Reconstruct itinerary

Time: O(T!/(N1!xN2!x…xNK!)); Space: O(T); medium

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.

Notes:

  • 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 form at least one valid itinerary.

Example 1:

Input: tickets = [[“MUC”, “LHR”], [“JFK”, “MUC”], [“SFO”, “SJC”], [“LHR”, “SFO”]]

Output: [“JFK”, “MUC”, “LHR”, “SFO”, “SJC”]

Example 2:

Input: tickets = [[“JFK”,“SFO”],[“JFK”,“ATL”],[“SFO”,“ATL”],[“ATL”,“JFK”],[“ATL”,“SFO”]]

Output: [“JFK”,“ATL”,“JFK”,“SFO”,“ATL”,“SFO”]

Explanation:

  • Another possible reconstruction is [“JFK”,“SFO”,“ATL”,“JFK”,“ATL”,“SFO”].

  • But it is larger in lexical order.

[1]:

import collections

class Solution1(object):
    """
    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)
    """
    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

[2]:

s = Solution1()

tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
assert s.findItinerary(tickets) == ["JFK", "MUC", "LHR", "SFO", "SJC"]

tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
assert s.findItinerary(tickets) == ["JFK","ATL","JFK","SFO","ATL","SFO"]