Flight Itinerary Sequel - Amazon Top Interview Questions


Problem Statement :


You are given a list of flights that were taken, represented as origin to destination airport pairs. Given that this list was shuffled, find all the airports that were visited in the correct order. If there's more than one valid itinerary, return the lexicographically smallest ones first.

Note: airports may be visited twice.

Example 1

Input

flights = [
    ["YYZ", "SEA"],
    ["JFK", "YYZ"],
    ["SEA", "JFK"]
]

Output

["JFK", "YYZ", "SEA", "JFK"]

Explanation

The three airports form a cycle so there's multiple answers but "JFK" is the lexicographically smallest answer.


Solution :



title-img 




                        Solution in C++ :

vector<string> solve(vector<vector<string>>& flights) {
    map<string, set<string>> graph;
    map<string, int> ind;
    for (auto& v : flights) {
        graph[v[0]].insert(v[1]);
        ind[v[1]]++;
        ind[v[0]];
    }

    set<pair<string, string>> used_edge;
    vector<string> ans;

    function<bool(const string&)> dfs = [&](const string& node) -> bool {
        ans.push_back(node);
        if (used_edge.size() == flights.size()) return true;
        for (auto& s : graph[node]) {
            if (used_edge.count({node, s})) continue;
            used_edge.insert({node, s});
            if (dfs(s)) return true;
            used_edge.erase({node, s});
        }
        ans.pop_back();
        return false;
    };

    bool check = false;
    for (auto& p : ind) {
        if (p.second == 0) {
            dfs(p.first);
            check = true;
            break;
        }
    }
    if (!check) {
        for (auto it = graph.begin(); it != graph.end(); it++)
            if (dfs(it->first)) break;
    }

    return ans;
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, flights):
        graph = defaultdict(list)
        degree = defaultdict(int)  # + in  - out

        for u, v in flights:
            graph[u].append(v)
            degree[u] -= 1
            degree[v] += 1

        for outgoing in graph.values():
            outgoing.sort(reverse=True)

        res = []
        path_start = min(degree, key=degree.__getitem__)
        circuit_start = min(graph)

        def dfs(u, graph):  # Hierholzer
            while graph[u]:
                v = graph[u].pop()
                dfs(v, graph)

            res.append(u)

        if degree[path_start] == -1:  # eulerian path ?
            dfs(path_start, graph)
        else:  # eulerian circuit
            dfs(circuit_start, graph)

        return res[::-1]


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