Cheapest Flights Within K Stops

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

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:

- `2 <= n <= 100`

- `0 <= flights.length <= (n * (n - 1) / 2)`

- `flights[i].length == 3`

- `0 <= from_i, to_i < n`

- `from_i != to_i`

- `1 <= price_i <= 10^4`

- There will not be any multiple flights between two cities.

- `0 <= src, dst, k < n`

- `src != dst`