<p>You are given a list of airline <code>tickets</code> where <code>tickets[i] = [from<sub>i</sub>, to<sub>i</sub>]</code> represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.</p>

<p>All of the tickets belong to a man who departs from <code>&quot;JFK&quot;</code>, thus, the itinerary must begin with <code>&quot;JFK&quot;</code>. If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string.</p>

<ul>
	<li>For example, the itinerary <code>[&quot;JFK&quot;, &quot;LGA&quot;]</code> has a smaller lexical order than <code>[&quot;JFK&quot;, &quot;LGB&quot;]</code>.</li>
</ul>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/itinerary1-graph.jpg" style="width: 382px; height: 222px;" />
<pre>
<strong>Input:</strong> tickets = [[&quot;MUC&quot;,&quot;LHR&quot;],[&quot;JFK&quot;,&quot;MUC&quot;],[&quot;SFO&quot;,&quot;SJC&quot;],[&quot;LHR&quot;,&quot;SFO&quot;]]
<strong>Output:</strong> [&quot;JFK&quot;,&quot;MUC&quot;,&quot;LHR&quot;,&quot;SFO&quot;,&quot;SJC&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/itinerary2-graph.jpg" style="width: 222px; height: 230px;" />
<pre>
<strong>Input:</strong> tickets = [[&quot;JFK&quot;,&quot;SFO&quot;],[&quot;JFK&quot;,&quot;ATL&quot;],[&quot;SFO&quot;,&quot;ATL&quot;],[&quot;ATL&quot;,&quot;JFK&quot;],[&quot;ATL&quot;,&quot;SFO&quot;]]
<strong>Output:</strong> [&quot;JFK&quot;,&quot;ATL&quot;,&quot;JFK&quot;,&quot;SFO&quot;,&quot;ATL&quot;,&quot;SFO&quot;]
<strong>Explanation:</strong> Another possible reconstruction is [&quot;JFK&quot;,&quot;SFO&quot;,&quot;ATL&quot;,&quot;JFK&quot;,&quot;ATL&quot;,&quot;SFO&quot;] but it is larger in lexical order.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= tickets.length &lt;= 300</code></li>
	<li><code>tickets[i].length == 2</code></li>
	<li><code>from<sub>i</sub>.length == 3</code></li>
	<li><code>to<sub>i</sub>.length == 3</code></li>
	<li><code>from<sub>i</sub></code> and <code>to<sub>i</sub></code> consist of uppercase English letters.</li>
	<li><code>from<sub>i</sub> != to<sub>i</sub></code></li>
</ul>

class Solution {
public:

    struct hashing{
            size_t operator() (const std::pair<std::string, std::string> & a) const{
                return std::hash<std::string>()(a.first) ^ std::hash<std::string>()(a.second)<<1;
            }
        };

    bool dfsForMinLexi(std::string next, std::vector<std::string> &currentWay, int travelLength, 
        std::unordered_map<std::pair<std::string, std::string>, int, hashing> & ticketsLeft,
        std::unordered_map<std::string, std::vector<std::string>> & connections){

            travelLength--;
            currentWay.push_back(next);
            if(travelLength==0){
                return true;
            }else{
                for(const auto & connection: connections[next]){
                    if(ticketsLeft[{next, connection}]>0){
                        ticketsLeft[{next, connection}]--;
                        if(dfsForMinLexi(connection, currentWay, travelLength, ticketsLeft, connections)){
                            return true;
                        }
                        ticketsLeft[{next, connection}]++;
                    }
                }
            }
            currentWay.pop_back();
            return false;
    }

    vector<string> findItinerary(vector<vector<string>>& tickets) {
        std::unordered_map<std::pair<std::string, std::string>, int, hashing> ticketsLeft;
        std::unordered_map<std::string, std::vector<std::string>> connections;

        for(const auto & ticket: tickets){
            ticketsLeft[{ticket[0], ticket[1]}] ++;
            if(ticketsLeft[{ticket[0], ticket[1]}] == 1){
                connections[ticket[0]].push_back(ticket[1]);
            }
        }

        for(auto & [key, c]: connections){
            sort(begin(c), end(c));
        }

        int travelLength = tickets.size()+1;
        std::vector<std::string> currentWay;
        dfsForMinLexi("JFK", currentWay, travelLength, ticketsLeft, connections);

        return currentWay;
    }
};
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