Reconstruct Itinerary Python Solution

LeetCode 332, Hard. Topics: Array, String, Depth-First Search, Graph Theory, Sorting, Heap (Priority Queue), Eulerian Circuit. View on LeetCode. Generate this problem as a practice environment: tested reference solution, 14 parametrized pytest cases, and a playground notebook:

lcpy gen -n 332   # by problem number
lcpy gen -s reconstruct_itinerary   # by problem name

Problem

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.

Examples

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

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

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(E log E)
    # Space: O(E)
    def find_itinerary(self, tickets: list[list[str]]) -> list[str]:
        graph: dict[str, list[str]] = {}
        for src, dst in tickets:
            graph.setdefault(src, []).append(dst)
        for src in graph:
            graph[src].sort(reverse=True)

        itinerary: list[str] = []

        def dfs(airport: str) -> None:
            destinations = graph.get(airport)
            while destinations:
                dfs(destinations.pop())
            itinerary.append(airport)

        dfs("JFK")
        itinerary.reverse()
        return itinerary

Complexity

Time	Space
O(E log E)	O(E)

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