Cheapest Flights Within K Stops

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- Bus Routes
- CheatSheet: Leetcode For Code Interview
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- Tag: #bfs, #dijkstra

There are n cities connected by m flights. Each fight starts from city u and arrives at v with a price w.

Now given all the cities and fights, together with starting city src and the destination dst, your task is to find the cheapest price from src to dst with up to k stops. If there is no such route, output -1.

Example 1: Input: n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]] src = 0, dst = 2, k = 1 Output: 200 Explanation: The graph looks like this:

The cheapest price from city 0 to city 2 with at most 1 stop costs 200, as marked red in the picture.

Example 2: Input: n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]] src = 0, dst = 2, k = 0 Output: 500 Explanation: The graph looks like this:

The cheapest price from city 0 to city 2 with at most 0 stop costs 500, as marked blue in the picture.

Note:

- The number of nodes n will be in range [1, 100], with nodes labeled from 0 to n – 1.
- The size of flights will be in range [0, n * (n – 1) / 2].
- The format of each flight will be (src, dst, price).
- The price of each flight will be in the range [1, 10000].
- k is in the range of [0, n – 1].
- There will not be any duplicated flights or self cycles.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## https://code.dennyzhang.com/cheapest-flights-within-k-stops ## Basic Ideas: BFS ## ## Complexity: Time O(n), Space O(n) class Solution: def findCheapestPrice(self, n, flights, src, dst, K): """ :type n: int :type flights: List[List[int]] :type src: int :type dst: int :type K: int :rtype: int """ import collections import sys d = collections.defaultdict(lambda: {}) for [start, end, price] in flights: m = d[start] m[end] = price price_map = {} queue = collections.deque() seen = set([]) queue.append(src) seen.add(src) price_map[src] = 0 level = 0 min_price = sys.maxsize while len(queue) != 0: level += 1 # print(level, K, src, dst, queue) if level>K+1: break for k in range(len(queue)): p = queue.popleft() # find the neighbors if p in d: m = d[p] for q in m: if q == dst: min_price = min(min_price, price_map[p]+m[q]) continue # not visited if (q not in seen): # print(price_map, p, q, m) price_map[q] = price_map[p]+m[q] queue.append(q) seen.add(q) # we have a better solution elif price_map[q] > price_map[p]+m[q]: queue.append(q) price_map[q] = price_map[p]+m[q] return min_price if min_price != sys.maxsize else -1