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787. Cheapest Flights Within K Stops

Difficulty Topics

Description

There are n cities connected by some number of flights. You are given an array flights where flights[i] = [fromi, toi, pricei] indicates that there is a flight from city fromi to city toi with cost pricei.

You are also given three integers src, dst, and k, return the cheapest price from src to dst with at most k stops. If there is no such route, return -1.

 

Example 1:

Input: n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1
Output: 700
Explanation:
The graph is shown above.
The optimal path with at most 1 stop from city 0 to 3 is marked in red and has cost 100 + 600 = 700.
Note that the path through cities [0,1,2,3] is cheaper but is invalid because it uses 2 stops.

Example 2:

Input: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
Output: 200
Explanation:
The graph is shown above.
The optimal path with at most 1 stop from city 0 to 2 is marked in red and has cost 100 + 100 = 200.

Example 3:

Input: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
Output: 500
Explanation:
The graph is shown above.
The optimal path with no stops from city 0 to 2 is marked in red and has cost 500.

 

Constraints:

  • 1 <= n <= 100
  • 0 <= flights.length <= (n * (n - 1) / 2)
  • flights[i].length == 3
  • 0 <= fromi, toi < n
  • fromi != toi
  • 1 <= pricei <= 104
  • There will not be any multiple flights between two cities.
  • 0 <= src, dst, k < n
  • src != dst

Solution

cheapest-flights-within-k-stops.py
class Solution:
    def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:
        pq = [(0, src, k + 1)]
        graph = defaultdict(list)
        seen = {}

        for a, b, price in flights:
            graph[a].append((b, price))

        while pq:
            weight, node, stops = heappop(pq)

            if node == dst: return weight

            if node in seen and seen[node] >= stops: continue

            seen[node] = stops

            if stops > 0:
                for adj, adjW in graph[node]:
                    heappush(pq, (weight + adjW, adj, stops - 1))

        return -1
cheapest-flights-within-k-stops.java
class Solution {

    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int K) {

        // Build the adjacency matrix
        int adjMatrix[][] = new int[n][n];
        for (int[] flight: flights) {
            adjMatrix[flight[0]][flight[1]] = flight[2];
        }

        // Shortest distances array
        int[] distances = new int[n];

        // Shortest steps array
        int[] currentStops = new int[n];
        Arrays.fill(distances, Integer.MAX_VALUE);
        Arrays.fill(currentStops, Integer.MAX_VALUE);
        distances[src] = 0;
        currentStops[src] = 0;

        // The priority queue would contain (node, cost, stops)
        PriorityQueue<int[]> minHeap = new PriorityQueue<int[]>((a, b) -> a[1] - b[1]);
        minHeap.offer(new int[]{src, 0, 0});

         while (!minHeap.isEmpty()) {

            int[] info = minHeap.poll();
            int node = info[0], stops = info[2], cost = info[1];

             // If destination is reached, return the cost to get here
            if (node == dst) {
                return cost;
            }

            // If there are no more steps left, continue 
            if (stops == K + 1) {
                continue;
            }

            // Examine and relax all neighboring edges if possible 
            for (int nei = 0; nei < n; nei++) {
                if (adjMatrix[node][nei] > 0) {
                    int dU = cost, dV = distances[nei], wUV = adjMatrix[node][nei];

                    // Better cost?
                    if (dU + wUV < dV) {
                        minHeap.offer(new int[]{nei, dU + wUV, stops + 1});
                        distances[nei] = dU + wUV;
                    }
                    else if (stops < currentStops[nei]) {
                        minHeap.offer(new int[]{nei, dU + wUV, stops + 1});          
                    }
                    currentStops[nei] = stops;
                }
            }
         }

        return distances[dst] == Integer.MAX_VALUE? -1 : distances[dst];
    }
}