#+PROPERTY: STUDY_DECK_02 * TODO 0787. Cheapest Flights Within K Stops :medium: :PROPERTIES: :NEETCODE: [[file:../../roadmap.org::*0787. Cheapest Flights Within K Stops][0787. Cheapest Flights Within K Stops]] :END: 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:* #+begin_src 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. #+end_src *Example 2:* #+begin_src 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. #+end_src *Example 3:* #+begin_src 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. #+end_src *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~ ** TODO Approach Write your approach here. ** TODO Python #+begin_src python class Solution: def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int: #+end_src ** TODO C++ #+begin_src cpp class Solution { public: int findCheapestPrice(int n, vector>& flights, int src, int dst, int k) { } }; #+end_src