# Time:  O(|E| + |V|log|V|)
# Space: O(|E| + |V|)

# 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 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 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.

import collections
import heapq


class Solution(object):
    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
        """
        adj = collections.defaultdict(list)
        for u, v, w in flights:
            adj[u].append((v, w))
        min_heap = [(0, src, K+1)]
        while min_heap:
            result, u, k = heapq.heappop(min_heap)
            if u == dst:
                return result
            if k > 0:
                for v, w in adj[u]:
                    heapq.heappush(min_heap, (result+w, v, k-1))
        return -1

# Time:  O((|E| + |V|) * log|V|)
# Space: O(|E| + |V|)

# 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 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 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.

import collections
import heapq


class Solution(object):
    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
        """
        adj = collections.defaultdict(list)
        for u, v, w in flights:
            adj[u].append((v, w))
        min_heap = [(0, src, K+1)]
        while min_heap:
            result, u, k = heapq.heappop(min_heap)
            if u == dst:
                return result
            if k > 0:
                for v, w in adj[u]:
                    heapq.heappush(min_heap, (result+w, v, k-1))
        return -1

Update cheapest-flights-within-k-stops.py