Problem Description

You need to take numCourses courses in total, numbered from 0 to numCourses - 1. Some courses have prerequisites that must be completed before you can take them.

You're given an array prerequisites where each element prerequisites[i] = [ai, bi] means that you must complete course bi before taking course ai. For instance, [0, 1] means you need to complete course 1 before you can take course 0.

The task is to determine whether it's possible to complete all courses given these prerequisite constraints. Return true if you can finish all courses, and false if it's impossible.

This problem essentially asks you to detect if there's a circular dependency in the course prerequisites. If course A requires course B, and course B requires course C, and course C requires course A, then it would be impossible to complete all courses. The solution uses topological sorting to detect such cycles by:

1. Building a graph where edges represent prerequisite relationships
2. Tracking the in-degree (number of prerequisites) for each course
3. Starting with courses that have no prerequisites (in-degree of 0)
4. Processing courses level by level, reducing in-degrees of dependent courses
5. Checking if all courses can be processed (no cycles exist)

If we can process all numCourses courses through this method, it means there are no circular dependencies and all courses can be completed.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem involves courses as nodes and prerequisites as directed edges between courses. Each prerequisite [a, b] represents a directed edge from course b to course a.

Is it a tree?

• No: The graph can have multiple paths between nodes and may contain cycles. A course can have multiple prerequisites, and a course can be a prerequisite for multiple other courses.

Is the problem related to directed acyclic graphs?

• Yes: We need to determine if the course dependency graph is acyclic. If there's a cycle in the prerequisites (e.g., course A requires B, B requires C, and C requires A), then it's impossible to complete all courses. The problem is essentially asking whether the directed graph is acyclic (DAG).

Topological Sort

• Yes: The flowchart leads us to Topological Sort, which is perfect for this problem.

Conclusion: The flowchart suggests using Topological Sort for the Course Schedule problem. While the solution shown uses BFS-based topological sorting (Kahn's algorithm), DFS can also be used to perform topological sorting by detecting cycles during the traversal. In DFS approach, we would mark nodes as visiting/visited and if we encounter a node that's currently being visited (in the current DFS path), we've found a cycle.

Intuition

The key insight is to recognize that course prerequisites form a directed graph where each course is a node and each prerequisite relationship is a directed edge. When course b is a prerequisite for course a, we draw an edge from b to a.

The core question becomes: can we find an order to take all courses while respecting prerequisites? This is only possible if there are no circular dependencies. Imagine if course A requires B, B requires C, and C requires A - we'd be stuck in a loop with no starting point.

This circular dependency is exactly what we call a cycle in graph theory. So the problem reduces to: does our directed graph contain a cycle?

To detect cycles, we can use the concept of topological sorting. The idea is simple: start with courses that have no prerequisites (in-degree of 0). These can be taken immediately. After "taking" these courses, we remove them from consideration and see which new courses now have all their prerequisites satisfied.

Think of it like peeling an onion layer by layer:

1. First layer: courses with no prerequisites
2. Second layer: courses whose only prerequisites were in the first layer
3. Third layer: courses whose prerequisites were all in the first two layers
4. And so on...

If we can successfully peel away all layers (process all courses), then there's no cycle and we can complete everything. But if at some point we can't find any course with all prerequisites satisfied (all remaining courses are waiting for each other), then we've detected a cycle.

The algorithm tracks this by counting how many prerequisites each course has (indeg). As we "complete" courses, we reduce this count for dependent courses. If a course's count reaches 0, it's ready to be taken. If we process all numCourses this way, we return true; otherwise, there's a cycle and we return false.

Solution Approach

Following the topological sorting approach, let's walk through the implementation step by step:

1. Build the Graph and Track In-degrees

First, we create an adjacency list representation of the graph and count the in-degrees (number of prerequisites) for each course:

g = [[] for _ in range(numCourses)]  # Adjacency list
indeg = [0] * numCourses             # In-degree array

for a, b in prerequisites:
    g[b].append(a)    # b -> a (course b is prerequisite for course a)
    indeg[a] += 1     # Course a has one more prerequisite

2. Find Starting Points

We identify all courses with no prerequisites (in-degree of 0) as our starting points:

q = [i for i, x in enumerate(indeg) if x == 0]

These courses can be taken immediately without any prerequisites.

3. Process Courses Layer by Layer

We process courses using BFS-like traversal (though we're using a list instead of a queue for simplicity):

for i in q:
    numCourses -= 1  # Mark this course as completed
    for j in g[i]:    # For each course that depends on course i
        indeg[j] -= 1 # Reduce its prerequisite count
        if indeg[j] == 0:  # If all prerequisites are satisfied
            q.append(j)     # Add it to be processed

The key insight here is that we're using the list q dynamically - as we iterate through it, we may append new elements, and the loop continues until we've processed all reachable courses.

4. Check if All Courses Were Processed

Finally, we check if we've successfully processed all courses:

return numCourses == 0

If numCourses reaches 0, it means we've successfully found an order to take all courses. If not, some courses remain unprocessed due to circular dependencies.

Time Complexity: O(V + E) where V is the number of courses and E is the number of prerequisites, as we visit each node once and traverse each edge once.

Space Complexity: O(V + E) for storing the graph and in-degree array.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: numCourses = 4, prerequisites = [[1,0], [2,0], [3,1], [3,2]]

This means:

• Course 1 requires course 0
• Course 2 requires course 0
• Course 3 requires course 1
• Course 3 requires course 2

Step 1: Build Graph and Calculate In-degrees

We create the adjacency list and count prerequisites:

• g[0] = [1, 2] (course 0 is prerequisite for courses 1 and 2)
• g[1] = [3] (course 1 is prerequisite for course 3)
• g[2] = [3] (course 2 is prerequisite for course 3)
• g[3] = [] (course 3 is not a prerequisite for any course)

In-degrees:

• indeg[0] = 0 (no prerequisites)
• indeg[1] = 1 (requires course 0)
• indeg[2] = 1 (requires course 0)
• indeg[3] = 2 (requires courses 1 and 2)

Step 2: Find Starting Points

Course 0 has in-degree of 0, so q = [0]

Step 3: Process Courses Layer by Layer

Processing course 0:

• Decrement numCourses to 3
• Check dependent courses: 1 and 2
  • indeg[1] becomes 0 → add course 1 to q
  • indeg[2] becomes 0 → add course 2 to q
• Now q = [0, 1, 2]

Processing course 1:

• Decrement numCourses to 2
• Check dependent course: 3
  • indeg[3] becomes 1 (still has course 2 as prerequisite)
• q = [0, 1, 2] (no new additions)

Processing course 2:

• Decrement numCourses to 1
• Check dependent course: 3
  • indeg[3] becomes 0 → add course 3 to q
• Now q = [0, 1, 2, 3]

Processing course 3:

• Decrement numCourses to 0
• No dependent courses

Step 4: Check Result

numCourses == 0, so we return true. All courses can be completed in the order: 0 → 1 → 2 → 3 (or 0 → 2 → 1 → 3).

Counter-example with a cycle:

If we had prerequisites = [[1,0], [0,1]] (course 1 requires 0, and course 0 requires 1):

• Both courses would have in-degree of 1
• q would be empty initially
• No courses could be processed
• numCourses would remain 2, returning false

Solution Implementation

Time and Space Complexity

Time Complexity: O(n + m)

The algorithm builds an adjacency list representation of the graph and performs topological sorting using Kahn's algorithm. Breaking down the operations:

• Building the graph and calculating in-degrees: O(m) - iterates through all prerequisites once
• Finding initial nodes with zero in-degree: O(n) - checks all courses once
• Processing the queue: Each course is added to and removed from the queue at most once (O(n)), and each edge is examined exactly once when decrementing in-degrees (O(m))

Total time complexity: O(n + m)

Space Complexity: O(n + m)

The space usage consists of:

• Graph adjacency list g: O(n + m) - n lists storing a total of m edges
• In-degree array indeg: O(n) - stores in-degree for each course
• Queue q: O(n) - in worst case, all courses could be in the queue

Total space complexity: O(n + m)

Where n represents the number of courses (numCourses) and m represents the number of prerequisites (length of prerequisites list).

Common Pitfalls

1. Misunderstanding the Prerequisite Direction

The Pitfall: One of the most common mistakes is reversing the prerequisite relationship. When given [a, b], many people incorrectly interpret it as "course a is a prerequisite for course b" instead of the correct interpretation: "course b is a prerequisite for course a" (you must take b before a).

Incorrect Implementation:

# WRONG: This reverses the dependency direction
for course, prerequisite in prerequisites:
    graph[course].append(prerequisite)  # Wrong direction!
    in_degree[prerequisite] += 1         # Wrong course!

Why This Causes Problems:

• The graph edges point in the wrong direction
• In-degrees are calculated for the wrong courses
• The topological sort will fail to detect actual cycles
• May return incorrect results for valid course schedules

The Fix: Always remember that [a, b] means "b → a" in the dependency graph (b must come before a):

# CORRECT: prerequisite must be taken before course
for course, prerequisite in prerequisites:
    graph[prerequisite].append(course)  # prerequisite points to course
    in_degree[course] += 1              # course has a prerequisite

2. Using a Traditional Queue Instead of Dynamic List Iteration

The Pitfall: Using a collections.deque and the traditional BFS pattern with while queue: and popleft():

from collections import deque
queue = deque([course for course, degree in enumerate(in_degree) if degree == 0])

while queue:
    current = queue.popleft()
    numCourses -= 1
    for dependent in graph[current]:
        in_degree[dependent] -= 1
        if in_degree[dependent] == 0:
            queue.append(dependent)

Why The Original Approach Works Better: While both approaches are correct, the list iteration approach (for i in queue:) is often cleaner and slightly more efficient in Python because:

• No need to import deque
• Python's list iteration with dynamic appending is optimized
• Simpler code with fewer operations
• The order of processing doesn't matter for cycle detection

3. Modifying the Original numCourses Parameter

The Pitfall: While the solution correctly uses numCourses -= 1 to count processed courses, this modifies the input parameter, which could be confusing or problematic if the value is needed elsewhere.

Better Practice: Use a separate counter variable:

def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
    # ... graph building code ...
  
    processed_count = 0  # Separate counter
  
    for current_course in queue:
        processed_count += 1  # Increment processed courses
      
        for dependent_course in graph[current_course]:
            in_degree[dependent_course] -= 1
            if in_degree[dependent_course] == 0:
                queue.append(dependent_course)
  
    return processed_count == numCourses  # Compare with original value

This makes the code more readable and preserves the original parameter value.