Cheapest Flights Within K Stops

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`