787_Cheapest_Flights_Within_K_Stops.py

"""
787. 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


"""

class Solution:
    def findCheapestPrice(self, n: int, flights: list[list[int]], src: int, dst: int, k: int) -> int:
        prices = [float("inf")] * n
        prices[src] = 0
        for _ in range(k + 1):
            tmpPrices = prices.copy()
            for s, d, p in flights:  # s - source, d = destination, p=price
                if prices[s] == float("inf"):
                    continue
                if prices[s] + p < tmpPrices[d]:
                    tmpPrices[d] = prices[s] + p
            prices = tmpPrices
        return -1 if prices[dst] == float("inf") else prices[dst]


"""
Runtime 183ms Beats23.50% of users with Python3
Memory 17.86MB Beats80.28% of users with Python3
"""

test = {
    "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
        }

print(Solution().findCheapestPrice(**test["Input"]) == test["Output"])