815. Bus Routes

Problem

You are given an array routes representing bus routes where routes[i] is a bus route that the ith bus repeats forever.

• For example, if routes[0] = [1, 5, 7], this means that the 0th bus travels in the sequence 1 -> 5 -> 7 -> 1 -> 5 -> 7 -> 1 -> ... forever.

You will start at the bus stop source (You are not on any bus initially), and you want to go to the bus stop target. You can travel between bus stops by buses only.

Return **the least number of buses you must take to travel from *source* to **target. Return -1 if it is not possible.

  Example 1:

Input: routes = [[1,2,7],[3,6,7]], source = 1, target = 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.

Example 2:

Input: routes = [[7,12],[4,5,15],[6],[15,19],[9,12,13]], source = 15, target = 12
Output: -1

  Constraints:

• 1 <= routes.length <= 500.

• 1 <= routes[i].length <= 10^5

• All the values of routes[i] are unique.

• sum(routes[i].length) <= 10^5

• 0 <= routes[i][j] < 10^6

• 0 <= source, target < 10^6

Solution

class Solution {
    public int numBusesToDestination(int[][] routes, int source, int target) {
        if (source == target) {
            return 0;
        }
        Set<Integer> targetRoutes = new HashSet<>();
        Queue<Integer> queue = new ArrayDeque<>();
        boolean[] taken = new boolean[routes.length];
        List<Integer>[] graph = buildGraph(routes, source, target, queue, targetRoutes, taken);
        if (targetRoutes.isEmpty()) {
            return -1;
        }
        int bus = 1;
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int route = queue.poll();
                if (targetRoutes.contains(route)) {
                    return bus;
                }
                for (int nextRoute : graph[route]) {
                    if (!taken[nextRoute]) {
                        queue.offer(nextRoute);
                        taken[nextRoute] = true;
                    }
                }
            }
            bus++;
        }
        return -1;
    }

    private List<Integer>[] buildGraph(
            int[][] routes,
            int source,
            int target,
            Queue<Integer> queue,
            Set<Integer> targetRoutes,
            boolean[] taken) {
        int len = routes.length;
        ArrayList[] graph = new ArrayList[len];
        for (int i = 0; i < len; i++) {
            Arrays.sort(routes[i]);
            graph[i] = new ArrayList<Integer>();
            int id = Arrays.binarySearch(routes[i], source);
            if (id >= 0) {
                queue.offer(i);
                taken[i] = true;
            }
            id = Arrays.binarySearch(routes[i], target);
            if (id >= 0) {
                targetRoutes.add(i);
            }
        }
        for (int i = 0; i < len; i++) {
            for (int j = i + 1; j < len; j++) {
                if (commonStop(routes[i], routes[j])) {
                    graph[i].add(j);
                    graph[j].add(i);
                }
            }
        }
        return graph;
    }

    private boolean commonStop(int[] routeA, int[] routeB) {
        int idA = 0;
        int idB = 0;
        while (idA < routeA.length && idB < routeB.length) {
            if (routeA[idA] == routeB[idB]) {
                return true;
            } else if (routeA[idA] < routeB[idB]) {
                idA++;
            } else {
                idB++;
            }
        }
        return false;
    }
}

Explain:

nope.

Complexity:

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