<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:
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        vector<string> itinerary;
        unordered_map<string, vector<pair<string, bool>>> map;
        for (const vector<string>& ticket : tickets)
            map[ticket[0]].push_back(make_pair(ticket[1], false));
        for (auto& airport : map)
            sort(airport.second.begin(), airport.second.end());
        int total = tickets.size();
        int count = 0;
        stack<int> stack;
        itinerary.push_back("JFK");
        while (count < total)
        {
            vector<pair<string, bool>>& airlines = map[itinerary.back()];
            auto first = find_if(begin(airlines), end(airlines), [](pair<string, bool> airline) { return !airline.second; });
            if (first == airlines.end())
            {
                bool found = false;
                while (!found)
                {
                    itinerary.pop_back();
                    --count;
                    vector<pair<string, bool>>& prev = map[itinerary.back()];
                    int top = stack.top();
                    stack.pop();
                    prev[top].second = false;
                    for (int i = top + 1; i < prev.size() && !found; ++i)
                    {
                        if (!prev[i].second && prev[i].first != prev[top].first)
                        {
                            found = true;
                            prev[i].second = true;
                            itinerary.push_back(prev[i].first);
                            ++count;
                            stack.push(i);
                        }
                    }
                }
                continue;
            }
            (*first).second = true;
            itinerary.push_back((*first).first);
            ++count;
            stack.push(first - airlines.begin());
        }
        return itinerary;
    }
};
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