## Reconstruct Itinerary

07/18/2016 Graph Depth First Search

## Question

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:

1. 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"].
2. All airports are represented by three capital letters (IATA code).
3. 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.

## Solution

Result: Accepted Time: 32 ms

Here should be some explanations.

class Solution {
public:
vector<string> findItinerary(vector<pair<string, string>> tickets) {
unordered_map<string, multiset<string>> graph;
for(const auto & x:tickets)
graph[x.first].insert(x.second);
vector<string> marching, itinerary;
marching.push_back("JFK");
while(marching.size())
{
auto from = marching.back();
if(graph.count(from) && graph[from].size()>0)
{
auto & to = graph[from];
marching.push_back(*to.begin());
to.erase(to.begin());
}
else
{
itinerary.push_back(from);
marching.pop_back();
}
}
reverse(itinerary.begin(), itinerary.end());
return itinerary;
}
};


Complexity Analytics

• Time Complexity: $O(n)$
• Space Complexity: $O(n)$