We have a list of bus routes. Eachroutes[i]is a bus route that the i-th bus repeats forever. For example ifroutes[0] = [1, 5, 7], this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->... forever.

We start at bus stopS(initially not on a bus), and we want to go to bus stopT. Travelling by buses only, what is the least number of buses we must take to reach our destination? Return -1 if it is not possible.

Example:
Input:

routes = [[1, 2, 7], [3, 6, 7]]
S = 1
T = 6

Output:
 2

Explanation:

The best strategy is take the first bus to the bus stop 7, then take the second bus to the bus stop 6.

Note:

• 1 < = routes.length < = 500 .
• 1 < = routes[i].length < = 500 .
• 0 < = routes[i][j] < 10 ^ 6 .

tag: BFS

class Solution {
    private class Point{
        int node;
        int depth;

        public Point(int node, int depth){
            this.node = node;
            this.depth = depth;
        }
    }

    public int numBusesToDestination(int[][] routes, int S, int T) {
        int n = routes.length;
        if (S == T) return 0;

        Map<Integer, Set<Integer>> graph = new HashMap<>();
        for (int i = 0; i < n; i++){
            Arrays.sort(routes[i]);
            graph.put(i, new HashSet<>());
        }

        for (int i = 0; i < n; i++){
            for (int j = i + 1; j < n; j++){
                if (intersect(routes[i], routes[j])){
                    graph.get(i).add(j);
                    graph.get(j).add(i);
                }
            }
        }

        Set<Integer> targets = new HashSet<>();
        Queue<Point> q = new LinkedList<>();
        Set<Integer> seen = new HashSet<>();

        for (int i = 0; i < n; i++){
            if (Arrays.binarySearch(routes[i], S) >= 0){
                seen.add(i);
                q.offer(new Point(i, 0));
            }
            if (Arrays.binarySearch(routes[i], T) >= 0){
                targets.add(i);
            }
        }


        while (!q.isEmpty()){
            Point cur = q.poll();
            int node = cur.node, depth = cur.depth;

            if (targets.contains(node)) return depth + 1;

            for (int nei : graph.get(node)){
                if (seen.contains(nei)) continue;

                seen.add(nei);
                q.offer(new Point(nei, depth + 1));
            }
        }

        return -1;
    }

    public boolean intersect(int[] A, int[] B) {
        int i = 0, j = 0;
        while (i < A.length && j < B.length) {
            if (A[i] == B[j]) return true;
            if (A[i] < B[j]) i++; else j++;
        }
        return false;
    }
}