Problem Description

In this LeetCode problem, we are given a scenario similar to course scheduling at a university. Each course is identified by a unique number, ranging from 0 to numCourses - 1. There are certain courses that must be completed before taking other courses. These prerequisites are expressed as an array of pairs, where each pair [a, b] indicates that course a must be taken before course b.

The prerequisites can have an indirect relationship as well. Meaning, if course a is a prerequisite for course b, and course b is a prerequisite for course c, then indirectly, course a is a prerequisite for course c.

We also have an array of queries. Each query [u,v] asks whether course u is a prerequisite for course v. The task is to write a function that returns a boolean array. Each boolean value corresponds to a query and determines whether the first course in the query is a prerequisite for the second one.

Flowchart Walkthrough

Let’s analyze problem leetcode 1462. Course Schedule IV using the algorithm flowchart available here. Here is a step-by-step breakdown to determine the suitable algorithm:

Is it a graph?

• Yes: The problem involves determining the prerequisites among courses, which can be represented as a graph where nodes are courses and edges indicate that one course is a prerequisite for another.

Is it a tree?

• No: The prerequisite relationships can form cycles (e.g., mutual prerequisites), therefore the course–prerequisite graph is not necessarily a tree.

Is the problem related to directed acyclic graphs (DAGs)?

• Yes: If we treat the prerequisites as directed edges (A → B means B is a prerequisite for A), managing these prerequisites implies a need to handle these relationships, commonly a directed acyclic graph barring any circular prerequisites.

Topological Sort:

• We reach the node suggesting Topological Sort. Topological sorting of a directed graph is a linear ordering of its vertices such that for every directed edge UV from vertex U to vertex V, U comes before V in the ordering. Topological sorting is relevant here as knowing the order could help in determining if it's possible to complete a course given its dependencies.

Conclusion:

The flowchart points towards utilizing Topological Sort for analyzing prerequisites in a DAG for Course Schedule IV. Often, topological sorting is implemented using Depth-First Search in cases where you need to explore prerequisites (dependencies) recursively. Therefore, using DFS along with careful checks for cycles (to verify if the graph indeed remains acyclic) is recommended.

Intuition

The key challenge of this problem is dealing with both direct and indirect prerequisites. To solve this, we need an efficient way to determine the relationship between courses.

One approach to solve this problem lies in using graph theory, where nodes represent courses, and directed edges represent prerequisite relationships. We build a directed graph such that, for each prerequisite pair [a,b], we have a directed edge from a to b, implying that one must take course a before course b.

With the graph built, we can determine the reachability of each course from every other course. This means that, once we establish that we can reach course b from course a, we can say that a is a prerequisite for b. To achieve this, we implement a modified Floyd-Warshall algorithm or use transitive closure on the graph to find all reachable vertices from a given vertex.

The solution posted uses an approach similar to topological sorting, a sorting method used on directed acyclic graphs (DAGs). It maintains a list, g, that holds the courses dependent on the current course and an array, indeg, to store the number of prerequisite courses needed before one can take the course. The matrix f[i][j] will be True if course i is a prerequisite of course j.

Here's a step-by-step breakdown:

1. Initialize a matrix f to record if one course is a prerequisite of another.
2. Create a graph g representing the course prerequisites and an indegree list indeg.
3. Use a queue q to keep track of courses that have no prerequisites (indegree of 0).
4. Begin processing the queue, and for each course i taken from q,
   • Mark f[i][j] to True for each course j that has i as a prerequisite.
   • Update the prerequisites for courses dependent on j to include the prerequisites for i.
   • Decrease the indegree for each course j and if it becomes 0, add it to the queue.
5. Now f contains information about all direct and indirect prerequisites.
6. Iterate through the queries and return the boolean results from the f matrix.

Learn more about Depth-First Search, Breadth-First Search, Graph and Topological Sort patterns.

Solution Approach

The implementation of this solution involves several key components typical to graph theory and breadth-first search (BFS) traversal, particularly adapted for the problem of determining prerequisites (which can be considered a form of transitive closure in a graph).

Here's how the solution approach works step-by-step:

1. Matrix Initialization: A 2D list f of size n x n is initialized to False, where n is the number of courses. This list will keep track of prerequisite relationships—f[i][j] being True represents that course i is a prerequisite for course j.

2. Graph Representation: A graph g is created as a list of empty lists, where g[i] will eventually contain all the courses that directly require course i as a prerequisite. The indeg list holds the indegree of each course, which counts the number of prerequisites each course has.

3. Building Graph and Indegree List: The prerequisites list is iterated over, and for each pair [a, b], course b is added to g[a], and the indegree of course b is incremented. This constructs the initial directed graph and the indegree list.

4. Queue Initialization for BFS: A queue q is initialized with all courses that have an indegree of 0, i.e., courses that can be taken without having taken any other courses first.

5. BFS Traversal:

   • While the queue is not empty, pop the front course i from the queue.
   • For each course j in g[i]:
     • Set f[i][j] to True, since i is a direct prerequisite for j.
     • Iterate over all courses h (from 0 to n-1):
       • Update f[h][j] to be the logical OR of its current value and f[h][i]. This line propagates the prerequisite information, meaning if h is a prerequisite for i and i for j, then h becomes a prerequisite for j.
     • Decrement the indegree of course j. If the indegree of j now becomes 0, it means that all its prerequisites have been accounted for, and it can be added to the queue for further processing.

6. Answering Queries: Finally, a simple list comprehension is used to iterate over each query in queries and to append the respective boolean value from f that indicates whether the first course in the query is a prerequisite for the second course [f[a][b] for a, b in queries].

By using this approach, we leverage a matrix to keep track of prerequisites while using a BFS approach to iteratively determine and propagate these relationships. This sidesteps complex recursive calls or deep searching algorithms that might be less efficient with large numbers of courses and prerequisites.

Example Walkthrough

To illustrate the solution approach, consider a small university with 4 courses (numbered from 0 to 3) and the following prerequisites: course 0 is a prerequisite for course 1 and course 2, and course 1 is a prerequisite for course 3. We also have queries asking if course 0 is a prerequisite for course 3, and if course 2 is a prerequisite for course 1.

Let's go through the steps to solve this:

1. Matrix Initialization: Initialize an n x n matrix f with all False, here n = 4. This will reflect that no courses are prerequisites of others to begin with.

2. Graph Representation: We create a graph represented as g = [[], [], [], []]. It's a list of lists where each outer list corresponds to a course and will eventually contain all the courses that directly require it as a prerequisite.

3. Building Graph and Indegree List: The prerequisites array maybe [[0,1], [0,2], [1,3]]. From this, we populate our graph g and an indegree list indeg as follows:

   • Add course 1 to g[0] and course 2 to g[0] (because course 0 is a prerequisite for both), increment indeg[1] and indeg[2].
   • Add course 3 to g[1] (because course 1 is a prerequisite for course 3) and increment indeg[3]. After this, g would be [[1,2], [3], [], []] and indeg would be [0,1,1,1].

4. Queue Initialization for BFS: Populate a queue q with courses that have an indegree of 0. In this case, course 0 has no prerequisites, so q = [0].

5. BFS Traversal: Process the queue as per the algorithm:

   • Pop course 0 from q. Its prerequisites will be courses 1 and 2.
   • Mark f[0][1] and f[0][2] to True.
   • Propagate dependencies: For dependent course 1, mark f[0][3] to True (since course 0 is a prerequisite for course 1 which in turn is for course 3). Similarly, if there were courses dependent on 2, propagate the prerequisites for 0 accordingly.
   • Decrease the indegree of courses 1, 2 by 1, and any course which now has an indegree 0 is added to q. Consequently, courses 1 and 2 are added to q.
   • Continue the process for courses 1 and 2. Since course 1 has 3 as a dependent, we mark f[1][3] to True, but also f[0][3] is already True so it remains so. Now, indegree of 3 becomes 0, and it is added to q to process although 3 has no dependents.

6. Answering Queries: For the queries, we simply check the f matrix: for the query [0, 3], we check f[0][3] which is True. For the query [2, 1], we check f[2][1] which is False. Therefore, the answer to the queries would be [True, False].

By progressing through the steps laid out in the solution approach using a small example with 4 courses and a simple prerequisite relationship, we have been able to establish a True or False value for each query, indicating whether one course is a prerequisite for another, considering direct and indirect prerequisite relationships.

Solution Implementation

Time and Space Complexity

Time Complexity

The given Python code performs multiple operations involving graphs and lists, so let's break down its time complexity:

1. Initialization of f and g: The two lists are initialized with sizes n x n and n, respectively. Initializations run in O(n^2) for f and O(n) for g.

2. Iterating over prerequisites list: The loop goes through the list of prerequisites, which in worst case will have O(n^2) elements (if every node is a prerequisite for every other node), and updates graph g and indeg. Each operation inside the loop takes constant time, so this step runs in O(E) where E is the number of edges or prerequisites.

3. Processing the graph: The while loop dequeues nodes with indegree of 0 and processes adjacent nodes. Each edge is considered once when its destination node's indegree becomes 0, and the for loop inside processes n possible predecessors. This results in O(n + E) operations for the while loop since every node and edge is considered once. But due to the nested for loops, we're actually considering n times the adjacent nodes for updating f which yields O(n * E).

4. Answering queries: This is a direct access operation for each query in f which takes O(1) time. There may be Q queries, thus this would take O(Q).

Combining all these, the overall time complexity is dominated by O(n * E), given that updating f can take up to O(n) time for each of the E edges in the worst-case scenario. Therefore, the time complexity is O(n^2 + E + n * E + Q) = O(n * E + Q).

Space Complexity

The space complexity of the algorithm can be analyzed as follows:

1. Space for f: O(n^2) since it stores boolean values for every possible pair of nodes.

2. Space for g and indeg: O(n) for each, since they store a list of adjacent nodes and indegree counts for each node, respectively.

3. Additional space for the queue q: In the worst case, this can store all n nodes, so O(n).

So, the overall space complexity is the sum of these which gives O(n^2) + 2 * O(n) + O(n) = O(n^2) as n^2 will dominate the space complexity for large n.

Learn more about how to find time and space complexity quickly using problem constraints.