Cheapest Flights Within K Stops

Medium

Problem Statement

There are n airports, labeled from 0 to n - 1, which are connected by some flights. You are given an array flights where flights[i] = [from_i, to_i, price_i] represents a one-way flight from airport from_i to airport to_i with cost price_i. You may assume there are no duplicate flights and no flights from an airport to itself.

You are also given three integers src, dst, and k where:

src is the starting airport
dst is the destination airport
src != dst
k is the maximum number of stops you can make (not including src and dst)

Return the cheapest price from src to dst with at most k stops, or return -1 if it is impossible.

Example 1:

Input: n = 4, flights = [[0,1,200],[1,2,100],[1,3,300],[2,3,100]], src = 0, dst = 3, k = 1

Output: 500

Explanation: The optimal path with at most 1 stop from airport 0 to 3 is shown in red, with total cost 200 + 300 = 500. Note that the path [0 -> 1 -> 2 -> 3] costs only 400, and thus is cheaper, but it requires 2 stops, which is more than k.

Example 2:

Input: n = 3, flights = [[1,0,100],[1,2,200],[0,2,100]], src = 1, dst = 2, k = 1

Output: 200

Explanation: The optimal path with at most 1 stop from airport 1 to 2 is shown in red and has cost 200.

Constraints:

1 <= n <= 100
fromi != toi
1 <= pricei <= 1000
0 <= src, dst, k < n

Recommended Time & Space Complexity

You should aim for a solution as good or better than O(n + (m * k)) time and O(n) space, where n is the number of cities, m is the number of flights, and k is the number of stops.

Hint 1

Consider this as a graph problem where the cities are nodes and the flights are edges connecting two cities, with the ticket cost as the edge weight. Can you think of a shortest path algorithm to solve the problem? Perhaps a better algorithm than Dijkstra's that can intuitively handle the k stops condition.

Hint 2

We can use the Bellman-Ford algorithm. Initialize a prices array of size n with Infinity, setting prices[source] = 0. These values describe the cost to reach a city from the source city. Iterate (k + 1) times (stops are 0-indexed), updating the cost to each city by extending paths from cities with valid costs. We only update the cost for a city if it is less than the previous cost. How would you implement this?

Hint 3

At each level of iteration, we go through the given flights and use them to update the price array with the minimum costs compared to the previous level. We use a temporary prices array at each level to store the updated costs. After completing all levels, we return the result stored in prices[dst]. If that value is Infinity, we return -1 instead.

Examples

1Example 1

Input:

{
  "n": 2,
  "flights": [
    [
      0,
      1,
      100
    ]
  ],
  "src": 0,
  "dst": 1,
  "k": 0
}

Output:

100

2Example 2

Input:

{
  "n": 3,
  "flights": [
    [
      0,
      1,
      50
    ],
    [
      1,
      2,
      50
    ]
  ],
  "src": 0,
  "dst": 2,
  "k": 1
}

Output:

100

3Example 3

Input:

{
  "n": 3,
  "flights": [
    [
      0,
      1,
      100
    ],
    [
      1,
      2,
      100
    ]
  ],
  "src": 0,
  "dst": 2,
  "k": 0
}

Output:

-1