LeetCode #332 — HARD

Reconstruct Itinerary

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a list of airline tickets where tickets[i] = [from_i, to_i] 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.

Constraints:

1 <= tickets.length <= 300
tickets[i].length == 2
from_i.length == 3
to_i.length == 3
from_i and to_i consist of uppercase English letters.
from_i != to_i

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]

Example 2

[["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #332: Reconstruct Itinerary
class Solution {
    private Map<String, List<String>> g = new HashMap<>();
    private List<String> ans = new ArrayList<>();

    public List<String> findItinerary(List<List<String>> tickets) {
        Collections.sort(tickets, (a, b) -> b.get(1).compareTo(a.get(1)));
        for (List<String> ticket : tickets) {
            g.computeIfAbsent(ticket.get(0), k -> new ArrayList<>()).add(ticket.get(1));
        }
        dfs("JFK");
        Collections.reverse(ans);
        return ans;
    }

    private void dfs(String f) {
        while (g.containsKey(f) && !g.get(f).isEmpty()) {
            String t = g.get(f).remove(g.get(f).size() - 1);
            dfs(t);
        }
        ans.add(f);
    }
}

// Accepted solution for LeetCode #332: Reconstruct Itinerary
func findItinerary(tickets [][]string) (ans []string) {
	sort.Slice(tickets, func(i, j int) bool {
		return tickets[i][0] > tickets[j][0] || (tickets[i][0] == tickets[j][0] && tickets[i][1] > tickets[j][1])
	})
	g := make(map[string][]string)
	for _, ticket := range tickets {
		g[ticket[0]] = append(g[ticket[0]], ticket[1])
	}
	var dfs func(f string)
	dfs = func(f string) {
		for len(g[f]) > 0 {
			t := g[f][len(g[f])-1]
			g[f] = g[f][:len(g[f])-1]
			dfs(t)
		}
		ans = append(ans, f)
	}
	dfs("JFK")
	for i := 0; i < len(ans)/2; i++ {
		ans[i], ans[len(ans)-1-i] = ans[len(ans)-1-i], ans[i]
	}
	return
}

# Accepted solution for LeetCode #332: Reconstruct Itinerary
class Solution:
    def findItinerary(self, tickets: List[List[str]]) -> List[str]:
        def dfs(f: str):
            while g[f]:
                dfs(g[f].pop())
            ans.append(f)

        g = defaultdict(list)
        for f, t in sorted(tickets, reverse=True):
            g[f].append(t)
        ans = []
        dfs("JFK")
        return ans[::-1]

// Accepted solution for LeetCode #332: Reconstruct Itinerary
use std::cmp::Reverse;
use std::collections::{BinaryHeap, HashMap};

impl Solution {
    pub fn find_itinerary(tickets: Vec<Vec<String>>) -> Vec<String> {
        let mut graph: HashMap<&str, BinaryHeap<Reverse<&str>>> = HashMap::new();
        for ticket in tickets.iter() {
            graph
                .entry(&ticket[0])
                .or_insert_with(BinaryHeap::new)
                .push(Reverse(&ticket[1]));
        }
        let mut answer: Vec<String> = Vec::with_capacity(tickets.len() + 1);
        let mut stack: Vec<&str> = vec!["JFK"];
        while let Some(src) = stack.last() {
            if let Some(dsts) = graph.get_mut(src) {
                if !dsts.is_empty() {
                    if let Some(dst) = dsts.pop() {
                        stack.push(dst.0);
                    }
                    continue;
                }
            }
            if let Some(last) = stack.pop() {
                answer.push(last.to_string());
            }
        }
        answer.reverse();
        answer
    }
}

// Accepted solution for LeetCode #332: Reconstruct Itinerary
function findItinerary(tickets: string[][]): string[] {
    const g: Record<string, string[]> = {};
    tickets.sort((a, b) => b[1].localeCompare(a[1]));
    for (const [f, t] of tickets) {
        g[f] = g[f] || [];
        g[f].push(t);
    }
    const ans: string[] = [];
    const dfs = (f: string) => {
        while (g[f] && g[f].length) {
            const t = g[f].pop()!;
            dfs(t);
        }
        ans.push(f);
    };
    dfs('JFK');
    return ans.reverse();
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × log m)

Space

O(m)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.