787 - Cheapest Flights Within K Stops.

Medium

There are n cities connected by some number of flights. You are given an array flights where <code>flightsi = from_i, to_i, price_i</code> indicates that there is a flight from city <code>from_i</code> to city <code>to_i</code> with cost <code>price_i</code>.

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

• <code>0 <= from_i, to_i< n</code>

• <code>from_i != to_i</code>

• <code>1 <= price_i<= 10⁴</code>

• There will not be any multiple flights between two cities.

• 0 <= src, dst, k < n

• src != dst