1462. Course Schedule IV

Problem

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 <= n - 1

• ai != bi

• All the pairs [ai, bi] are unique.

• The prerequisites graph has no cycles.

• 1 <= queries.length <= 10^4

• 0 <= ui, vi <= n - 1

• ui != vi

Solution (Java)

class Solution {
    public List<Boolean> checkIfPrerequisite(
            int numCourses, int[][] prerequisites, int[][] queries) {
        Map<Integer, List<Integer>> m = new HashMap<>();
        int[] ind = new int[numCourses];
        for (int[] p : prerequisites) {
            m.computeIfAbsent(p[1], o -> new ArrayList<>()).add(p[0]);
            ind[p[0]]++;
        }
        boolean[][] r = new boolean[numCourses][numCourses];
        Queue<Integer> q = new LinkedList<>();
        for (int i = 0; i < numCourses; i++) {
            if (ind[i] == 0) {
                q.add(i);
            }
        }
        while (!q.isEmpty()) {
            int j = q.poll();
            for (int k : m.getOrDefault(j, new ArrayList<>())) {
                r[k][j] = true;
                for (int l = 0; l < r.length; l++) {
                    if (r[j][l]) {
                        r[k][l] = true;
                    }
                }
                ind[k]--;
                if (ind[k] == 0) {
                    q.offer(k);
                }
            }
        }
        List<Boolean> a = new ArrayList<>();
        for (int[] qr : queries) {
            a.add(r[qr[0]][qr[1]]);
        }
        return a;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).