Reconstruct Itinerary

The problem belongs to the domain of Graph Theory. This is because the tickets can be represented as edges in a graph, where the ‘from’ and ’to’ airports are nodes.

What Components

1. List of airline tickets, each represented by a pair of ‘from’ and ’to’ airports.
2. Requirement to start the itinerary from “JFK.”
3. The necessity to use each ticket exactly once.
4. If multiple itineraries are possible, the smallest lexical order is preferred.

Classification

This is a Pathfinding Problem combined with a Sequencing Problem.

Pathfinding Problem

You need to find a valid path that uses all the tickets. The requirement to start at “JFK” and use each ticket exactly once maps directly onto finding a specific kind of path in a graph.

Sequencing Problem

You are not just looking for any path; you are looking for the lexicographically smallest path. This involves sequencing the “from” and “to” pairs in a specific order.

In summary, you are looking for a specific kind of path in a graph (Pathfinding Problem) while also maintaining a certain sequence to satisfy lexical conditions (Sequencing Problem).

Clarification Questions

1. Can a ticket have the same ‘from’ and ’to’ airports? The problem states fromi != toi, but it’s good to confirm.

2. Is it guaranteed that a valid itinerary exists for all given inputs?

3. Can multiple tickets have the same ‘from’ and ’to’ airport pairs?

4. What should the output be if there are no valid itineraries? The problem assumes at least one exists, but asking clarifies the edge case.

5. Is the list of tickets sorted in any specific order, or is it random?

6. Is there a limit on the number of airports, or could any airport appear on a ticket?

7. Is the lexical order based purely on ASCII values of the characters in the airport codes, or is there another rule for lexical ordering?

8. Will all airport codes always be exactly three uppercase English letters?

9. Are all tickets one-way, or could there be round-trip tickets in the list?

10. Are computational efficiency or runtime constraints a concern for solving this problem?

Asking these questions can remove ambiguities and make it clearer what assumptions you should or shouldn’t make while solving the problem.

Identifying Problem Isomorphism

The problem “Reconstruct Itinerary” can be mapped to “Eulerian Path Problem.”

Reasoning

Both problems involve finding a path in a graph where each edge is used exactly once. In “Reconstruct Itinerary,” the airports act as vertices and the tickets as directed edges. In the Eulerian Path Problem, you have a graph with nodes and edges where the task is to traverse every edge exactly once. The requirement to start at “JFK” corresponds to specifying a start vertex in an Eulerian Path.

Complexity

The Eulerian Path Problem is simpler. It lacks the lexicographical ordering constraint that “Reconstruct Itinerary” has.

10 Prerequisite LeetCode Problems

For the Reconstruct Itinerary, the following is a good preparation:

1. Number of Islands - Teaches you basic graph traversal using DFS.
2. Course Schedule - Another problem that deals with graph traversal and dependency resolution.
3. Clone Graph - Helps understand graph construction and traversal, which are fundamental to the given problem.
4. All Paths from Source to Target - Enhances your skills in traversing all possible paths in a directed graph.
5. Detect Cycle in a Directed Graph - Knowing how to detect cycles will help you understand edge cases for your problem.
6. Topological Sort - Gives a different perspective on ordering elements, which is useful for the lexicographical constraint.
7. Implement Trie - Trie can be helpful for storing airport codes and quickly sorting them lexicographically.
8. Minimum Height Trees - Helps with understanding tree-like structures within graphs.
9. Network Delay Time - Practices solving problems related to weighted edges, although the original problem doesn’t have weights.
10. Longest Increasing Path in a Matrix - Teaches you dynamic programming in the context of a grid, which can be considered a special case of a graph.

These problems cover various graph theory fundamentals, and practicing them can provide a strong foundation for tackling the Reconstruct Itinerary problem.

Problem Analysis and Key Insights

Key Insights from Analyzing “Reconstruct Itinerary”:

1. Directed Graph: The problem can be represented as a directed graph where airports are vertices and tickets are directed edges.

2. Start Vertex: The problem specifies that the journey must begin at “JFK,” setting a fixed starting point for the graph traversal.

3. Lexicographical Order: If multiple itineraries are possible, the one with the smallest lexical order is required. This adds a sorting constraint to the graph traversal.

4. One-time Use: Each ticket can be used only once, so each edge in the graph should be traversed exactly once.

5. Total Tickets: The constraint that all tickets must be used ensures that the graph traversal must cover all edges.

6. Valid Itinerary Exists: It is stated that at least one valid itinerary exists, meaning the graph will have a valid Eulerian path starting from “JFK.”

Understanding these insights can guide the problem-solving approach, especially towards using graph algorithms and dealing with sorting constraints.

Problem Boundary

The scope of the “Reconstruct Itinerary” problem is primarily focused on:

1. Graph Theory: Specifically, finding a path through a directed graph that covers all edges exactly once.

2. String Comparison: It includes lexicographical sorting for scenarios where multiple itineraries are possible.

3. Constraints: There are limitations such as starting from “JFK,” using all tickets once, and the string length of airport codes.

This problem does not involve other areas like dynamic programming, greedy algorithms, or arithmetic computations. It is constrained to graph traversal and sorting.

To establish the boundary of the “Reconstruct Itinerary” problem, consider the following factors:

1. Input Constraints: The problem clearly states the number of tickets, the length of airport names, and the requirement that they be in uppercase. These set the input boundaries.

2. Starting Point: The journey starts at “JFK,” thus limiting the starting vertex for traversal.

3. Problem Objective: The goal is to find an itinerary that covers all tickets exactly once. This rules out scenarios where some tickets can be ignored or reused.

4. Sorting Requirement: Among multiple valid itineraries, the one with the smallest lexical order should be selected. This sets a boundary on how the problem is solved, requiring attention to lexicographic ordering.

5. Existence of Solution: The problem statement assures that at least one valid itinerary exists. This eliminates the need to handle cases where no solution is possible.

6. Graph Nature: The problem can be mapped to a directed graph traversal, and it does not require other techniques like dynamic programming or backtracking, thus defining the algorithmic boundary.

7. Output Format: The problem specifies the output should be a list of airport codes that constitute the itinerary.

Understanding these boundaries allows you to focus on the relevant techniques and constraints while solving the problem.

Distilling the Problem to Its Core Elements

Fundamental Concept

The fundamental concept here is Graph Traversal. Specifically, you’re dealing with a directed graph where each ticket is an edge between two vertices (airports), and you have to find a path that uses all edges exactly once.

Simplest Description

Imagine you have a bunch of one-way airplane tickets from one city to another. You have to use all tickets and start your journey from New York (JFK). The task is to find the order in which to use these tickets so that you visit every city.

Core Problem

The core problem is to find a valid path through all the given tickets starting from JFK, such that if there are multiple paths, you choose the one that comes first in alphabetical order.

Key Components

1. Graph Construction: Create a graph from the given list of tickets.
2. Graph Traversal: Traverse the graph starting from JFK, using each edge exactly once.
3. Lexicographic Order: When multiple choices are available, pick the one that is lexicographically smallest.
4. Output Formatting: Format the final path as a list of airport codes.

Minimal Set of Operations

1. Sort Tickets: Sort the destination airports for each source in lexicographic order. This helps in picking the smallest when you have a choice.
2. Initialize Stack and Output List: Use a stack to keep track of the current path, and an output list for the final itinerary.
3. Traverse: Start from JFK and move to the next airport based on the tickets, updating the stack and output list as you go.

By understanding these components and operations, you can build an effective strategy to solve the problem.

Visual Model of the Problem

A directed graph is an effective way to visualize this problem. Each airport is a vertex, and each ticket is a directed edge from the departure airport to the arrival airport.

Steps to Visualize:

1. Vertices: Write down all unique airport codes mentioned in the tickets as vertices.
2. Directed Edges: For each ticket [from, to], draw a directed edge from the from vertex to the to vertex.
3. Multiple Edges: If there are multiple tickets between the same pair of airports, add that many number of directed edges.
4. Start Vertex: Highlight or color the vertex “JFK” differently, as that’s your starting point.

Legend:

• Vertex: Airport
• Directed Edge: A ticket from one airport to another
• Highlighted Vertex: JFK

After setting up the graph, you can manually trace possible paths to get a concrete understanding of the task. Make sure to follow the rules:

• Start from JFK.
• Choose the lexicographically smaller destination if you have multiple choices.
• Use each ticket exactly once.

This visualization will give you a concrete understanding of what the problem is asking and how you can approach it.

Problem Restatement

You have a list of one-way flight tickets, each showing departure and arrival airports. You start your journey from JFK airport. Your task is to use all the tickets to form a trip itinerary. If you have options for your next flight, choose the one with the smaller alphabetical order. You have to use all the tickets exactly once, and you can assume that at least one valid itinerary exists. Return the complete itinerary.

Abstract Representation of the Problem

In abstract terms, you’re given a directed graph where each edge represents a one-way ticket from one node (departure airport) to another (arrival airport). You start at a designated node (“JFK”) and must traverse all edges exactly once. When multiple choices exist, opt for the edge leading to the lexicographically smaller node. The objective is to find such a traversal path.

Terminology

1. Directed Graph: A set of nodes connected by edges, where each edge has a direction from one node to another. In this problem, each airport is a node, and each ticket is a directed edge from the departure to the arrival airport.

2. Lexicographical Order: A way to arrange words based on their alphabetical sequence. Here, it’s used to prioritize among multiple outgoing edges from a node.

3. Traversal: The act of moving through all nodes in a particular order. You’re required to traverse all edges exactly once to form the itinerary.

4. Starting Node: The initial point from which the traversal begins. In this case, it’s always the JFK airport.

5. Eulerian Path: A path in a directed graph that visits every edge exactly once. This problem essentially asks for an Eulerian path with a specific starting point and lexicographical constraints.

Each of these terms plays a critical role in defining the problem’s constraints and objectives. Understanding them is key to formulating an effective solution.

Problem Simplification and Explanation

Imagine you have a bunch of train tickets. Each ticket has a departure station and an arrival station. You need to use all the tickets to travel from one station to another, but you must start from a specific station, let’s say “Station A.”

Your job is to figure out the exact order in which to use these tickets so that you can visit every station and end up back where you started. Also, if you have more than one ticket from the same departure station, you should use the one with the destination that comes first alphabetically.

Key Concepts:

1. Starting Point: You always begin your journey at “Station A” (JFK in the problem).
2. Ticket Use: You have to use all the tickets exactly once.
3. Order: If multiple tickets can be used next, pick the one that goes to the station with the name that comes first in the alphabet.

How they interact: You start at “Station A,” look at all the tickets you can use from there, pick the one with the earliest alphabetical arrival station, and move on. You keep doing this until you’ve used all the tickets.

Metaphor: Think of this like solving a jigsaw puzzle where each piece can only connect in a particular way and you always start with a specific corner piece. The challenge is to lay down the pieces in such a way that you follow a certain order (alphabetically) when you have multiple options.

Constraints

1. Starting Point: The problem always starts at JFK. This fixed starting point means we don’t have to search for a starting point; we can directly begin processing from there.

2. Ticket Use Once: Each ticket is used exactly once. This eliminates the possibility of cycles and simplifies the problem to a directed graph traversal.

3. Lexicographical Order: If there are multiple choices, we have to choose the one that comes first alphabetically. This means we can pre-sort the tickets for each departure point, making it quicker to select the next destination.

4. Constraints:

• The number of tickets is limited to 300.
• Station names are uppercase and 3 characters long.

These constraints indicate that the data sets are not exceedingly large, so sorting or hashing would be efficient.

5. Ticket Pairing: Each ticket is a pair of departure and arrival points. These can be naturally represented as edges in a graph, and the problem essentially becomes a graph traversal problem.

By identifying these characteristics, you can guide your problem-solving approach towards graph-based solutions, specifically using techniques optimized for directed graphs. The lexicographical order constraint also suggests sorting or priority queue data structures could be beneficial.

1. Upper Limit: The constraint that tickets.length <= 300 means that computational complexity isn’t a significant concern. Most algorithms of O(n^2) or better will be efficient enough for this problem.

2. String Length: Since the length of airport codes is fixed at 3 uppercase characters, string comparison, sorting, or hashing operations will be very quick.

3. No Cycles: The constraint of using each ticket only once eliminates the possibility of cycles in the graph, simplifying the graph traversal strategy.

4. Directed Edges: The constraint that ‘fromi’ and ’toi’ are different means that we are working with a directed graph. This narrows down the types of graph algorithms that can be applied.

These constraints provide information about the scale of the problem, the characteristics of the graph, and the nature of the strings involved. They allow for more focused algorithm and data structure choices.

Case Analysis

Let’s explore various test cases that will help you understand the problem’s intricacies.

1. Minimum Input Size

• Input: [["JFK", "SFO"]]
• Output: ["JFK", "SFO"]
• Analysis: One ticket, hence only one way to go. This represents the minimum size constraint.

2. Multiple Ways - Lexicographically Smaller First

• Input: [["JFK", "SFO"], ["JFK", "ATL"], ["SFO", "ORD"], ["ATL", "ORD"]]
• Output: ["JFK", "ATL", "ORD"]
• Analysis: Two ways to go from JFK to ORD, but lexicographically smaller path is chosen.

3. Single Source, Multiple Destinations

• Input: [["JFK", "SFO"], ["JFK", "ATL"], ["JFK", "ORD"]]
• Output: ["JFK", "ATL"] or ["JFK", "ORD"] or ["JFK", "SFO"]
• Analysis: One start point, multiple end points. Lexicographically smaller one is chosen.

4. Multi-Step Itinerary

• Input: [["JFK", "SFO"], ["SFO", "ATL"], ["ATL", "ORD"]]
• Output: ["JFK", "SFO", "ATL", "ORD"]
• Analysis: Multiple hops required to reach the final destination.

5. Disconnected Graph

• Input: [["JFK", "SFO"], ["ATL", "ORD"]]
• Output: Invalid
• Analysis: JFK and ATL are not connected; it violates the problem statement of a single itinerary.

Edge Cases:

1. Minimum Input Size: Helps to check if the code can handle lower bounds.
2. Disconnected Graph: Even though the problem statement assumes a valid path, it’s good to handle this scenario to ensure robustness.
3. Multiple Ways - Lexicographically Smaller First: Verifies that the lexicographically smaller route is chosen when multiple valid itineraries exist.

These test cases cover different aspects of graph traversal, lexicographic sorting, and edge conditions. They help ensure that your solution will be both correct and robust.

Visualizing these cases can be helpful for better understanding. We can use directed graphs where each node represents an airport and each edge represents a flight from one airport to another. Below is the visualization of each case:

1. Minimum Input Size

• Nodes: JFK, SFO
• Edges: JFK -> SFO
• Graph: JFK —> SFO

2. Multiple Ways - Lexicographically Smaller First

• Nodes: JFK, SFO, ATL, ORD
• Edges: JFK -> SFO, JFK -> ATL, SFO -> ORD, ATL -> ORD
• Graph:

      ORD  
     /   \  
  SFO ---- ATL  
     \   /  
      JFK  
  

3. Single Source, Multiple Destinations

• Nodes: JFK, SFO, ATL, ORD
• Edges: JFK -> SFO, JFK -> ATL, JFK -> ORD
• Graph:

  SFO --  
         \  
  ATL --- JFK  
         /  
  ORD --  
  

4. Multi-Step Itinerary

• Nodes: JFK, SFO, ATL, ORD
• Edges: JFK -> SFO, SFO -> ATL, ATL -> ORD
• Graph:

  ORD  
   |  
  ATL  
   |  
  SFO  
   |  
  JFK  
  

5. Disconnected Graph

• Nodes: JFK, SFO, ATL, ORD
• Edges: JFK -> SFO, ATL -> ORD
• Graph:

  ORD   SFO  
   |     |  
  ATL   JFK  
  

In these graphs, the arrows represent the direction of flight, indicating the from and to of each ticket.

By visualizing the cases in this manner, you get a clear idea of how the airports (nodes) and tickets (edges) are connected. It helps in conceptualizing the problem and considering edge cases.

Analyzing different cases offers several key insights:

1. Minimum Input Size: The problem can be as simple as a single flight. In such a case, the itinerary would just follow the single available route.

2. Multiple Ways - Lexicographically Smaller First: When multiple valid itineraries exist, priority must be given to the lexicographically smaller destination airport.

3. Single Source, Multiple Destinations: When multiple flights are departing from the same airport, sorting becomes important for lexicographically smaller routes.

4. Multi-Step Itinerary: For multi-step routes, it’s important to ensure that each ticket is used exactly once, highlighting the need for tracking visited routes.

5. Disconnected Graph: Some airports might not connect to any other airport. While the problem states there will be at least one valid itinerary, it’s crucial to make sure all tickets are used.

These insights will guide the problem-solving process, particularly in handling edge cases and developing a robust, efficient solution.

Identification of Applicable Theoretical Concepts

The problem can be tackled using Graph Theory. Specifically, it is a variation of the Eulerian Path problem. An Eulerian path is a path in a graph that visits every edge exactly once. The tickets can be represented as directed edges in a graph, and the airports can be represented as vertices. The constraint of using all tickets exactly once maps directly to finding an Eulerian Path in the graph.

Graph traversal algorithms, such as Depth-First Search (DFS), can be effective here. A Min-Heap or Priority Queue can be used to ensure lexical ordering when there are multiple edges from a single vertex, i.e., when there are multiple choices of destination from a particular airport.

By identifying this as an Eulerian Path problem, we can immediately apply existing algorithms and data structures to achieve a more efficient solution. This also allows us to leverage established properties of Eulerian Paths to guide our algorithm design. For example, we know that a directed graph has an Eulerian Path if and only if at most one vertex has (out-degree - in-degree) = 1, and at most one vertex has (in-degree - out-degree) = 1, and all other vertices have equal in-degree and out-degree. This property can help in quickly determining if a valid itinerary exists.

Simple Explanation

Imagine you have a bunch of flight tickets. Each ticket takes you from one airport to another. You start your journey at JFK airport. Your goal is to use all the tickets you have, so that you go through a series of airports and end up back at JFK.

The twist is, if you have more than one option to fly out from an airport, you should pick the airport that comes first in alphabetical order. So, if you have a ticket from JFK to Atlanta and another from JFK to Boston, you’d go to Atlanta first because ‘A’ comes before ‘B’.

It’s like having a set of puzzle pieces where each piece has a starting point and an ending point written on it. You start with the JFK piece and try to connect all the pieces in a way that you end up back at JFK, but also make sure you pick the pieces in alphabetical order when you have choices.

So, the problem is figuring out how to use all your tickets following these rules.

Problem Breakdown and Solution Methodology

To solve this problem, think of it as arranging dominos in a line where each domino’s end must match with the next domino’s start. But here, the dominos are flight tickets, and the matching involves departure and arrival airports. Here’s the step-by-step approach:

1. Organize Tickets: Create a ‘map’ where each departure airport points to a list of arrival airports. Sort each list in alphabetical order.

2. Start at JFK: We know we start at JFK. Place JFK as the first element in the final itinerary.

3. Recursive Path Finding:

   • Look at your current airport.
   • If it has outgoing flights, take the one that is alphabetically first and add it to your itinerary. Remove this ticket from your map.
   • Repeat this for the next airport.

4. Backtrack if Stuck: If you reach a point where there are no more outgoing flights and you haven’t used all tickets, backtrack. That means removing the last airport you added to your itinerary and trying the next alphabetical airport from the previous step.

5. Complete the Loop: Once you’ve used all tickets, you’ll have your final itinerary.

Effect of Parameters

• More Tickets: If the number of tickets increases, the solution might take longer but the approach remains the same.
• Multiple Choices: The more choices you have at each airport, the more backtracking you might have to do.

Example

Let’s say we have tickets represented by pairs: JFK->ATL, ATL->JFK, JFK->SFO, SFO->ATL.

1. Organize Tickets:

   • JFK: [ATL, SFO]
   • ATL: [JFK]
   • SFO: [ATL]

2. Start at JFK:

   • Itinerary = [JFK]

3. Recursive Path Finding:

   • JFK has two choices: ATL and SFO. ATL is alphabetically first.
   • Itinerary = [JFK, ATL]
   • ATL has only JFK.
   • Itinerary = [JFK, ATL, JFK]
   • Now, JFK has only SFO.
   • Itinerary = [JFK, ATL, JFK, SFO]
   • SFO has only ATL.
   • Itinerary = [JFK, ATL, JFK, SFO, ATL]

We used all tickets, so we’re done. The final itinerary is [JFK, ATL, JFK, SFO, ATL].

By following these steps, you can solve the problem in an organized manner.

Inference of Problem-Solving Approach from the Problem Statement

1. Itinerary: This is the end result, a list of airports visited in order. Understanding that we need to create an itinerary informs the need for a data structure to hold this sequence.

2. Tickets: These are the pairs of [from, to] airports. Knowing we have these tickets guides us to create a mapping from each “from” airport to its possible “to” airports.

3. Departure and Arrival Airports (from, to): These are the elements in each ticket. They are the keys and values in our mapping.

4. JFK: This is the starting point for the itinerary. Knowing this helps us initialize our itinerary list and mapping.

5. Smallest Lexical Order: This constraint informs us to sort the “to” airports for each “from” airport so that we can simply follow the first available option when building the itinerary.

6. Use all Tickets: This condition tells us we need to visit all the airports on the tickets and we can’t leave any out. This will inform our exit condition for any recursive or iterative approach we take.

7. Multiple Valid Itineraries: Knowing there can be more than one valid answer allows us to prepare for backtracking if we reach an invalid state.

8. Backtracking: This is an algorithmic concept where if you reach an invalid state, you backtrack to the previous state and try a different path. Given the constraints, backtracking is an ideal strategy for this problem.

9. Recursive Path Finding: This suggests using a recursive function to explore each path from the current airport until either the itinerary is complete or you need to backtrack.

Each keyword or concept informs a particular part of the solution, whether it’s the data structure used, the algorithm employed, or how to initialize and terminate the process.

1. Itinerary: Create a table with a single row. As you construct the itinerary, keep adding airport codes in the order they are visited.

   JFK	MUC	LHR	SFO	SJC

2. Tickets: Make a table where each row represents a ticket with a “from” and “to” airport.

   From	To
   JFK	MUC
   MUC	LHR
   LHR	SFO
   SFO	SJC

3. Departure and Arrival Airports: You can enhance the Tickets table with additional columns for “Is Departure” and “Is Arrival.”

   From	To	Is Departure	Is Arrival
   JFK	MUC	Yes	No
   MUC	LHR	No	Yes
   LHR	SFO	No	Yes

4. JFK: Highlight JFK in your tables or diagrams, as it’s your starting point. You could use a color or a marker to do this.

5. Smallest Lexical Order: When listing “to” airports for each “from” airport in a table or graph, sort them lexically.

   From	To Options
   JFK	MUC, LHR (sorted)

6. Use all Tickets: In your Tickets table, add a column for “Used” and mark it as you go along.

   From	To	Used
   JFK	MUC	Yes
   MUC	LHR	No

7. Multiple Valid Itineraries: When you hit a branching point in your diagrams where you could go to more than one airport next, represent it clearly, like a tree diagram.

8. Backtracking: Whenever you backtrack in your tree diagram, mark that branch as invalid or erase it.

9. Recursive Path Finding: When exploring a path, extend it from the existing nodes in your tree diagram.

These visual aids can help you understand the different components and constraints of the problem, making it easier to arrive at a solution.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Stepwise Refinement

1. Stepwise Refinement of Approach:

   • Step 1: Data Preparation

     1. Read the given list of tickets.
     2. Organize tickets into a map where the key is the “from” airport and the value is a list of “to” airports.
     3. Sort each list of “to” airports lexicographically.

   • Step 2: Initialization

     1. Create an empty list to store the final itinerary.
     2. Add “JFK” as the starting point of the itinerary.

   • Step 3: DFS (Depth First Search)

     1. Start the DFS algorithm from “JFK”.
     2. For each airport (starting from JFK), explore its “to” options in lexical order.
     3. Backtrack if a path does not use all tickets.

   • Step 4: Final Itinerary

     1. Once DFS is complete and all tickets are used, the itinerary list should be the answer.

2. Distill into Granular, Actionable Steps:

   • Data Preparation

     1. Use a hash table to organize tickets.
     2. Use built-in sorting functions to sort “to” airports.

   • Initialization

     1. Initialize an empty Python list or array.
     2. Append “JFK” to the list.

   • DFS

     1. Create a function for DFS that takes the current airport and the itinerary as parameters.
     2. Use a loop to iterate over “to” airports.
     3. Use conditional statements to decide when to backtrack.

   • Final Itinerary

     1. Return the itinerary list.

3. Independent Parts:

   • Sorting each “to” list can be done independently.
   • Once data preparation is complete, the DFS step is also independent.

4. Repeatable Patterns:

   • The DFS step is recursive and thus a repeatable pattern.
   • The sorting of “to” airports is another repeatable pattern that can be abstracted.

By breaking down the problem in this way, you can focus on each component, making it easier to tackle the full problem.

Solution Approach and Analysis

Detailed Approach to Solving the Problem

1. Data Preparation

First, we need to arrange the given list of tickets in a way that’s easy to navigate. Think of this like organizing a bunch of keys on a keyring, each labeled with its corresponding door.

Steps:

• Create an empty hash table.
• Loop through each ticket in the list.
• Insert the “from” airport as a key and “to” airports as values in a list.

2. Initialization

Once our data is prepared, we establish our starting point. Imagine you’re at a train station, and all trips must start from Platform 1.

Steps:

• Create an empty list to store the final route.
• Append “JFK” as the first element since it’s our starting point.

3. Depth First Search (DFS)

Next, we perform a Depth-First Search (DFS) from JFK. Think of DFS like exploring a maze. You pick one path and follow it as far as it goes before you backtrack.

Steps:

• Initiate DFS at “JFK”.
• For each “from” airport, visit its “to” airports in lexical order (due to sorting in the first step).
• If a path doesn’t consume all tickets, backtrack and try the next option.
• If a path does consume all tickets, store it as a possible solution.

4. Final Itinerary

After DFS completes, we’ll have our final itinerary.

Steps:

• Return the itinerary list as the answer.

Changes in Problem Parameters and Effects

• Number of Tickets: Increasing the number of tickets will increase the time complexity.
• Lexical Ordering: Different lexical orders could produce different valid itineraries, but the smallest lexicographic itinerary will be chosen due to sorting.

Example Cases

Let’s look at one example for demonstration:

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

1. Data Preparation:

   • {"JFK":["LHR","MUC"], "MUC":["SFO"], "LHR":["SFO"], "SFO":["JFK"]}

2. Initialization:

   • ["JFK"]

3. DFS:

   • Start from JFK -> Go to LHR (smallest lexicographic option)
   • From LHR -> Go to SFO
   • From SFO -> Go back to JFK
   • From JFK -> Go to MUC
   • From MUC -> Go to SFO

4. Final Itinerary:

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

This way, each step contributes to finding the final itinerary by navigating through the ‘from’ and ’to’ airports in a DFS manner.

Identify Invariant

The invariant in this problem is that the journey starts from “JFK” and each “from” airport must have a corresponding “to” airport in the given list of tickets. In other words, once you land at an airport, you must have a ticket to fly out of it, except for the final destination. This constraint remains true throughout the problem and guides the traversal through the list of tickets.

Identify Loop Invariant

In the context of this problem, if you’re using a loop to implement Hierholzer’s algorithm, the loop invariant could be:

“At the start of each iteration of the loop, all vertices in the stack represent a sequence of edges that form a valid path from the starting vertex ‘JFK’ to the current vertex on the top of the stack.”

This loop invariant essentially ensures that as you iterate through your loop to traverse the graph, you maintain a stack that keeps track of a valid sequence of flights from the starting airport to wherever the stack’s top element is at that point in time.

Maintaining this invariant helps ensure that each iteration of the loop builds on a correct partial solution, allowing you to eventually pop elements off the stack to form the complete itinerary in reverse order.

In this problem, the terms “invariant” and “loop invariant” could refer to different things depending on the context in which they are used.

The “loop invariant” is specific to the loop structure within the code. It is a condition that holds before and after each iteration of the loop, ensuring the correctness of the algorithm.

On the other hand, an “invariant” in a broader sense could refer to a condition or set of conditions that remain constant throughout the execution of the entire algorithm or program, not just within a loop.

In the case of Hierholzer’s algorithm for this problem, if you’re not using a loop, then talking about a loop invariant wouldn’t make sense. You’d discuss the algorithm’s invariant in a broader context.

However, if you’re using a loop to implement the algorithm, the loop invariant becomes relevant, and it would be a specific instance of a broader algorithmic invariant.

Identify Recursion Invariant

In the context of this problem, if you’re using a recursive function to implement Hierholzer’s algorithm, the recursion invariant would be:

“At any given recursive call, all edges (tickets) that have been traversed from the starting vertex to the current vertex form a valid part of the Eulerian path.”

In other words, the partial path built up to the point of any recursive call should adhere to the rules of forming an Eulerian path, i.e., each edge is used exactly once, and the in-degree and out-degree balance rule is observed for each visited vertex.

Maintaining this invariant ensures that you will ultimately build the complete Eulerian path by the time your base case is reached. It also guarantees that each recursive call is working on a valid “sub-problem” of the larger problem, adhering to the constraints and requirements set by the problem statement.

In the context of Hierholzer’s algorithm for finding an Eulerian path, the term “invariant” could refer to a condition that holds true throughout the algorithm’s execution. On the other hand, an “invariant during recursion” is a specific type of invariant that holds true before and after each recursive call.

If the algorithm is implemented recursively, then discussing an “invariant during recursion” would be relevant. This would be a condition that must hold true every time a recursive call is made and returned, ensuring that the algorithm works as expected.

In summary, while both are types of invariants, they serve different roles depending on the structure of the algorithm—whether it is implemented using loops, recursion, or some other construct.

Thought Process

Cues in the Problem Statement

1. All journeys start from “JFK.”
2. If there are multiple itineraries, pick the lexicographically smaller one.
3. Use all the tickets exactly once.

Direction for Approach

1. It’s a graph problem: airports are nodes, and tickets are directed edges.
2. The graph traversal should start from “JFK.”
3. Lexicographically smaller paths should be preferred.

Insights

• Since the lexicographically smaller path should be preferred, a priority queue could be useful to maintain the available options at each airport.

Step-by-Step Approach

1. Create an adjacency list with priority queues to represent the graph.
2. Start the traversal from “JFK.”
3. Use Depth First Search (DFS) and backtrack to find the path.
4. Return the reconstructed path.

Code

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from collections import defaultdict
import heapq

def findItinerary(tickets):
    graph = defaultdict(list)
    
    # Step 1: Build the graph
    for from_airport, to_airport in tickets:
        heapq.heappush(graph[from_airport], to_airport)
    
    path = []
    
    # Step 3: DFS Function
    def dfs(airport):
        while graph[airport]:
            next_airport = heapq.heappop(graph[airport])
            dfs(next_airport)
        path.append(airport)
        
    # Step 2: Start DFS from "JFK"
    dfs("JFK")
    
    # Step 4: Return the path in reverse order
    return path[::-1]

# Test the function
tickets1 = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
print(findItinerary(tickets1))

tickets2 = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
print(findItinerary(tickets2))

This code should correctly solve the problem by following the defined steps and using the DFS method for graph traversal.

Establishing Preconditions and Postconditions

Parameters

1. Inputs: The input is a list of lists, where each inner list contains two strings.
2. Types: List of lists of strings.
3. Representation: Each inner list represents a ticket with a “from” airport and a “to” airport.

Preconditions

1. State of Program: None specific.
2. Constraints: Each list should contain exactly two airports represented as strings. The graph formed must be semi-Eulerian.
3. Program State: No specific state requirements.

Method Functionality

1. Expected Action: The method is expected to find an itinerary that uses all the given tickets exactly once, starting from “JFK.”
2. Interaction: It builds a graph from the input tickets and then performs DFS to find the path.

Postconditions

1. State of Program: A path is identified, and the graph data structure might be emptied in the process.
2. Return Value: A list of strings representing the itinerary.
3. Side Effects: The graph data structure is modified during execution.

Error Handling

1. Preconditions Not Met: The method does not explicitly handle errors for invalid inputs or non-semi-Eulerian graphs.
2. Response: No specific error handling is built in; an incorrect or incomplete itinerary may be returned if the preconditions are not met.

Problem Decomposition

Problem Understanding

1. In My Words: Given a list of flight tickets represented as “from-to” pairs, find a path that uses every ticket once and starts from “JFK.”
2. Key Components: Tickets list, starting point (“JFK”), path formation.

Initial Breakdown

1. Major Parts:
   • Building a graph from tickets.
   • Traversing the graph to form the itinerary.

Subproblem Refinement

1. Build Graph:
   • Sort destination airports.
   • Create adjacency lists.
2. Traverse Graph:
   • Initialize starting point (“JFK”).
   • Use DFS to find the path.
   • Handle branching (multiple destinations).

Task Identification

1. Repeated Tasks:
   • Graph traversal (DFS) could be repetitive when dealing with branching.

Task Abstraction

1. Clarity and Reusability:
   • Build and sort graph: Could be reused in other graph problems.
   • DFS traversal: Reusable in other graph-based problems.

Method Naming

1. Task Names:
   • buildGraph(): for graph formation.
   • findItinerary(): for DFS traversal and path formation.

Subproblem Interactions

1. Order and Dependencies:
   • Build graph first, then traverse.
   • Traversal depends on the graph being built and sorted.

From Brute Force to Optimal Solution

Brute Force Solution

1. Enumerate all possible paths: Starting from “JFK,” try every possible combination of flights.
2. Validation: Check which of these paths use all the tickets.
3. Lexicographical Order: Sort all the valid paths and pick the smallest one.

Inefficiencies

• Time Complexity: Generating all permutations would take O(n!), which is infeasible for large numbers of tickets.
• Space Complexity: Storing all these paths would also take up a significant amount of space, O(n!).

Optimization Steps

Step 1: Use Graph Data Structure

• Reasoning: The problem represents a graph, where the airports are nodes and the tickets are edges.
• Improvement: This reduces the problem space by representing the relationships more naturally.
• Impact: Time and Space complexity are now potentially O(V + E), where V is the number of vertices (airports) and E is the number of edges (tickets).

Step 2: Sort the Destinations

• Reasoning: We are interested in the lexicographically smallest path, so sorting the destinations helps.
• Improvement: Once sorted, you don’t need to compare paths at the end.
• Impact: Sorting each adjacency list would take O(V * log(V)) for each airport, but it’s a one-time cost.

Step 3: Depth-First Search (DFS) for Traversal

• Reasoning: You only need to consider each ticket once, and DFS is well-suited for such scenarios.
• Improvement: DFS allows us to find the path in linear time relative to the number of tickets.
• Impact: The time complexity remains O(V + E).

Step 4: Backtracking

• Reasoning: If a path turns out to be a dead-end (i.e., not all tickets are used), we need to backtrack.
• Improvement: We will only keep valid paths in memory, further optimizing space usage.
• Impact: DFS with backtracking still works in O(V + E) time, and space is further optimized.

Final Complexity

• Time Complexity: O(V * log(V) + E), dominated by sorting and DFS.
• Space Complexity: O(V + E) for storing the graph and the stack for DFS.

The optimized approach is much more efficient than the brute-force method in both time and space.

Code Explanation and Design Decisions

1. Initial Parameters

• Graph: A hash table representing our flight routes.
• Start: The starting airport, initially “JFK.”
• Path: An empty list to hold the final itinerary.

These parameters help us keep track of airports, available routes, and our current path.

2. Primary Loop

The primary loop is a Depth-First Search (DFS) that iterates through each destination from a given airport. Each iteration signifies one possible extension of the current itinerary.

3. Conditions and Branches

There are two main conditions:

1. Check for Dead-End: If we reach an airport with no available outgoing flights, we’ve hit a dead-end.
2. All Tickets Used: If the length of the path equals the number of tickets plus one (for the starting point), we’ve found a valid path.

These conditions align with the problem constraints: finding a valid itinerary that uses all tickets and starts from “JFK.”

4. Updates to Parameters

Within the loop, we remove a ticket once it’s used and add its destination to the path. We then call the DFS recursively. These updates reflect the state changes in our path and available tickets.

5. Invariant

The invariant is that the path list always contains a valid, albeit possibly incomplete, itinerary. This ensures we are always working towards the problem’s objective.

6. Significance of Final Output

The final output is the lexicographically smallest valid itinerary that uses all the tickets starting from “JFK.” It satisfies all the constraints and requirements laid out in the problem statement.

Coding Constructs

1. High-Level Strategies

The code uses Depth-First Search (DFS) to explore all possible itineraries. It leverages a graph data structure to represent the flight connections and employs backtracking to revert steps and explore alternative routes.

2. Purpose to a Non-Programmer

Imagine you have a handful of flight tickets, each labeled with a departure and arrival airport. This code helps you stitch these tickets together into a complete travel itinerary that uses all tickets and starts from a specified airport, JFK, in the lexicographically smallest order.

3. Logical Elements

• Loop: To go through each destination for a given airport.
• Conditional Statements: To check for end conditions or dead-ends.
• Recursion: To explore each route deeply before backtracking.
• Data Structures: Hash table for the graph, list for the path.

4. Algorithmic Approach in Plain English

Start at JFK airport. Look at all the places you can go to next. Choose the destination with the smallest name and fly there. Once there, repeat the process. If you reach a point where you can’t fly anywhere, backtrack to the previous airport and try the next smallest destination. Keep doing this until you have used all your tickets.

5. Key Steps or Operations

1. Initialize: Set up the graph and the path.
2. Sort Destinations: Lexicographically sort destinations for each airport in the graph.
3. DFS Loop: Start from JFK, and for each destination:
   • Use the ticket (remove it from the graph).
   • Add the destination to the path.
   • Recursively perform DFS.
   • If stuck, backtrack (undo step).

6. Algorithmic Patterns

• Graph Traversal: Using DFS for exploring all routes.
• Backtracking: Reverting steps when hitting a dead-end.
• Sorting: To achieve the lexicographically smallest path.
• Recursion: To implement the DFS and backtracking logic.

Q&A

The problem can be solved using Depth-First Search (DFS). You can think of the tickets as a directed graph where each ticket is an edge from “from” to “to”. The goal is to traverse the graph in such a way that all the edges are visited exactly once.

Here’s a step-by-step guide to the solution:

1. Create the Graph: Create a graph as an adjacency list. Since we need the lexicographically smallest itinerary, sort the destinations for each source.

2. DFS Traversal: Perform DFS traversal starting from “JFK”. As you traverse each edge, remove it from the graph to mark it as visited. Use a stack to store the current itinerary.

3. Handle Dead Ends: If you reach a dead end (no more unvisited destinations from the current airport), backtrack by popping airports from the stack and adding the current airport to the result.

4. Build Result: Once all the edges are visited, the stack will contain the reversed itinerary. Reverse it to get the final result.

Here’s the code:

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from collections import defaultdict

class Solution:
    def findItinerary(self, tickets):
        # Create a graph with sorted destinations for each source
        graph = defaultdict(list)
        for from_airport, to_airport in sorted(tickets, reverse=True):
            graph[from_airport].append(to_airport)

        # Perform DFS
        result = []
        stack = ["JFK"]
        while stack:
            while graph[stack[-1]]:
                stack.append(graph[stack[-1]].pop())
            result.append(stack.pop())

        # Return the reversed result
        return result[::-1]

Key Takeaways

• The problem is modeled as a graph where a DFS is used to find the itinerary.
• Backtracking helps in handling dead ends and ensures all tickets are used.
• The code leverages a stack to keep track of the itinerary and constructs the result efficiently.
• The time complexity of this solution is (O(E \log E)), where (E) is the number of edges (tickets), and the space complexity is (O(E)).

Here’s the implementation of the findItinerary method:

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from collections import defaultdict

class Solution:
    def findItinerary(self, tickets):
        def dfs(node):
            while graph[node]:
                dfs(graph[node].pop(0))
            result.append(node)

        graph = defaultdict(list)
        for fromi, toi in tickets:
            graph[fromi].append(toi)

        for key in graph:
            graph[key].sort()  # Sort destinations in lexical order

        result = []
        dfs("JFK")

        return result[::-1]  # Reverse the result to get the correct order

This implementation uses a depth-first search (DFS) approach to reconstruct the itinerary. The graph dictionary is used to store destinations for each departure airport. Sorting the destinations in lexical order ensures the smallest lexical order for valid itineraries. The DFS starts from “JFK” and builds the itinerary. Finally, the result is reversed to get the correct order.

Identifying Problem Isomorphism

“Reconstruct Itinerary” involves creating an itinerary from a list of tickets, each represented as a pair of source and destination airports. This problem deals with graph traversal and topological sorting and requires maintaining lexical order for multiple valid paths.

A simpler problem employing related concepts is “Course Schedule”. In this problem, you need to determine whether it’s possible to finish all courses given a list of prerequisites. It is simpler because it only requires a yes/no answer, rather than constructing an itinerary.

An approximately isomorphic problem is “All Paths From Source to Target”. This problem requires finding all possible paths from source to target in a directed acyclic graph (DAG), akin to creating an itinerary. However, it does not require maintaining lexical order.

A more complex problem is “Find Itinerary from Flights”. In this problem, given a list of flights, you need to reconstruct the sequence of flights into a continuous itinerary. What makes it more complex is the possibility of multiple connections from a given city and the need to make optimal decisions.

Arranged from simpler to more complex:

1. “Course Schedule” - Determine if it’s possible to finish all courses given a list of prerequisites.
2. “Reconstruct Itinerary” - Build an itinerary from a list of flights, maintaining lexical order for multiple valid paths.
3. “All Paths From Source to Target” - Similar to the original problem but requires finding all possible paths in a DAG.
4. “Find Itinerary from Flights” - More complex due to the need to deal with multiple connections and make optimal decisions.

10 Prerequisite LeetCode Problems

“332. Reconstruct Itinerary” requires depth-first search and graph traversal. Here are some problems you should solve before tackling this one:

1. 207. Course Schedule: A fundamental problem that introduces the concept of directed graph traversal using depth-first search.

2. 210. Course Schedule II: A sequel to Course Schedule, this problem requires not only checking for possible traversal, but also returning a valid order, which is similar to Reconstruct Itinerary.

3. 261. Graph Valid Tree: This problem tests the ability to identify whether a graph is a tree using depth-first search, which requires understanding of basic graph traversal techniques.

4. 269. Alien Dictionary: This problem involves constructing an ordering from a list of strings, which is similar to the itinerary reconstruction.

5. 133. Clone Graph: This problem involves depth-first search in a graph and can help to understand the concept of graph traversal.

6. 743. Network Delay Time: This problem involves finding shortest paths in a graph, similar to finding the lexicographically smallest itinerary.

7. 886. Possible Bipartition: This problem requires assigning nodes in a graph to two distinct sets, which is a variant on the typical graph traversal problem.

8. 684. Redundant Connection: This problem involves identifying and removing edges in a graph to eliminate cycles, requiring a good understanding of graph structure.

9. 547. Number of Provinces: This problem tests the ability to identify connected components in a graph using depth-first search.

10. 490. The Maze: This problem involves searching through a maze to find a target location, which will further refine your depth-first search skills and help you understand more complex traversal techniques.

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class Solution:
def findItinerary(self, tickets: List[List[str]]) -> List[str]:
    graph = {}

    for src, dst in tickets:
        if src in graph:
            graph[src].append(dst)
        else:
            graph[src] = [dst]

    for src in graph.keys():
        graph[src].sort(reverse=True)

    stack = []
    res = []
    stack.append("JFK")

    while len(stack) > 0:
        elem = stack[-1]
        if elem in graph and len(graph[elem]) > 0: 

            stack.append(graph[elem].pop())
        else:
            res.append(stack.pop())

    return res[::-1]

Problem Classification

The task is to reconstruct a journey from a list of tickets. This is a Journey Reconstruction Problem.

Language Agnostic Coding Drills

1. Understanding the data structure - Dictionary: The code makes use of dictionaries to create an adjacency list of airports (source to destination). The task is to learn how to create a dictionary, add elements, check if a key is in a dictionary, and modify values of a key.

2. Understanding data structure - Lists and List operations: The code involves extensive usage of lists. List operations such as appending elements, popping elements, and reversing a list are utilized. Learn about lists and their operations.

3. Understanding of Sorting: The code uses sorting to arrange destinations in lexicographical order. Understand how sorting works in the language of your choice.

4. Understanding the data structure - Stack: The solution uses the stack data structure to keep track of the itinerary. It’s crucial to understand how stack works including operations like push (append in python) and pop.

5. Working with loops and conditionals: The code contains while loops and if-else conditions. Understand how to write and control loops and how to use conditional statements to guide the program flow.

6. Problem-Specific Concept - Depth-First Search (DFS) and Hierholzer’s algorithm: The problem is essentially finding an Eulerian path in the graph. Hierholzer’s algorithm is used for this purpose. DFS is used for traversal in the graph. This requires an understanding of graphs, DFS and Hierholzer’s algorithm.

Following this incremental learning path will equip learners with all the necessary skills required to solve this problem and similar ones. Understanding these steps is crucial for devising the final solution. Start by building the graph from given tickets. Once the graph is built, start from “JFK” and do a DFS, visiting destinations in lexicographical order. The key point to remember here is that during the DFS, if a node is found that has no outgoing edge (i.e., no further travel is possible), it is added to the final itinerary. Since we’ve traversed the destinations in lexicographical order, the final itinerary obtained (after reversing) will also maintain the same order among all possible itineraries.

Targeted Drills in Python

1. Drill 1 - Creating a Dictionary:

   This is the fundamental step in this solution, where we are creating a graph represented by a dictionary.

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   def create_dict(tickets): graph = {} for src, dst in tickets: if src in graph: graph[src].append(dst) else: graph[src] = [dst] return graph

2. Drill 2 - Sorting a Dictionary:

   Once we have the dictionary created, we need to sort the destinations of each source in reverse order. This is necessary for later steps.

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   def sort_dict(graph): for src in graph.keys(): graph[src].sort(reverse=True) return graph

3. Drill 3 - Depth-First Search:

   Here we are going to perform the depth-first search algorithm using a stack. We are going to pop the last destination from the destination list and append it to the stack until there are no more destinations left. Then, we’ll append the stack elements to the result list.

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   def dfs(graph): stack = [] res = [] stack.append("JFK") while len(stack) > 0: elem = stack[-1] if elem in graph and len(graph[elem]) > 0: stack.append(graph[elem].pop()) else: res.append(stack.pop()) return res

4. Drill 4 - Integrating all drills:

   Here we integrate all the drills into one function to get the final solution.

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   class Solution: def findItinerary(self, tickets): graph = self.create_dict(tickets) graph = self.sort_dict(graph) res = self.dfs(graph) return res[::-1]

Here each drill helps to build up to the final solution, and the concepts involved are creating a dictionary, sorting a dictionary, and performing a depth-first search on the graph. This can be combined to solve the problem as shown in the final solution.

Q&A

Input: [[“JFK”,“KUL”],[“JFK”,“NRT”],[“NRT”,“JFK”]] Output: [“JFK”,“KUL”] Expected: [“JFK”,“NRT”,“JFK”,“KUL”]

How could “NRT” have a higher lexical ordering than “KUL”?

The output is incorrect because the solution didn’t account for the case where going to a lexicographically larger destination (“NRT” in this case) would actually allow using all the tickets.

In this example, if you go from “JFK” to “KUL” first, you won’t be able to use the [“NRT”, “JFK”] ticket. The lexicographically smaller choice traps you into a suboptimal itinerary.

So, while it’s tempting to always pick the lexicographically smaller airport, it’s essential to consider the entire itinerary to make sure all tickets are used. This case highlights the need for backtracking or other methods to explore multiple paths.

When input is [ [“JFK”,“LHR”] ,[“JFK”,“MUC”] ]

the output expected is [“JFK”, “LHR”,“MUC” ]

this is wrong because the person does not have a ticket from LHR to MUC at all! where did the concept of “from” and “to” go ?

You’re correct. The expected output [“JFK”, “LHR”, “MUC”] is incorrect based on the problem statement. According to the given tickets, you can only fly from “JFK” to either “LHR” or “MUC.” There is no ticket allowing a flight from “LHR” to “MUC,” so the expected output violates the problem’s constraints.

The concept of “from” and “to” must be strictly adhered to when reconstructing the itinerary. If there’s no ticket for a specific leg, that path is not valid. Based on the input [[“JFK”,“LHR”],[“JFK”,“MUC”]], the only valid itineraries are [“JFK”, “LHR”] or [“JFK”, “MUC”], not [“JFK”, “LHR”, “MUC”].

Input: [[“JFK”,“KUL”],[“JFK”,“NRT”],[“NRT”,“JFK”]] Output: [“JFK”,“KUL”] Expected: [“JFK”,“NRT”,“JFK”,“KUL”]

As KUL has a smaller lexical order than NRT, that is chosen. Isn’t that expected? Why does the expected output have JFK, NRT?

The problem statement specifies that if there are multiple valid itineraries, you should return the itinerary with the smallest lexical order. However, lexical order applies when you have choices at a single airport.

In this case, you start at “JFK” and have two choices: “KUL” and “NRT.” Lexically, “KUL” comes first. But choosing “KUL” first doesn’t allow you to use all the tickets. That violates the constraint of using all tickets exactly once.

On the other hand, if you choose “NRT” first, you can go from “NRT” back to “JFK,” and then from “JFK” to “KUL,” thus using all tickets. Therefore, even though “KUL” has a smaller lexical order, the constraint of using all tickets takes precedence. That’s why the expected output is [“JFK”,“NRT”,“JFK”,“KUL”].

Could anyone help me understand why does Hierholzer’s algorithm work in this problem?? I see that the problme mentions there exists an answer, is that the valid reason, why if we keep traversing and backtracking on a dead end, would give us a valid answer??

I’m looking for more formal explaination/logic than simple words, could anyone help?

Hierholzer’s algorithm is typically used to find an Eulerian path or circuit in a graph. An Eulerian path is a trail in a graph that visits every edge exactly once. The algorithm works well in this problem because of a few key properties:

1. Existence of Eulerian Path: The problem guarantees that an Eulerian path exists, i.e., you can travel every ’ticket’ exactly once. This is confirmed by the problem’s condition that an answer always exists.

2. Directed Graph: The tickets can be considered as directed edges in a graph. You can only go from the “from” airport to the “to” airport as specified on each ticket. Hierholzer’s algorithm can easily adapt to directed graphs.

3. Degree of Vertices: In an Eulerian path, each vertex (except possibly for the start and end vertex) has an equal in-degree and out-degree. This means you can ’enter’ and ’leave’ each vertex the same number of times. In the problem, the number of tickets to enter an airport would be the same as the number to leave, except for JFK (the start) and possibly another airport if the path is not a circuit.

4. Dead Ends and Backtracking: Hierholzer’s algorithm naturally handles dead ends via backtracking. When you hit a dead-end, you know you’ve traversed all outgoing edges (tickets) for that vertex (airport). You can then backtrack and continue traversing other unexplored edges, eventually forming the Eulerian path.

5. Lexicographic Ordering: While Hierholzer’s doesn’t inherently give a lexicographically minimal path, you can achieve this by always choosing the smallest available vertex to visit next, which aligns well with the problem’s constraints.

The formal reasoning behind why Hierholzer’s works is fundamentally based on the properties of Eulerian paths and the constraints of the problem. Given the existence of an Eulerian path and the balanced in-degree and out-degree for each vertex, Hierholzer’s will construct the path by traversing each edge exactly once, thus providing a valid answer.