Parallel Courses III

class Solution:
    def minimumTime(self, n: int, relations: List[List[int]], time: List[int]) -> int:
        graph = defaultdict(list)
        inDegree = [0] * n
        for previous, nekst in relations:
            previous, nekst = previous - 1, nekst - 1  # convert into zero-based index
            inDegree[nekst] += 1

        q = deque([])
        distance = [0] * n
        for u in range(n):
            if inDegree[u] == 0:
                distance[u] = time[u]

        while q:
            u = q.popleft()
            for v in graph[u]:
                distance[v] = max(distance[u] + time[v], distance[v])  # Update `dist[v]` using the maximum dist of the predecessor nodes
                inDegree[v] -= 1
                if inDegree[v] == 0:
        return max(distance)

Problem Classification

Based on the analysis of the given problem statement, here are the categorizations:

Domain: Graph theory, algorithms

What components:

  • Directed acyclic graph (DAG) representing courses and prerequisites
  • Nodes represent courses numbered 1 to n
  • Edges represent prerequisite relationships
  • time[] array represents time to complete each course
  • Need to find minimum total time to complete all courses

Problem classification:

  • Topological sorting problem on a DAG
  • Need to find longest path in DAG considering edge weights (course times)
  • Graph traversal (BFS) to generate topological ordering
  • Dynamic programming to calculate minimum time for each course based on prerequisites

So in summary, this is a graph theory and algorithms problem that involves topological sorting, graph traversal, and dynamic programming to find the longest path in a directed acyclic graph. The core techniques used would be BFS, topological ordering, DP tabulation to handle prerequisite constraints and calculate optimal times.

Visual Model of the Problem

We can visualize the problem statement as follows:

  1. Draw a directed acyclic graph (DAG) with n nodes representing the courses numbered 1 to n.

  2. Add directed edges between nodes representing the prerequisite relationships from relations[]. An edge from node A to B denotes A is a prerequisite for B.

  3. Add weights on each node representing the time taken to complete that course from time[].

  4. The objectives are:

    • We need to find an ordering of courses that satisfy prerequisites (topological order of DAG)

    • We want to minimize the total time required to complete all courses (longest path in DAG considering edge weights)

  5. Maintain a table that stores minimum time needed to complete each course considering prerequisites. Initialize all times to 0.

  6. Do a topological traversal and whenever visiting a node, update its minimum time as max(current time, maximum prerequisite time) + its own time.

  7. The final minimum time needed will be the maximum time in the table after traversing all nodes.

So in summary, we can visualize this problem as finding the longest path in a weighted DAG while satisfying precedence constraints. The topological ordering gives a valid course schedule and the table stores optimal times respecting prerequisites.

Problem Restatement

Here is how I would paraphrase the problem statement in my own words:

We are given a set of n courses numbered from 1 to n. These courses have prerequisite relationships, meaning some courses need to be completed before others can be taken. These prerequisite relationships are provided as pairs of courses in the relations 2D array, where relations[i] = [A, B] means course A needs to be completed before course B.

Additionally, each course takes a specific number of months to complete. This is given by the time array, where time[i] represents the months needed to complete the (i+1)th course.

Our goal is to find the minimum total number of months required to complete all the n courses. The constraints are:

  • We can take multiple courses in parallel, as long as the prerequisites of those courses have been met.

  • We can start a course at any time, as long as its prerequisites have already been completed.

  • The given relationships form a directed acyclic graph, meaning there are no cyclic dependencies and an ordering of courses satisfying all prerequisites exists.

So in summary, we need to schedule the courses in a optimal way such that we minimize the total time required, while satisfying all prerequisite constraints. This involves using the relations array to create a DAG, then finding a topological ordering of courses and calculating the critical path through the DAG that gives the minimum months required.

Abstract Representation of the Problem

Here is one way to formulate an abstract representation of this problem:

We are given a directed acyclic graph (DAG) G = (V, E) where:

  • V is the set of vertices representing tasks to be completed
  • E is the set of directed edges representing dependencies between tasks
  • Each vertex v has a weight w(v) representing the time required to complete that task

We are also given a source vertex s and sink vertex t in the graph.

The problem is to find the longest path from s to t in G, where the length of a path is defined as the sum of the weights of the vertices along that path.

This can be formulated abstractly as:


  • DAG G = (V, E)
  • Weight function w: V → N
  • Source vertex s
  • Sink vertex t

Output: Length of longest path from s to t where length is defined as:

length(path) = Σ w(v) for all v in path

To summarize, this abstraction focuses on the graph structure, dependencies, and weights, while extracting away the specifics of courses and prerequisites. It generalizes the problem as finding the longest path in a weighted DAG.


Here are some key technical concepts relevant to this problem:

  • Directed Acyclic Graph (DAG): A graph with directed edges and no cycles. Used to model prerequisite relationships.

  • Topological ordering: An ordering of vertices in a DAG such that if there is a path from vertex A to vertex B, then A appears before B in the ordering. Needed to validate ordering of courses satisfies prerequisites.

  • Indegree/Outdegree: Number of incoming/outgoing edges for a vertex. Used to find vertices with no prerequisites.

  • Adjacency list: Compact graph representation as a map from vertices to their adjacent vertices. Enables prerequisite lookup.

  • BFS traversal: Breadth-first search algorithm to traverse a graph. Used to generate topological ordering.

  • Dynamic programming: Method to solve problems by combining pre-computed solutions. Helps optimize total time based on prerequisite times.

  • Longest path problem: Finding the longest path between two vertices in a graph. Equivalent to minimizing total course time.

  • Critical path: The longest path from start to end in a weighted DAG. Gives optimal schedule length.

The key ideas are using DAG structure to model prerequisites, topological ordering to validate course order, BFS traversal for ordering, DP to optimize time with dependencies, and longest path to find minimum duration.

Problem Simplification and Explanation

Here is one way to break down this course scheduling problem into simpler terms and concepts:

Key concepts:

  • Courses = Nodes
  • Prerequisites = Edges
  • Months to complete = Node weights
  • Total minimum months = Longest path


Think of the courses as different cities on a map. The prerequisites are directional roads connecting the cities. The months to complete each course are the distances between those cities.

Your goal is to travel from your starting city (course 1) to the destination city (course n) visiting all cities in between only once. However, you must follow the directional roads to respect the prerequisites.

What’s the shortest route that allows you to visit all cities exactly once, respecting the directional roads? This is like finding the longest path in the course prerequisite graph to minimize total months.

Simplified steps:

  1. Model courses and prerequisites as a directed graph
  2. Assign months to complete as node weights
  3. Topologically sort graph to get valid course order
  4. Calculate longest path in graph using weights

So in summary, we simplify the problem by focusing on modeling it as a graph, topological ordering, and calculating the longest path on that graph. The city travel analogy helps relate it to a more common scenario.


Here are some specific characteristics in the problem statement that could help design an efficient solution:

  • Directed Acyclic Graph structure: Since prerequisites form a DAG, we can topological sort the graph to get a valid ordering of courses. This avoids backtracking.

  • Edge constraints: The number of edges is <= n(n-1)/2 and <= 5*104. This means the graph is sparse. So an adjacency list will be efficient vs an adjacency matrix.

  • Vertex constraints: Number of nodes/courses is <= 5*104. This means brute force approaches may still be feasible.

  • Integer weights: The time to complete each course is an integer between 1 and 104. This small discrete range allows preprocessing the weights for fast lookups.

  • Unique edges: All prerequisite pairs are unique. This avoids duplicate edges which could complicate the graph structure.

  • Total nodes known: The number of courses/nodes n is given. So we can allocate data structures like adjacency list, indegree arrays etc of size n upfront.

  • DAG guarantees no cycles: We don’t need cycle detection logic. The topological ordering is guaranteed to succeed.

So in summary, the DAG structure, small discrete weights, sparse unique edges, and known number of nodes allow efficient data structures like adjacency list, no cycle logic, and potential dynamic programming optimizations.

Here are the key insights from analyzing the constraints of this problem:

  1. Directed Acyclic Graph structure:
  • Since the prerequisites form a DAG, we are guaranteed there will be no cycles. This means a topological ordering of courses that satisfies prerequisites must exist.

  • We can use a graph algorithm like BFS to traverse and topo sort the DAG. No need for cycle detection.

  1. Bounded constraints on n, edges:
  • The upper bounds on number of courses n and max edges means the graph will be sparse. So adjacency list is efficient.

  • Bounded n also allows allocating data structures like indegree array of size n upfront.

  1. Unique edges:
  • All prerequisite relationships are unique. This avoids duplicate edges and simplifies the graph structure.
  1. Discrete time values:
  • Times per course are integers in a small range. This allows preprocessing weights for fast lookup.
  1. Can take multiple courses in parallel:
  • No limit on number of courses in a semester. Allows optimizing using longest path.
  1. Can start course anytime if prerequisites done:
  • Enables modeling as DAG with topological ordering.

In summary, the constraints on size, edges, discrete times, parallel courses, and DAG structure allow efficient data structures, graph algorithms, preprocessing, and optimizations for this problem. The key is exploiting the sparse DAG structure.

Case Analysis

Here are some additional examples covering different aspects and conditions of the problem:

Example 1: DAG with long chain of prerequisites

Input: n = 10 relations = [[1,2],[2,3],[3,4],[4,5],[5,6],[6,7],[7,8],[8,9],[9,10]] time = [1,1,1,1,1,1,1,1,1,1]

Output: 10

Reasoning: This tests a long chain of prerequisites. The total time will be the number of courses since they are all sequential.

Example 2: Star DAG with one root prerequisite

Input: n = 10 relations = [[1,2],[1,3],[1,4],[1,5],[1,6],[1,7],[1,8],[1,9],[1,10]] time = [2,1,1,1,1,1,1,1,1,1]

Output: 11

Reasoning: One course is prerequisite for all other courses. Total time depends on the root course time.

Example 3: Completely parallel courses

Input: n = 10 relations = []
time = [1,2,3,4,5,6,7,8,9,10]

Output: 10

Reasoning: No prerequisites so all courses can be done in parallel based on max time course.

Example 4: Cyclic prerequisites

Input: n = 3 relations = [[1,2],[2,3],[3,1]] time = [1,1,1]

Output: Not feasible (cyclic prerequisites)

Reasoning: Tests case where prerequisites have cycles so infeasible.

This covers chains, star patterns, parallel, cyclic cases and different time values. Computing on paper helps validate expected outputs.

Identification of Applicable Theoretical Concepts

Here are some mathematical and algorithmic concepts that can help simplify and solve this course scheduling problem more efficiently:

  • Graph theory: The prerequisites form a directed acyclic graph (DAG). We can leverage graph algorithms and theorems.

  • Topological ordering: As the graph is a DAG, we are guaranteed to be able to generate a topological ordering of vertices that satisfies dependencies. This gives a valid course order.

  • BFS traversal: We can use breadth-first search, a common graph algorithm, to traverse the DAG and generate the topological ordering.

  • Adjacency list: Storing the graph as an adjacency list (map from vertex to its neighbors) allows efficient prerequisite lookups and iterations.

  • Indegree/Outdegree: Calculating indegree and outdegree helps identify courses with no prerequisites to start traversing the graph.

  • Dynamic programming: We can optimize the total time needed in a bottom-up manner using DP. Store optimal subproblem solutions and build on them.

  • Longest path algorithms: The problem equates to finding the longest path in a weighted DAG. We can adapt algorithms like Dijkstra’s to calculate longest path length.

So in summary, we can leverage graph theory, graph algorithms like BFS and topological sort, data structures like adjacency list, and dynamic programming to develop an efficient solution.

Problem Breakdown and Solution Methodology

Here is a step-by-step approach to solving this course scheduling problem:

  1. Build the course prerequisite graph:
  • Represent each course as a vertex and each prerequisite as a directed edge. This forms a DAG.

  • Store the graph using an adjacency list structure for efficient lookups.

  1. Initialize indegree and topological order list:
  • Indegree counts number of prerequisites for each course. Helps find courses to start with.

  • Topological order list will store the final valid ordering of courses.

  1. Find all courses with indegree 0 and add them to a queue:
  • Courses with 0 indegree have no prerequisites. They can start first.
  1. Repeat until queue is empty:
  • Pop a vertex from queue and append it to topological order list.

  • Decrease indegree of all its neighbors by 1.

  • If a neighbor’s indegree becomes 0, add to queue.

  1. Calculate longest path:
  • Once topological order is generated, calculate longest path from first to last course using course times and dynamic programming.

  • Store optimal time for each previous subproblem. Build on it to get final solution.

Let’s look at a simple example:

Courses: A, B, C, D Prereqs: A -> B, B -> C, A -> D Times: A = 2, B = 4, C = 3, D = 1

We first create the DAG. Then topological sort gives [A, D, B, C]. Finally, longest path is A(2) -> D(1) -> C(3) -> B(4). Total time = 2 + 1 + 3 + 4 = 10.

If times change, only the longest path calculation is affected. If prerequisites change, the topological order and longest path need to be recomputed.

Inference of Problem-Solving Approach from the Problem Statement

How did you infer from the problem statement that this problem can be solved using ?

Stepwise Refinement

Here is a stepwise refinement for solving this course scheduling problem:

High-level approach:

  1. Model courses and prerequisites as a directed graph
  2. Topologically sort graph to get valid ordering
  3. Calculate longest path in graph to minimize total time

Granular steps:

  1. Build graph 1.1 Create adjacency list to represent prerequisites 1.2 Add nodes for courses 1.3 Add directed edges for prerequisites

  2. Topological sort 2.1 Calculate indegrees of each node 2.2 Add nodes with 0 indegree to queue 2.3 While queue not empty: 2.3.1 Pop node from queue 2.3.2 Append popped node to topological order 2.3.3 Reduce indegree of all child nodes 2.3.4 Add child nodes with 0 indegree to queue

  3. Longest path 3.1 Initialize dist[] array with time for each node 3.2 For each node v in topological order: 3.2.1 For each neighbor u of v: dist[u] = max(dist[u], dist[v] + time[u])

Independent parts:

  • Building graph
  • Topological sorting
  • Calculating longest path

These can be developed and tested separately.

Repeating patterns:

  • Graph traversal (BFS)
  • Dynamic programming to calculate optimal values
  • Using topological order for valid sequencing

So in summary, we break down the solution into modular steps that can be developed independently. We leverage repeating patterns like graph algorithms, DP, and topological sorting.

Solution Approach and Analysis

Here is a detailed step-by-step approach to solve this course scheduling problem:

  1. Build dependency graph:
  • Represent courses as vertices and prerequisites as directed edges.

  • An adjacency list stores each vertex’s outgoing edges efficiently.

  • Analogous to cities and directional roads on a map.

  1. Topologically sort graph:
  • Calculates indegree of each vertex (number of prerequisites).

  • Courses with indegree 0 are added to a queue to start traversal.

  • Like identifying origin cities with no incoming roads.

  • Traverse graph using BFS, reducing indegree of child vertices.

  • Append each vertex to topological order after visit.

  • Ensures prerequisites are satisfied before next course.

  1. Calculate longest path:
  • Dynamic programming to calculate longest path using course times and topological order.

  • Maintain array dist[] storing optimal time to reach each vertex.

  • Iterate vertices in topological order:

    • For each neighbor, update its dist if longer path found.
  • Analogous to calculating optimal travel times between cities.

  • dist[last] gives final minimum time needed.

Let’s walk through an example:

Courses: A, B, C, D
Prereqs: A->B, B->C, A->D Times: A = 2, B = 4, C = 3, D = 1


A->B A->D B->C

Topological order: A, D, B, C

dist[] initialized to course times. For A, update dist[B] = max(dist[B], dist[A] + time[B]) = 6 update dist[D] = max(dist[D], dist[A] + time[D]) = 3

For D, no updates needed

For B, update dist[C] = max(dist[C], dist[B] + time[C]) = 7

Final min time = dist[C] = 7

The steps are modular, allow incremental development, and can be modified independently. Example illustrates the approach concretely.

Thought Process

Here is my thought process for solving this course scheduling problem step-by-step:

Problem Statement Cues:

  • Courses as nodes, prerequisites as directed edges => Represent as directed graph
  • Find minimum months to complete all courses => Think shortest/longest paths
  • Can take multiple courses together => No dependencies between parallel courses
  • Take course anytime if prerequisites done => Topological ordering will give valid schedule

Approach Direction:

  1. Represent courses and prerequisites as a directed acyclic graph (DAG)

  2. Topologically sort the DAG to generate a valid ordering of courses that satisfies prerequisites

  3. Calculate the longest path in the DAG from the first to last course to minimize total months


from collections import defaultdict

# Build graph
graph = defaultdict(list)
for pre, next in relations:

# Topological sort  
def topoSort(graph):
  indegree = {i:0 for i in range(1, n+1)}
  for i in graph:
    for j in graph[i]:
      indegree[j] += 1

  queue = []  
  for i in range(1, n+1):
    if indegree[i] == 0:

  order = []

  while queue:
    curr = queue.pop(0)

    for neighbor in graph[curr]:
      indegree[neighbor] -= 1
      if indegree[neighbor] == 0:

  return order

# Calculate longest path 
dist = [0] * (n+1)
order = topoSort(graph)

for i in order:
  if dist[i] + time[i-1] > dist[i]:
    dist[i] += time[i-1]
  for j in graph[i]:
    dist[j] = max(dist[j], dist[i] + time[j-1])

return dist[n]  

This follows the step-by-step approach derived from the problem statement cues. The topological sort gives valid order and longest path gives minimum months.


From Brute Force to Optimal Solution

Here is a brute force approach and how we can optimize it:

Brute force:

  • Try all possible orderings of courses
  • For each ordering, check if it satisfies prerequisites
  • Calculate total months for valid orderings
  • Return minimum total months

This evaluates n! orderings in worst case.


  1. Prune invalid orderings:
  • Build a prerequisite graph
  • Only consider orderings that satisfy graph edges

Reduces orderings from n! to valid topological orderings.

  1. Efficiently get valid orderings:
  • Use BFS traversal on prerequisite graph
  • Append visited course to topological ordering

Generates valid orderings without having to check all permutations.

  1. Calculate months efficiently:
  • Use dynamic programming
  • Store results of subproblems
  • Reuse instead of recomputing

Avoids recalculating month counts, improves to linear time.

  1. Use adjacency list for graph:
  • Speeds up neighbor lookups
  • More efficient than adjacency matrix for sparse graph
  1. Preprocess indegrees:
  • Initialize indegree counts once
  • Speeds up finding courses to start with

These optimizations improve the complexity from O(n! * n) to:

Time: O(V+E) for BFS traversal + O(V) for DP Space: O(V+E) for graph and DP table

By exploiting the underlying graph structure and using algorithms like topological sort, BFS, DP, and adjacency list, we get a much faster solution.

Coding Constructs

Here are my answers to the high-level analysis of this course scheduling code:

  1. It uses graph algorithms, dynamic programming, and topological sorting as the high-level problem-solving techniques.

  2. I would explain that this code models the course prerequisites as a graph, topologically sorts the graph to get a valid ordering, and uses dynamic programming to calculate the fastest way to complete all courses.

  3. The logical constructs used are graphs, queues, adjacency lists, arrays, loops, and recursion.

  4. The algorithm follows these steps at a high level: build a directed graph representing prerequisites, topologically sort the graph using BFS to get a valid ordering, calculate the longest path in the graph using dynamic programming to minimize total time.

  5. The key steps are building an adjacency list to represent the graph, initializing indegrees and queues, reducing indegrees and enqueuing nodes during BFS traversal, generating the topological order, initializing the DP array, and iteratively updating the optimal times for each node by considering incoming edges. This achieves both a valid ordering and optimal schedule length.

  6. The algorithmic patterns and strategies are topological sorting of a DAG using BFS, calculating indegrees, tabular dynamic programming to find longest path, and using an adjacency list for efficient representation of a sparse graph.

Overall, the code cleverly combines common graph algorithms and DP to solve the optimization problem of minimizing course schedule length given precedence constraints.

Language Agnostic Coding Drills

Here is a deconstruction of the key concepts in this code:

Coding Concepts in Increasing Order of Difficulty:

  1. Array initialization - Creating arrays to store vertex distances and indegrees. Easy.

  2. Adjacency list - Storing graph as nested hashmap from node to neighbors. Easy.

  3. Queue - Adding/removing nodes from queue for BFS traversal. Easy.

  4. Looping - Iterating over nodes and edges. Easy.

  5. Variable assignment - Storing intermediate values. Easy.

  6. Indegree calculation - Counting incoming edges per node. Medium.

  7. BFS traversal - Visiting nodes level-by-level to topo sort. Medium.

  8. Topological ordering - Appending visited nodes to output list. Medium.

  9. Recursion - Updating adjacent nodes during traversal. Medium.

  10. Dynamic programming - Storing subproblem solutions to optimize final result. Hard.

  11. Longest path calculation - Updating optimal distance incrementally. Hard.

Problem-solving approach:

  1. Initialize graph, distances, indegrees.

  2. Calculate indegrees of each node.

  3. Add nodes of 0 indegree to queue.

  4. Start BFS traversal:

    • Decrement neighbor indegrees and enqueue if 0.
    • Append visited nodes to topological order.
  5. Iterate over nodes in topological order:

    • Update optimal distances of adjacent nodes.
  6. Return longest distance value.

So in summary, we break down the key coding concepts, order them by difficulty, and use them in a step-by-step problem-solving approach focused on graph algorithms and dynamic programming.

Targeted Drills in Python

Here are some Python coding drills for the key concepts:

  1. Array initialization
distances = [0] * num_nodes 
indegrees = [0] * num_nodes
  1. Adjacency list
graph = defaultdict(list)
  1. Queue
from collections import deque

q = deque()
  1. Looping
for i in range(num_nodes):

for child in graph[node]:
  # process child 
  1. Variable assignment
child = graph[node][0]
total = 0
  1. Indegree calculation
for node in graph:
  for child in graph[node]:
    indegrees[child] += 1
  1. BFS traversal
visited = set()

while q:
   node = q.popleft()

   for child in graph[node]:
  1. Topological ordering
order = []

for node in visited:
  1. Dynamic programming
dp = [0] * num_nodes

for node in order:
   for child in graph[node]:
      dp[child] = max(dp[child], dp[node] + weight) 

These can be integrated in order to build the full solution using graph algorithms and dynamic programming for optimizing the schedule.


Similar Problems

Given the problem , identify and list down 10 similar problems on LeetCode. These should cover similar concepts or require similar problem-solving approaches as the provided problem. Please also give a brief reason as to why you think each problem is similar to the given problem.