332 Reconstruct Itinerary

🔀 Pick One

You are given a list of airline tickets where tickets[i] = [fromi, toi] represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.

All of the tickets belong to a man who departs from "JFK", thus, the itinerary must begin with "JFK". 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"].

You may assume all tickets form at least one valid itinerary. You must use all the tickets once and only once.

Example 1:

MUC --> LHR --> SFO
 ^               |
 |               v
JFK             SJC

Input: tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
Output: ["JFK","MUC","LHR","SFO","SJC"]

Example 2:

SFO <-.
 ^   \ \
 |     \ \
 |  --> v |
JFK <-- ATL

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.

Constraints:

  • 1 <= tickets.length <= 300
  • tickets[i].length == 2
  • fromi.length == 3
  • toi.length == 3
  • fromi and toi consist of uppercase English letters.
  • fromi != toi
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from collections import defaultdict

class Solution:
    def findItinerary(self, tickets: List[List[str]]) -> List[str]:
        self.graph = defaultdict(list)

        for src, des in tickets:
            self.graph[src].append(des)

        self.visit_map = {}

        # sort the itinerary based on the lexical order
        for origin, destinations in self.graph.items():
            # Note that we could have multiple identical flights,
            # i.e. same origin and destination.
            destinations.sort()
            self.visit_map[origin] = [False] * len(destinations)

        self.n = len(tickets)
        self.result = []
        route = ['JFK']
        self.backtracking('JFK', route)

        return self.result

    def backtracking(self, src, route):
        if len(route) == self.n + 1:
            self.result = route
            # since the itineraries are sorted, we can stop
            # once the first one is found
            return True

        for i, next_des in enumerate(self.graph[src]):
            if not self.visit_map[src][i]:
                # mark the visit before the next recursion
                self.visit_map[src][i] = True
                ret = self.backtracking(next_des, route + [next_des])
                self.visit_map[src][i] = False
                if ret:
                    return True
        return False