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# Time: O(t! / (n1! * n2! * ... nk!)), t is the total number of tickets,

# ni is the number of the ticket which from is city i,

# k is the total number of cities.

# Space: O(t)

# Given a list of airline tickets represented by pairs of departure

# and arrival airports [from, to], reconstruct the itinerary in order.

# All of the tickets belong to a man who departs from JFK.

# Thus, the itinerary must begin with JFK.

#

# Note:

# If there are multiple valid itineraries, you should return the itinerary

# that has the smallest lexical order when read as a single string.

# For example, the itinerary ["JFK", "LGA"] has a smaller lexical

# order than ["JFK", "LGB"].

# All airports are represented by three capital letters (IATA code).

# You may assume all tickets may form at least one valid itinerary.

# Example 1:

# tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]

# Return ["JFK", "MUC", "LHR", "SFO", "SJC"].

# Example 2:

# tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]

# Return ["JFK","ATL","JFK","SFO","ATL","SFO"].

# Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].

# But it is larger in lexical order.

class Solution(object):

def findItinerary(self, tickets):

"""

:type tickets: List[List[str]]

:rtype: List[str]

"""

def route_helper(origin, ticket_cnt, graph, ans):

if ticket_cnt == 0:

return True

for i, (dest, valid) in enumerate(graph[origin]):

if valid:

graph[origin][i][1] = False

ans.append(dest)

if route_helper(dest, ticket_cnt - 1, graph, ans):

return ans

ans.pop()

graph[origin][i][1] = True

return False

graph = collections.defaultdict(list)

for ticket in tickets:

graph[ticket[0]].append([ticket[1], True])

for k in graph.keys():

graph[k].sort()

origin = "JFK"

ans = [origin]

route_helper(origin, len(tickets), graph, ans)

return ans