cpp-flashcards/org/study_deck_02/dsa/advanced-graphs/0332-reconstruct-itinerary.org

#+PROPERTY: STUDY_DECK_02
* TODO 0332. Reconstruct Itinerary :hard:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*0332. Reconstruct Itinerary][0332. Reconstruct Itinerary]]
:END:

You are given a list of airline ~tickets~ where ~tickets[i] = [from_{i}, to_{i}]~ 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:*


#+begin_src
Input: tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
Output: ["JFK","MUC","LHR","SFO","SJC"]
#+end_src


*Example 2:*


#+begin_src
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.
#+end_src


*Constraints:*

 - ~1 <= tickets.length <= 300~

 - ~tickets[i].length == 2~

 - ~from_{i}.length == 3~

 - ~to_{i}.length == 3~

 - ~from_{i}~ and ~to_{i}~ consist of uppercase English letters.

 - ~from_{i} != to_{i}~

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python
class Solution:
    def findItinerary(self, tickets: List[List[str]]) -> List[str]:
#+end_src

** TODO C++
#+begin_src cpp
class Solution {
public:
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        
    }
};
#+end_src