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


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


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


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

Solution :


                        Solution in C++ :

vector<string> solve(vector<vector<string>>& flights) {
    map<string, set<string>> graph;
    map<string, int> ind;
    for (auto& v : flights) {

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

    function<bool(const string&)> dfs = [&](const string& node) -> bool {
        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});
        return false;

    bool check = false;
    for (auto& p : ind) {
        if (p.second == 0) {
            check = true;
    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:
            degree[u] -= 1
            degree[v] += 1

        for outgoing in graph.values():

        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)


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

        return res[::-1]

View More Similar Problems

Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

View Solution →

Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →

Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

View Solution →

Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

View Solution →

Inserting a Node Into a Sorted Doubly Linked List

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function

View Solution →

Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

View Solution →