Leetcode Reconstruct Itinerary Java Solution

// SOLVING THIS WITH AN AI ASSISTANT (2026)

If you are working through this problem with an AI coding assistant — Claude, ChatGPT, Cursor chat, Gemini, GitHub Copilot, Aider, or any agent — the goal isn’t to ask for the answer. It is to use the tool to understand the pattern. The prompt sequence I’d run:

Spec it back to me first. “In your own words, what is this problem actually testing? What’s the smallest example that fails the naive approach?”
Brute-force first, optimize after. “Write the simplest correct solution, even if it’s O(n²). Don’t optimize. Just make it correct, with comments explaining each step.”
Ask for the upgrade. “Now show me the optimal solution. What insight makes it possible? What pattern is this an instance of?”
Stress-test it. “Generate 10 edge cases — empty input, single element, duplicates, max size, sorted, reverse-sorted. Run my solution against each.”

The pattern matters more than the answer. If the agent just hands you optimized code, you’ve trained yourself to lose interviews.

Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK. Thus, the itinerary must begin with JFK.

Note:

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"].
All airports are represented by three capital letters (IATA code).
You may assume all tickets form at least one valid itinerary.
One must use all the tickets once and only once.

Example 1:

Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]

Example 2:

Input: [["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.

Solution:
We will use the Eulerian Path algorithm.

Step 1: Build Graph using HashMap and store destinations using PriorityQueue, to maintain the lexical order of destinations.

Step 2: Perform a depth-first search and add the final destination first to the result and return the result.

class HackerHeap {
    HashMap<String, PriorityQueue<String>> map = new HashMap<String,PriorityQueue<String>>();

    LinkedList<String> result = new LinkedList<String>();
    
    public List<String> findItinerary(List<List<String>> tickets) {
       // Building The Graph
        for(List<String> ticket: tickets){
            if(!map.containsKey(ticket.get(0))) {
                PriorityQueue<String> q = new PriorityQueue<String>();
                map.put(ticket.get(0),q);
            }
            map.get(ticket.get(0)).offer(ticket.get(1));
        }
        dfs("JFK");
        return result;
    }
    
    public void dfs(String s) {
        PriorityQueue<String> q = map.get(s);
        
        while(q!=null && !q.isEmpty()) {
            dfs(q.poll());
        }
        result.addFirst(s);
    }
}

For the AI-native engineering side of HackerHeap — building MCP servers, comparing agents (Claude Code, Cursor, Windsurf, Codex, Gemini, Copilot), and weekly working code — see the Friday Build newsletter and the MCP archive.

rajendra