cpp-flashcards/org/study_deck_02/dsa/advanced-graphs/0787-cheapest-flights-within-k-stops.org

#+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<vector<int>>& flights, int src, int dst, int k) {
        
    }
};
#+end_src