cheapest-flight-within-k-stops.py

Author: Chris Wu

cheapest-flights-within-k-stops.py

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+import heapq
+import collections
+
+"""
+We start from src and only got K+1 stops to use
+Each time, we choose the cheapest place to go.
+
+If the city we popout is dst, then the price must be lowest
+Since we always pick the lowest place to go.
+
+If we still have stops left (stops>1), we put its neighbor to the priority queue.
+So the city in the priority queue must be within the stops limit.
+
+Making the graph takes O(E)
+The size of priority queue is O(V), since we might put all the cities in it.
+So for every pop, it is O(LogV). Total is O(VLogV).
+For every edge we call an heappush, so that is ELogV
+O(E+ (V+E)LogV) -> O((V+E)LogV)
+V is the number of cities within range K stops.
+"""
+#Dijkstra
+class Solution1(object):
+	def findCheapestPrice(self, n, flights, src, dst, K):
+		graph = collections.defaultdict(list)
+		pq = []
+
+		for u, v, w in flights: graph[u].append((w, v))
+
+		heapq.heappush(pq, (0, K+1, src))
+		while pq:
+			price, stops, city = heapq.heappop(pq)
+
+			if city is dst: return price
+			if stops>0:
+				for price_to_nei, nei in graph[city]:
+					heapq.heappush(pq, (price+price_to_nei, stops-1, nei))
+		return -1
+
+"""
+This is mostly straight forward BFS.
+When we are out of stops, or price is greater than min_price, we stop adding cities to the queue.
+Every time we encounter dst we compare the price and set it to the min.
+
+Making the graph takes O(E)
+BFS every node in adjacent list takes O(V+E)
+V is the number of cities within range K stops.
+"""
+#BFS
+class Solution2(object):
+	def findCheapestPrice(self, n, flights, src, dst, K):
+		graph = collections.defaultdict(list)
+		q = collections.deque()
+		min_price = float('inf')
+
+		for u, v, w in flights: graph[u].append((w, v))
+		q.append((src, 0, 0))
+		while q:
+			city, stops, price = q.popleft()
+			if city==dst:
+				min_price = min(min_price, price)
+				continue
+
+			if stops<=K and price<=min_price:
+				for price_to_nei, nei in graph[city]:
+					q.append((nei, stops+1, price+price_to_nei))
+
+		return min_price if min_price!=float('inf') else -1
+