(int i = 0; i < n; i++)
free(adj[i + 1]);
return maxTime;
}
int main() {
// Example usage
int n = 3;
int relations[2][2] = {{1,3},{2,3}};
int* relationsPtrs[2] = {relations[0], relations[1]};
int time[] = {3, 2, 5};
int result = minMonths(n, 2, relationsPtrs, time);
printf("%d\n", result);
return 0;
}
``````csharp
using System;
using System.Collections.Generic;
public class ParallelCourses {
public static int MinTime(int n, int[][] relations, int[] time) {
var adj = new List<int>[n + 1];
var indegree = new int[n + 1];
var dp = new int[n + 1];
for (int i = 0; i <= n; i++)
adj[i] = new List<int>();
foreach (var relation in relations) {
int u = relation[0], v = relation[1];
adj[u].Add(v);
indegree[v]++;
}
var queue = new Queue<int>();
for (int i = 1; i <= n; i++) {
if (indegree[i] == 0) {
queue.Enqueue(i);
dp[i] = time[i - 1];
}
}
while (queue.Count > 0) {
int u = queue.Dequeue();
foreach (var v in adj[u]) {
dp[v] = Math.Max(dp[v], dp[u] + time[v - 1]);
indegree[v]--;
if (indegree[v] == 0) {
queue.Enqueue(v);
}
}
}
int maxTime = 0;
for (int i = 1; i <= n; i++)
maxTime = Math.Max(maxTime, dp[i]);
return maxTime;
}
public static void Main() {
int n = 5;
int[][] relations = new int[][] { new int[] { 1,5 }, new int[] { 2,5 }, new int[] { 3,5 }, new int[] { 3,4 }, new int[] { 4,5 } };
int[] time = new int[] { 1, 2, 3, 4, 5 };
Console.WriteLine(MinTime(n, relations, time)); // Output: 12
}
}
```This Python solution uses collections to define adjacency lists and maintains indegrees for each node. It efficiently processes courses using a deque for queue operations and updates the time required using a dp array, finally finding the maximum time needed to complete all courses.
Time Complexity: O(n + E), where the algorithm works through each course and relation once.
Space Complexity: O(n + E), for storing graph details and additional data arrays.
Approach 2: DFS with Memorization
This approach uses Depth-First Search (DFS) with memorization to efficiently compute the completion time for courses.
- Model the courses and prerequisites as a directed graph.
- Utilize DFS to explore courses, where each exploration path calculates the time required as we resolve prerequisites recursively.
- Utilize memorization to store already computed completion times, avoiding redundant calculations and ensuring optimal performance.
```python
def dfs(u, adj, dp, time):
if dp[u] != 0:
return dp[u]
dp[u] = time[u - 1]
for v in adj[u]:
dp[u] = max(dp[u], dfs(v, adj, dp, time) + time[u - 1])
return dp[u]
def minTimeDFS(n, relations, time):
from collections import defaultdict
adj = defaultdict(list)
for u, v in relations:
adj[u].append(v)
dp = [0] * (n + 1)
maxTime = 0
for i in range(1, n + 1):
maxTime = max(maxTime, dfs(i, adj, dp, time))
return maxTime
# Example usage:
n = 5
relations = [[1, 5], [2, 5], [3, 5], [3, 4], [4, 5]]
time = [1, 2, 3, 4, 5]
print(minTimeDFS(n, relations, time)) # Output: 12
``````java
import java.util.*;
public class ParallelCoursesDFS {
private static int dfs(int u, List<Integer>[] adj, int[] dp, int[] time) {
if (dp[u] != 0) return dp[u];
dp[u] = time[u - 1];
for (int v : adj[u]) {
dp[u] = Math.max(dp[u], dfs(v, adj, dp, time) + time[u - 1]);
}
return dp[u];
}
public static int minTimeDFS(int n, int[][] relations, int[] time) {
List<Integer>[] adj = new List[n + 1];
for (int i = 0; i <= n; i++)
adj[i] = new ArrayList<>();
for (int[] rel : relations)
adj[rel[0]].add(rel[1]);
int[] dp = new int[n + 1];
int maxTime = 0;
for (int i = 1; i <= n; i++)
maxTime = Math.max(maxTime, dfs(i, adj, dp, time));
return maxTime;
}
public static void main(String[] args) {
int n = 5;
int[][] relations = {{1,5},{2,5},{3,5},{3,4},{4,5}};
int[] time = {1, 2, 3, 4, 5};
System.out.println(minTimeDFS(n, relations, time));
}
}
``````cpp
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int dfs(int u, const vector<vector<int>>& adj, vector<int>& dp, const vector<int>& time) {
if (dp[u] != 0) return dp[u];
dp[u] = time[u - 1];
for (int v : adj[u]) {
dp[u] = max(dp[u], dfs(v, adj, dp, time) + time[u - 1]);
}
return dp[u];
}
int minTimeDFS(int n, vector<vector<int>>& relations, vector<int>& time) {
vector<vector<int>> adj(n + 1);
for (const auto& rel : relations) {
adj[rel[0]].push_back(rel[1]);
}
vector<int> dp(n + 1, 0);
int maxTime = 0;
for (int i = 1; i <= n; ++i) {
maxTime = max(maxTime, dfs(i, adj, dp, time));
}
return maxTime;
}
int main() {
int n = 5;
vector<vector<int>> relations = {{1,5},{2,5},{3,5},{3,4},{4,5}};
vector<int> time = {1, 2, 3, 4, 5};
cout << minTimeDFS(n, relations, time) << endl;
return 0;
}
``````javascript
function dfs(u, adj, dp, time) {
if (dp[u] !== 0) return dp[u];
dp[u] = time[u - 1];
for (const v of adj[u]) {
dp[u] = Math.max(dp[u], dfs(v, adj, dp, time) + time[u - 1]);
}
return dp[u];
}
function minTimeDFS(n, relations, time) {
const indegree = new Array(n + 1).fill(0);
const adj = Array.from({ length: n + 1 }, () => []);
const dp = new Array(n + 1).fill(0);
for (const [u, v] of relations) {
adj[u].push(v);
}
let maxTime = 0;
for (let i = 1; i <= n; i++) {
maxTime = Math.max(maxTime, dfs(i, adj, dp, time));
}
return maxTime;
}
// Example usage:
const n = 5;
const relations = [[1, 5], [2, 5], [3, 5], [3, 4], [4, 5]];
const time = [1, 2, 3, 4, 5];
console.log(minTimeDFS(n, relations, time)); // Output: 12
``````c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 50001
int dfs(int u, int** adj, int* adjSize, int* dp, int* time) {
if (dp[u] != 0) return dp[u];
dp[u] = time[u - 1];
for (int i = 0; i < adjSize[u]; i++) {
int v = adj[u][i];
dp[u] = (dp[u] > dfs(v, adj, adjSize, dp, time) + time[u - 1]) ? dp[u] : dfs(v, adj, adjSize, dp, time) + time[u - 1];
}
return dp[u];
}
int minMonths_dfs(int n, int relationsSize, int** relations, int* time) {
int* adj[n+1];
int adjSize[n+1];
int dp[n+1];
memset(dp, 0, sizeof(dp));
memset(adjSize, 0, sizeof(adjSize));
for(int i = 0; i <= n; i++)
adj[i] = (int*)malloc((n+1) * sizeof(int));
for (int i = 0; i < relationsSize; i++) {
int u = relations[i][0];
int v = relations[i][1];
adj[u][adjSize[u]++] = v;
}
int maxTime = 0;
for (int i = 1; i <= n; i++) {
maxTime = (maxTime > dfs(i, adj, adjSize, dp, time)) ? maxTime : dfs(i, adj, adjSize, dp, time);
}
for (int i = 0; i <= n; i++)
free(adj[i]);
return maxTime;
}
int main() {
int n = 5;
int relations[5][2] = {{1,5},{2,5},{3,5},{3,4},{4,5}};
int* relationsPtrs[5] = {relations[0], relations[1], relations[2], relations[3], relations[4]};
int time[] = {1, 2, 3, 4, 5};
int result = minMonths_dfs(n, 5, relationsPtrs, time);
printf("%d\n", result);
return 0;
}
``````csharp
using System;
using System.Collections.Generic;
public class ParallelCoursesDFS {
private static int DFS(int u, List<int>[] adj, int[] dp, int[] time) {
if (dp[u] != 0) return dp[u];
dp[u] = time[u - 1];
foreach (var v in adj[u]) {
dp[u] = Math.Max(dp[u], DFS(v, adj, dp, time) + time[u - 1]);
}
return dp[u];
}
public static int MinTimeDFS(int n, int[][] relations, int[] time) {
List<int>[] adj = new List<int>[n + 1];
for (int i = 0; i <= n; i++)
adj[i] = new List<int>();
foreach (var relation in relations)
adj[relation[0]].Add(relation[1]);
int[] dp = new int[n + 1];
int maxTime = 0;
for (int i = 1; i <= n; i++)
maxTime = Math.Max(maxTime, DFS(i, adj, dp, time));
return maxTime;
}
public static void Main() {
int n = 5;
int[][] relations = new int[][] { new int[] { 1,5 }, new int[] { 2,5 }, new int[] { 3,5 }, new int[] { 3,4 }, new int[] { 4,5 } };
int[] time = new int[] { 1, 2, 3, 4, 5 };
Console.WriteLine(MinTimeDFS(n, relations, time));
}
}
```This Python implementation effectively uses DFS with a dictionary adj to trace course dependencies, and a dp array to memorize computed outcomes. The result is determined by calculating the maximum time taken from these evaluated paths.
Time Complexity: O(n + E), considering the traversal of the nodes and edges with memoized results.
Space Complexity: O(n + E), related to graph storage and recursion instances within the stack.
Approach 3: Topological Sorting + Dynamic Programming
First, we construct a directed acyclic graph based on the given prerequisite course relationships, perform topological sorting on this graph, and then use dynamic programming to find the minimum time required to complete all courses according to the results of the topological sorting.
We define the following data structures or variables:
- Adjacency list
gstores the directed acyclic graph, and an arrayindegstores the in-degree of each node; - Queue
qstores all nodes with an in-degree of0; - Array
fstores the earliest completion time of each node, initiallyf[i] = 0; - Variable
ansrecords the final answer, initiallyans = 0;
When q is not empty, take out the head node i in turn, traverse each node j in g[i], update f[j] = max(f[j], f[i] + time[j]), update ans = max(ans, f[j]) at the same time, and reduce the in-degree of j by 1. If the in-degree of j is 0 at this time, add j to the queue q;
Finally, return ans.
The time complexity is O(m + n), and the space complexity is O(m + n). Where m is the length of the array relations.
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;
}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
}Complexity comparison
| Approach | Complexity |
|---|---|
| Topological Sort with Dynamic Programming | Time Complexity: O(n + E), where the algorithm works through each course and relation once. |
| DFS with Memorization | Time Complexity: O(n + E), considering the traversal of the nodes and edges with memoized results. |
| Topological Sorting + Dynamic Programming | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Topological Sort with Dynamic Programming | O(n + m) | O(n + m) | Best general solution for DAG scheduling problems with prerequisites |
| DFS with Memoization | O(n + m) | O(n + m) | Useful when modeling the problem as longest path in a DAG using recursion |
Frequently asked questions
Is Parallel Courses III easy or hard?
Parallel Courses III is classified as a Hard problem because it combines multiple concepts: graph modeling, topological sorting, and dynamic programming on DAGs. The difficulty comes from recognizing that the task is essentially finding the longest weighted path in a directed acyclic graph.
Parallel Courses III Python/Java solution
Most implementations follow the same structure across languages: build an adjacency list, compute in-degrees, perform topological sort using a queue, and update a DP array for completion times. The logic is identical in Python, Java, C++, C#, and JavaScript with only syntax differences.
How to solve Parallel Courses III in O(n + m)?
Model the courses as a DAG and perform topological sorting using Kahn’s algorithm. Maintain a DP array storing the earliest completion time for each course. When processing an edge from course u to v, update dp[v] = max(dp[v], dp[u] + time[v]). After processing all nodes, the maximum value in the DP array represents the minimum time to finish all courses.
What is the best approach for Parallel Courses III?
Topological Sort combined with Dynamic Programming is the most efficient and widely accepted solution. Process courses in topological order and maintain the earliest completion time for each course. Each edge updates the dependent course's finish time using the maximum prerequisite completion time. The algorithm runs in O(n + m) time where n is the number of courses and m is the number of prerequisite relations.
Is Parallel Courses III asked at Google/Amazon/Meta?
Graph scheduling and topological sort problems frequently appear in interviews at companies like Google, Amazon, Meta, and Microsoft. Variants of course scheduling and dependency resolution are common system and algorithm questions. This problem tests understanding of DAG processing and dynamic programming.
What data structure is used in Parallel Courses III?
The solution typically uses an adjacency list to represent the graph, a queue for topological sorting, and an array for dynamic programming. The adjacency list efficiently stores prerequisite relationships, while the queue processes nodes with zero in-degree during Kahn’s algorithm.
What is the time complexity of Parallel Courses III?
The optimal solutions run in O(n + m) time and O(n + m) space. Both the topological sort approach and the DFS with memoization approach visit each course and prerequisite relation once. This linear complexity is possible because the course dependency graph forms a directed acyclic graph.