1462. Course Schedule IV

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course ai first if you want to take course bi.

• For example, the pair [0, 1] indicates that you have to take course 0 before you can take course 1.

Prerequisites can also be indirect. If course a is a prerequisite of course b, and course b is a prerequisite of course c, then course a is a prerequisite of course c.

You are also given an array queries where queries[j] = [uj, vj]. For the jth query, you should answer whether course uj is a prerequisite of course vj or not.

Return a boolean array answer, where answer[j] is the answer to the jth query.

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]]
Output: [false,true]
Explanation: The pair [1, 0] indicates that you have to take course 1 before you can take course 0.
Course 0 is not a prerequisite of course 1, but the opposite is true.

Example 2:

Input: numCourses = 2, prerequisites = [], queries = [[1,0],[0,1]]
Output: [false,false]
Explanation: There are no prerequisites, and each course is independent.

Example 3:

Input: numCourses = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]]
Output: [true,true]

Constraints:

• 2 <= numCourses <= 100
• 0 <= prerequisites.length <= (numCourses * (numCourses - 1) / 2)
• prerequisites[i].length == 2
• 0 <= ai, bi <= numCourses - 1
• ai != bi
• All the pairs [ai, bi] are unique.
• The prerequisites graph has no cycles.
• 1 <= queries.length <= 104
• 0 <= ui, vi <= numCourses - 1
• ui != vi

Approach 01:

class Solution {
 public:
  vector<bool> checkIfPrerequisite(int numCourses,
                                   vector<vector<int>>& prerequisites,
                                   vector<vector<int>>& queries) {
    vector<bool> result;
    // isPrerequisiteMatrix[i][j] := true if course i is a prerequisite of course j
    vector<vector<bool>> isPrerequisiteMatrix(numCourses, vector<bool>(numCourses));

    // Populate the initial prerequisites
    for (const vector<int>& prerequisitePair : prerequisites) {
      const int prerequisiteCourse = prerequisitePair[0];
      const int dependentCourse = prerequisitePair[1];
      isPrerequisiteMatrix[prerequisiteCourse][dependentCourse] = true;
    }

    // Use the Floyd-Warshall algorithm to determine all prerequisite relationships
    for (int intermediateCourse = 0; intermediateCourse < numCourses; ++intermediateCourse)
      for (int courseA = 0; courseA < numCourses; ++courseA)
        for (int courseB = 0; courseB < numCourses; ++courseB)
          isPrerequisiteMatrix[courseA][courseB] =
              isPrerequisiteMatrix[courseA][courseB] ||
              (isPrerequisiteMatrix[courseA][intermediateCourse] &&
               isPrerequisiteMatrix[intermediateCourse][courseB]);

    // Check each query
    for (const vector<int>& queryPair : queries) {
      const int courseA = queryPair[0];
      const int courseB = queryPair[1];
      result.push_back(isPrerequisiteMatrix[courseA][courseB]);
    }

    return result;
  }
};

class Solution:
    def checkIfPrerequisite(self, numCourses: int, prerequisites: List[List[int]], queries: List[List[int]]) -> List[bool]:
        # isPrerequisiteMatrix[i][j] := True if course i is a prerequisite of course j
        isPrerequisiteMatrix = [[False] * numCourses for _ in range(numCourses)]

        # Populate the initial prerequisites
        for prerequisitePair in prerequisites:
            prerequisiteCourse = prerequisitePair[0]
            dependentCourse = prerequisitePair[1]
            isPrerequisiteMatrix[prerequisiteCourse][dependentCourse] = True

        # Use the Floyd-Warshall algorithm to determine all prerequisite relationships
        for intermediateCourse in range(numCourses):
            for courseA in range(numCourses):
                for courseB in range(numCourses):
                    isPrerequisiteMatrix[courseA][courseB] = (
                        isPrerequisiteMatrix[courseA][courseB]
                        or (
                            isPrerequisiteMatrix[courseA][intermediateCourse]
                            and isPrerequisiteMatrix[intermediateCourse][courseB]
                        )
                    )

        # Check each query
        result = []
        for queryPair in queries:
            courseA = queryPair[0]
            courseB = queryPair[1]
            result.append(isPrerequisiteMatrix[courseA][courseB])

        return result

Time Complexity:

• Floyd-Warshall Algorithm:

  The algorithm involves three nested loops, each iterating over the total number of courses, \( numCourses \). Thus, the time complexity for this part is \( O(numCourses^3) \).

• Processing Queries:

  Each query is checked in \( O(1) \), and for \( q \) queries, this step has a time complexity of \( O(q) \).

• Overall Time Complexity:

  The total time complexity is \( O(numCourses^3 + q) \), where \( q \) is the number of queries.

Space Complexity:

• Prerequisite Matrix:

  The algorithm uses a \( numCourses \times numCourses \) matrix to store prerequisite relationships, resulting in a space complexity of \( O(numCourses^2) \).

• Auxiliary Space:

  Other variables and data structures (e.g., vectors for the result) require \( O(q) \) space, where \( q \) is the number of queries.

• Overall Space Complexity:

  The total space complexity is \( O(numCourses^2 + q) \).