332.md

332. Reconstruct Itinerary

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 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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

my thoughts:

1. BFS/ DFS?

2. graph dfs

my solution:

********** TLE
class Solution:
    def findItinerary(self, tickets):
        """
        :type tickets: List[List[str]]
        :rtype: List[str]
        """
        res = []
        cur = ["JFK"]
        l = len(tickets)
        check = [True] * l
        
        def dfs(cur, check, l):
            if len(cur) == l + 1:
                res.append(' '.join(cur))
                return
            found = False
            for i in range(l):
                if check[i] and tickets[i][0] == cur[-1]:
                    found = True
                    cur.append(tickets[i][1])
                    check[i] = False
                    dfs(cur, check, l)
                    cur.pop()
                    check[i] = True
            if not found:
                return
        
        dfs(cur, check, l)
        res.sort()
        return res[0].split(' ')

********** 68 ms
class Solution:
    def findItinerary(self, tickets):
        """
        :type tickets: List[List[str]]
        :rtype: List[str]
        """
        res = []
        d = {}
        for t in tickets:
            if t[0] in d:
                d[t[0]].append(t[1])
            else:
                d[t[0]] = [t[1]]
                
        l = len(tickets) + 1
        
        for k in d:
            d[k].sort(reverse = True)
        
        def visit(a):
            while a in d and d[a]:
                visit(d[a].pop())
            res.append(a)
        
        visit("JFK")
        
        return res[::-1]

my comments:

from other ppl's solution:

1. All the airports are vertices and tickets are directed edges. Then all these tickets form a directed graph.

The graph must be Eulerian since we know that a Eulerian path exists.

Thus, start from “JFK”, we can apply the Hierholzer’s algorithm to find a Eulerian path in the graph which is a valid reconstruction.

Since the problem asks for lexical order smallest solution, we can put the neighbors in a min-heap. In this way, we always visit the smallest possible neighbor first in our trip.

public class Solution {

    Map<String, PriorityQueue<String>> flights;
    LinkedList<String> path;

    public List<String> findItinerary(String[][] tickets) {
        flights = new HashMap<>();
        path = new LinkedList<>();
        for (String[] ticket : tickets) {
            flights.putIfAbsent(ticket[0], new PriorityQueue<>());
            flights.get(ticket[0]).add(ticket[1]);
        }
        dfs("JFK");
        return path;
    }

    public void dfs(String departure) {
        PriorityQueue<String> arrivals = flights.get(departure);
        while (arrivals != null && !arrivals.isEmpty())
            dfs(arrivals.poll());
        path.addFirst(departure);
    }
}

79 / 79 test cases passed.
Status: Accepted
Runtime: 11 ms