Here, We see Reconstruct Itinerary LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Reconstruct Itinerary LeetCode Solution

Problem Statement

You are given a list of airline tickets where tickets[i] = [fromi, toi] represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.

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

• For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].

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

Example 1:

Input: tickets = [[“MUC”,”LHR”],[“JFK”,”MUC”],[“SFO”,”SJC”],[“LHR”,”SFO”]]
Output: [“JFK”,”MUC”,”LHR”,”SFO”,”SJC”]

Example 2:

Input: tickets = [[“JFK”,”SFO”],[“JFK”,”ATL”],[“SFO”,”ATL”],[“ATL”,”JFK”],[“ATL”,”SFO”]]
Output: [“JFK”,”ATL”,”JFK”,”SFO”,”ATL”,”SFO”]
Explanation: Another possible reconstruction is [“JFK”,”SFO”,”ATL”,”JFK”,”ATL”,”SFO”] but it is larger in lexical order.

Reconstruct Itinerary LeetCode Solution C++

class Solution {
public:
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        unordered_map<string, vector<string>> graph;
        for (auto& ticket : tickets) {
            graph[ticket[0]].push_back(ticket[1]);
        }
        for (auto& [_, destinations] : graph) {
            sort(destinations.rbegin(), destinations.rend());
        }
        vector<string> itinerary;
        function<void(const string&)> dfs = [&](const string& airport) {
            while (!graph[airport].empty()) {
                string next = graph[airport].back();
                graph[airport].pop_back();
                dfs(next);
            }
            itinerary.push_back(airport);
        };
        dfs("JFK");
        reverse(itinerary.begin(), itinerary.end());
        return itinerary;        
    }
};

Reconstruct Itinerary LeetCode Solution Java

class Solution {
    public List<String> findItinerary(List<List<String>> tickets) {
        Map<String, PriorityQueue<String>> graph = new HashMap<>();
        for (List<String> ticket : tickets) {
            graph.putIfAbsent(ticket.get(0), new PriorityQueue<>());
            graph.get(ticket.get(0)).add(ticket.get(1));
        }
        LinkedList<String> itinerary = new LinkedList<>();
        dfs("JFK", graph, itinerary);
        return itinerary;
    }
    private void dfs(String airport, Map<String, PriorityQueue<String>> graph, LinkedList<String> itinerary) {
        PriorityQueue<String> nextAirports = graph.get(airport);
        while (nextAirports != null && !nextAirports.isEmpty()) {
            dfs(nextAirports.poll(), graph, itinerary);
        }
        itinerary.addFirst(airport);
    }
}Code language: JavaScript (javascript)

Reconstruct Itinerary Solution JavaScript

var findItinerary = function(tickets) {
    const graph = {};
    for (const [src, dst] of tickets) {
        if (!graph[src]) graph[src] = [];
        graph[src].push(dst);
    }
    for (const key in graph) {
        graph[key].sort().reverse();
    }
    const itinerary = [];
    function dfs(airport) {
        while (graph[airport] && graph[airport].length > 0) {
            dfs(graph[airport].pop());
        }
        itinerary.push(airport);
    }
    dfs("JFK");
    return itinerary.reverse();
};Code language: JavaScript (javascript)

Reconstruct Itinerary Solution Python

class Solution(object):
    def findItinerary(self, tickets):
        graph = defaultdict(list)
        for src, dst in sorted(tickets, reverse=True):
            graph[src].append(dst)
        itinerary = []
        def dfs(airport):
            while graph[airport]:
                dfs(graph[airport].pop())
            itinerary.append(airport)
        dfs("JFK")
        return itinerary[::-1]