LeetCode - 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]

BFS TLE:

class Solution {
    public List<Boolean> checkIfPrerequisite(int n, int[][] prerequisites, int[][] queries) {
        //initiate adjacency list
        Map<Integer, List<Integer>> adj = new HashMap<>();
        for (int i = 0; i < n; i++) {
            adj.put(i, new ArrayList<Integer>());
        }
        //insert nodes' neighbors
        for (int[] prereq : prerequisites) {
            adj.get(prereq[0]).add(prereq[1]);
        }
        List<Boolean> res = new ArrayList<>();
        Map<Integer, Set<Integer>> cache = new HashMap<>();
        for(int[] query : queries) {
            Queue<Integer> queue = new LinkedList<>();
            queue.add(query[0]);
            boolean found = false;
            while(!queue.isEmpty()) {
                Integer node = queue.poll();
                if(cache.containsKey(node)) {
                    if (cache.get(node).contains(query[1])) {
                        found = true;
                        break;
                    }
                }
                for (int next : adj.get(node)) {
                    if (next == query[1]) {
                        found = true;
                        break;
                    }
                    Set<Integer> set = cache.getOrDefault(query[0], new HashSet<Integer>());
                    set.add(next);
                    cache.put(query[0], set);
                    queue.add(next);
                }
                if (found) {
                    break;
                }
            }
            res.add(found);
        }
        return res;
    }

}

 BFS With Caching:

class Solution {
    public List<Boolean> checkIfPrerequisite(int n, int[][] prerequisites, int[][] queries) {
        //initiate adjacency list
        Map<Integer, List<Integer>> adj = new HashMap<>();
        for (int i = 0; i < n; i++) {
            adj.put(i, new ArrayList<Integer>());
        }
        //insert nodes' neighbors
        for (int[] prereq : prerequisites) {
            adj.get(prereq[0]).add(prereq[1]);
        }
        boolean[][] isReachable = new boolean[n][n];
        for (int i = 0; i < n; i++) {
            isReachable[i][i] = true;
            Queue<Integer> queue = new LinkedList<>();
            Set<Integer> visited=new HashSet();
            queue.add(i);
            while(!queue.isEmpty()) {
                Integer node = queue.poll();
                for (int nb : adj.get(node)) {
                    if (!visited.contains(nb)) {
                        isReachable[i][nb] = true;
                        queue.add(nb);
                        visited.add(nb);
                    }
                }
            }
        }
        List<Boolean> res = new ArrayList<>();
        for (int[] query : queries) {
            res.add(isReachable[query[0]][query[1]]);
        }
        return res;
    }

}