# 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 :

```                        ```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].insert(v);
ind[v]++;
ind[v];
}

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]```
```

## 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

## 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 -

## 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

## Find Merge Point of Two Lists

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## 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

## 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.