Cheapest Flights Within K Stops

There are n cities connected by m flights. Each flight starts from city u and arrives at v with a price w.

Now given all the cities and flights, together with starting city src and the destination dst, your task is to find the cheapest price from src to dst with up to k stops. If there is no such route, output -1.

Example 1:
Input: 
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1
Output: 200
Explanation: 
The graph looks like this:

The cheapest price from city 0 to city 2 with at most 1 stop costs 200, as marked red in the picture.

Example 2:
Input: 
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 0
Output: 500
Explanation: 
The graph looks like this:

The cheapest price from city 0 to city 2 with at most 0 stop costs 500, as marked blue in the picture.

Constraints:

• The number of nodes n will be in range [1, 100], with nodes labeled from 0 to n - 1.

• The size of flights will be in range [0, n * (n - 1) / 2].

• The format of each flight will be (src, dst, price).

• The price of each flight will be in the range [1, 10000].

• k is in the range of [0, n - 1].

• There will not be any duplicated flights or self cycles.

class Solution {
    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int k) {
        Map<Integer, Map<Integer, Integer>> graph = new HashMap<>();
        for (int[] edge : flights) {
            if (!graph.containsKey(edge[0]))
                graph.put(edge[0], new HashMap<>());
            graph.get(edge[0]).put(edge[1], edge[2]);
        }
        // PQ based on cost
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        // Cost , Node , Stops allowed
        pq.add(new int[] { 0, src, k + 1 });
        while (!pq.isEmpty()) {
            int[] top = pq.remove();
            int price = top[0];
            int city = top[1];
            int stops = top[2];
            if (city == dst)
                return price;
            // Because there could be routes which their length is shorter but pass more
            // stops, and those routes don't necessarily constitute the best route in the
            // end. To deal with this, rather than maintain the optimal routes for each
            // node, the solution simply put all possible routes into the
            // priority queue, so that all of them has a chance to be processed.
            if (stops > 0) {
                // Going through all the neighbors
                Map<Integer, Integer> adjNodes = graph.getOrDefault(city, new HashMap<>());
                for (int node : adjNodes.keySet()) {
                    // Adding all the neighbor nodes
                    // (irrespective of whether they already are used or not with a different cost)
                    pq.add(new int[] { price + adjNodes.get(node), node, stops - 1 });
                }
            }
        }
        return -1;
    }
}