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

• 0787. Cheapest Flights Within K Stops
  • Approach
  • Python
  • C++

TODO 0787. Cheapest Flights Within K Stops   medium

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

TODO Approach

Write your approach here.

TODO Python

class Solution:
    def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:

TODO C++

class Solution {
public:
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int k) {
        
    }
};