Course Schedule IV

There are a total of n courses you have to take, labeled from 0 to n-1.

Some courses may have direct prerequisites, for example, to take course 0 you have first to take course 1, which is expressed as a pair: [1,0]

Given the total number of courses n, a list of direct prerequisite pairs and a list of queries pairs.

You should answer for each queries[i] whether the course queries[i][0] is a prerequisite of the course queries[i][1] or not.

Return a list of boolean, the answers to the given queries.

Please note that 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.

Example 1:

Input: n = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]]
Output: [false,true]
Explanation: course 0 is not a prerequisite of course 1 but the opposite is true.

Example 2:

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

Example 3:

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

Example 4:

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

Example 5:

Input: n = 5, prerequisites = [[0,1],[1,2],[2,3],[3,4]], queries = [[0,4],[4,0],[1,3],[3,0]]
Output: [true,false,true,false]

Constraints:

• 2 <= n <= 100

• 0 <= prerequisite.length <= (n * (n - 1) / 2)

• 0 <= prerequisite[i][0], prerequisite[i][1] < n

• prerequisite[i][0] != prerequisite[i][1]

• The prerequisites graph has no cycles.

• The prerequisites graph has no repeated edges.

• 1 <= queries.length <= 10^4

• queries[i][0] != queries[i][1]

class Solution {
    public List<Boolean> checkIfPrerequisite(int n, int[][] prerequisites, int[][] queries) {
        // Creating directed graph
        boolean[][] connected = new boolean[n][n];
        for (int[] p : prerequisites)
            connected[p[0]][p[1]] = true; // p[0] -> p[1]
        for (int k = 0; k < n; k++)
            for (int i = 0; i < n; i++)
                for (int j = 0; j < n; j++)
                    connected[i][j] = connected[i][j] || connected[i][k] && connected[k][j];
        List<Boolean> ans = new ArrayList<>();
        for (int[] q : queries)
            ans.add(connected[q[0]][q[1]]);
        return ans;
    }
}