Clone Graph
Steps for Iterative Algorithm
Define Problem
 What is the problem to be solved?
 What is the precondition? Where are you starting?
 What is the post condition? Where is your destination?
Define Step
What basic stpes will head you in the correct direction?
Measure of Progress
You must define a measure of progress.
Define Loop Invariant
Define a loop invariant that gives you a picture of the computation state when it is at the top of the main loop.
Main Steps
Consider a typical step to be taken during the middle of the computation. Write the pseudocode to take a single step within the loop.
Make Progress
Each iteration of main step must make progress according to measure of progress.
Maintain Loop Invariant
Each iteration of main step must ensure that the loop invariant is true again when the computation gets back to the top of the loop.
Establish the Loop Invariant
Write the pseudocode before you enter the loop to establish the loop invariant.
Exit Condition
Write the condition that causes the computation to break out of the loop.
Ending
 How does the exit condition together with the invariant ensure that the problem is solved?
 How do you product the required output?
 Write the pseudocode after the loop ends and to return the required output.
10 Prerequisite LeetCode Problems
Identify 10 LeetCode simpler problems, excluding the problem itself that I should solve as preparation for tackling . Include the name of the given problem in the response before the list. Do not add double quotes for the items in the list. Include the reason why that problem is relevant. The format of the response must be:
For the , the following is a good preparation:
Problem Classification
The problem belongs to the domain of Graph Theory and Data Structures. It specifically deals with graph traversal and object cloning.
What
 A connected, undirected graph is given.
 Each node in the graph has an integer value and a list of neighbors.
 A deep copy (clone) of the graph needs to be returned.
Further Classification
 Graph Structure: The graph is undirected and connected.
 Node Structure: Each node has a value and a list of neighboring nodes.
 Objective: Create a deep copy of the graph.
 Constraints: Unique node values, no selfloops, and no repeated edges. The graph can contain between 0 and 100 nodes.
This is a graph cloning problem, requiring understanding of graph traversal algorithms like DepthFirst Search (DFS) or BreadthFirst Search (BFS) for a complete solution. It also tests objectoriented programming concepts like object copying.
Clarification Questions
 Is the graph strictly connected, or could there be isolated nodes?
 Are node values always positive integers?
 Do we have to preserve the node values in the cloned graph, or can we renumber them?
 What should be returned if the graph is empty?
 Is the given adjacency list the only format the graph will be presented in?
 Will the graph have weighted edges, or are all edges considered to be of equal weight?
 Do we need to validate that the input graph satisfies all the given constraints, or can we assume it will always be valid?
 Will the list of neighbors for each node always be unordered, or could they be sorted in some way?
 Can the graph have cycles, or is it a treelike structure without any cycles?
 Is the memory usage a concern while cloning the graph, or are we only focusing on correctness?
Asking these questions can help clarify assumptions and constraints, leading to a more accurate solution.
Identifying Problem Isomorphism
“Clone Graph” can be mapped to “Copy List with Random Pointer”.
Reasoning:
Both problems deal with the concept of deepcopying a data structure with interconnected nodes. In the first problem, you are deepcopying a graph with an adjacency list, and in the second, you are deepcopying a linked list where each node has a random pointer to another node in the list.
The “Copy List with Random Pointer” problem is simpler because it involves a linear data structure (linked list), which is generally easier to manage than the more complex structure of a graph in “Clone Graph”.
Problem Analysis and Key Insights
Graph Representation: The problem deals with an undirected graph, represented through adjacency lists.
Node Structure: Each node has a value and a list of neighbors. The value is unique and acts like an identifier for the node.
Deep Copy: The goal is to create a clone of the original graph, meaning that new memory should be allocated for each cloned node.
Uniqueness: Node values are unique and range from 1 to 100, making them reliable keys if a mapping technique is to be used.
Constraints:
 No repeated edges
 No selfloops
 The graph is connected
Initial Node: The starting node for the traversal is always given, and it is the node with the value 1.
Output: The output should also be a node reference, presumably of the node with value 1 in the cloned graph.
Test Case Format: The test case format specifies the graph using adjacency lists, making it clear how the graph’s edges are defined.
Empty Graph: Special cases include empty graphs or nodes without neighbors. These edge cases need to be handled explicitly.
Objective: The main objective is to return the clone of the given node as a reference to the cloned graph.
These insights can guide the design of an efficient algorithm for solving the problem.
Problem Boundary
The scope of the problem is confined to:
Graph Theory: The main domain is graph theory, focusing on undirected graphs.
Data Structure Manipulation: It involves manipulating and cloning nodes, each containing a value and a list of neighbors.
Traversal Algorithms: Graph traversal methods like Depth First Search (DFS) or Breadth First Search (BFS) are likely to be used for cloning.
Memory Management: Requires the creation of new nodes for the cloned graph, emphasizing memory allocation.
Edge Cases: Includes handling special or edge cases like empty graphs or nodes without neighbors.
Uniqueness Constraint: Requires maintaining the unique value constraint while cloning.
Connectivity: Assumes the graph is connected, meaning that all nodes are reachable from the given starting node.
Algorithm Complexity: The problem could be solved in varying time and space complexities, but optimizing those would be within the problem’s scope.
Output: The final output is a reference to the cloned graph, specifically the node with value 1.
The problem does not cover:
Directed Graphs: Only undirected graphs are considered.
Weighted Edges: The problem statement does not include weighted edges, so algorithms dealing with weights are out of scope.
Cycle Detection: Since the graph is connected and without selfloops or repeated edges, cycle detection is not necessary.
Disjoint Sets: Since the graph is connected, there’s no need to consider disjoint sets or components.
By understanding the scope, you can better focus on the essential parts of the problem, thereby making the problemsolving process more efficient.
To establish the boundary of the problem, consider the following aspects:
Input Constraints:
 Number of nodes ranges from 0 to 100.
 Node values are unique and range from 1 to 100.
 No selfloops or repeated edges.
Functional Requirements:
 Perform a deep copy of the graph.
 Must return the clone of the node with value 1.
Domain:
 Limited to undirected graphs.
Algorithmic Complexity:
 While the problem doesn’t specify time or space constraints, an efficient algorithm would be desirable.
Data Structure:
 Nodes have a value and a list of neighbors.
Graph Properties:
 Graph is connected.
 Nodes have 1based index values.
Output Requirements:
 Return a reference to the first node of the cloned graph.
Edge Cases:
 Empty graph.
 Node without neighbors.
External Libraries:
 The problem is expected to be solved using standard programming libraries, without the aid of specialized graph libraries.
By identifying these aspects, you set the boundary of the problem. This aids in understanding what to focus on and what can be disregarded during the problemsolving process.
Distilling the Problem to Its Core Elements
Fundamental Concept:
 This problem is fundamentally about graph traversal and cloning. It explores how to make an independent copy of a given graph data structure, which can be crucial in various computational contexts.
Simplest Description:
 Imagine you have a web of interconnected dots. Each dot knows which other dots it’s connected to. You have to create a new web that is exactly the same but independent of the original.
Core Problem:
 The core problem is to create an identical but independent copy of a given interconnected web of nodes.
Key Components:
 Nodes with unique values.
 Edges connecting these nodes.
 The need for deep cloning, meaning the new graph should be independent of the original.
Minimal Set of Operations:
 Visit each node in the graph exactly once.
 Create a new node for each visited node, with the same value.
 Map the relationships between the original node and its neighbors to the new node.
 Store a reference between original and new nodes to avoid duplicates.
By identifying these elements, you have a clearer vision of the core problem, the essential operations needed to solve it, and the underlying concept it is based upon.
Visual Model of the Problem
Visualizing this problem can be effective using a graph representation. Here’s how to go about it:
Nodes as Dots: Think of each node in the graph as a dot on a paper. Label them with their respective values for easier identification.
Edges as Lines: Draw lines between the dots to represent connections (edges) between nodes. These lines represent the neighbors of each node.
Color Coding: You could use different colors for nodes to show their status during the traversal. For example, white for unvisited, grey for currently being processed, and black for visited and copied.
Clone Graph: Next to the original graph, start drawing a clone graph. As you walk through the algorithm, add the respective nodes and edges in the clone graph, establishing the onetoone mapping visually.
Mapping Arrows: Draw arrows or dashed lines between the nodes of the original graph and their respective clones in the new graph. This will help you visualize the cloning process.
Traversal Path: You can also visually represent the traversal path, either using numbers to indicate the order in which nodes were visited or drawing arrows to indicate the direction of traversal.
Adjacency List/Table: Additionally, you can maintain a table or list on the side, jotting down how each node is connected to its neighbors. Update this for the cloned graph as well.
By doing this, you can get a spatial understanding of both the structure of the graph and the cloning process. It helps clarify the concept of “deep copy” and the relationships between nodes in the original and cloned graphs.
Problem Restatement
You have a graph made of nodes and edges, where each node has a unique integer value and a list of its neighboring nodes. Your task is to make an exact copy of this graph, including both nodes and their connections. The input graph is connected, meaning you can reach any node from any other node. There are no duplicate edges or selfloops. You are given a reference to the starting node of the graph (the one with the value of 1), and you should return a reference to the starting node of the cloned graph.
Abstract Representation of the Problem
The problem is about creating an isomorphic copy of a graph. In abstract terms, you have a connected, undirected graph (G) with (N) vertices, each having a unique identifier. The graph is defined by its adjacency list representation, specifying the neighbors for each vertex. Your task is to create another graph (G’) that is isomorphic to (G), preserving the unique identifiers and the adjacency relationships between vertices. You are given a reference to a specific vertex (v) in (G), and you should return a reference to the corresponding vertex (v’) in (G’).
Terminology
Graph: A collection of vertices (nodes) and edges connecting these vertices. The graph is the primary data structure in this problem.
Vertex (Node): An individual element in a graph. In this problem, each vertex has a unique identifier (
val
) and a list of neighboring vertices (neighbors
).Edge: A connection between two vertices. In this problem, edges are undirected, meaning the order of the vertices does not matter.
Connected Graph: A graph in which there’s a path between any two vertices. The problem specifies that the given graph is connected.
Undirected Graph: A graph in which edges have no direction. This means if there’s an edge from vertex ( A ) to ( B ), there’s also an edge from ( B ) to ( A ).
Adjacency List: A way to represent a graph by listing each vertex’s neighbors. The problem provides the graph in this form.
Isomorphic Graphs: Two graphs that are structurally identical, meaning there’s a onetoone mapping between their vertices and edges. The task is to create such a graph.
Deep Copy: Creating a new object that is an exact copy of an existing object, including the nested structures like lists of neighbors in this case.
Reference: A way to access or point to an object in memory. In the problem, you’re given a reference to a node, and you must return a reference to its clone in the new graph.
Understanding these terms is crucial for comprehending the problem’s requirements and constraints, as well as for formulating a solution.
Problem Simplification and Explanation
Certainly. The problem asks you to make an exact copy of a given network of friends. Imagine each person in this network is holding hands with their friends. This holding of hands creates a complex web. Your job is to replicate this web of friendships exactly, including who is friends with whom.
Here are the key concepts:
 Person (Node): Each individual in this friendship network.
 Friendship (Edge): The connection between two people.
 Network (Graph): The entire collection of friends and their connections.
 Deep Copy: You need to create a new network that looks exactly the same but consists of entirely new individuals.
Metaphor: Think of the friendship network as a complex piece of origami. You have to fold another piece of paper in exactly the same way to recreate the same shape. The new origami piece is separate but looks identical to the original.
In simpler terms:
 You’re given one person from a group of friends.
 Each friend knows certain other friends directly.
 You have to create a new group of friends that have the same friendship pattern as the original group.
 Finally, you have to return one person from this new group, and this person’s friendship connections should guide us through the entire new network.
By understanding these key elements and their interactions, you’ll be better equipped to solve the problem.
Constraints
Here are some specific characteristics and conditions in the problem statement that can be exploited for an efficient solution:
Node Uniqueness: Each node’s value is unique. This characteristic can be useful when trying to identify or map nodes between the original and cloned graphs. A hash table can be an ideal data structure for this.
No Repeated Edges or SelfLoops: This simplifies the problem as you don’t have to worry about duplicate connections or nodes being connected to themselves. You can confidently traverse the graph without doublecounting.
Connected Graph: The graph is connected, meaning you can reach any node starting from the given node. This tells us that we don’t have to worry about disconnected components and that a single traversal starting from the initial node will reach every node in the graph.
Number of Nodes Range [0, 100]: The upper limit on the number of nodes is not very high, which indicates that a solution with a time complexity that is quadratic or better will likely be acceptable.
Adjacency List Representation: The problem employs an adjacency list, a spaceefficient way to represent a graph, which allows for straightforward traversal and cloning operations.
Node Value Range [1, 100]: The values are bounded, but since they are unique, the range doesn’t provide a direct advantage.
These patterns and numerical ranges can be strategically used to form an algorithmic solution that is both correct and efficient.
Analyzing the constraints provides several key insights that can guide the problemsolving approach:
Connected Graph: Knowing that the graph is connected simplifies traversal. A single depthfirst or breadthfirst search will visit all nodes, avoiding the need to find isolated components.
Unique Node Values: This simplifies identification and mapping between the original and cloned nodes. It also allows us to use data structures like hash tables for efficient lookups.
No SelfLoops or Duplicate Edges: This makes the graph simpler to navigate and eliminates the need to handle edge cases related to duplicate edges or loops.
Bounded Number of Nodes: The number of nodes is constrained to be between 0 and 100, implying that even a somewhat inefficient algorithm could still perform well within these bounds. However, the constraint also indicates that we should aim for a solution that is at least O(n^2) or better in terms of time complexity.
Adjacency List: This suggests that the graph is sparse, allowing for efficient traversal and making it easier to clone nodes and their edges.
Node Values 1Indexed and Equal to Index: This could simplify some operations or lookups but generally doesn’t provide a significant computational advantage.
Low Upper Bound for Node Values: Knowing that node values won’t exceed 100 can simplify data storage but doesn’t directly impact the algorithm’s efficiency.
These insights derived from the constraints help us narrow down our algorithmic choices and reduce the possibility of edge cases, making it easier to develop an efficient solution.
Number of Nodes: The graph can have between 0 and 100 nodes.
Number of Edges: In a connected, undirected graph with N nodes and no selfloops or multiple edges, the maximum number of edges is N * (N  1) / 2. For N = 100, this would be 100 * 99 / 2 = 4950 edges max.
Edge Weights: The problem statement does not mention edge weights, so we assume that they are not a factor in this problem.
Loops: The problem specifies that there are no selfloops in the graph. Negativesum loops are not mentioned but are irrelevant as there are no weights.
Graph Direction: The graph is specified to be undirected.
Multiple Edges and SelfLoops: The problem specifies that there are no repeated edges and no selfloops.
By understanding these characteristics, we can make more accurate assumptions about the scope and constraints of the problem. This helps in choosing the appropriate data structures and algorithms for solving it.
Case Analysis
Certainly, let’s look at some examples to cover a range of conditions for the “deep copy of a graph” problem:
Test Case 1: Empty Graph (Edge Case)
 Input:
[]
 Output:
[]
 Analysis: The graph is empty. No nodes exist. The clone should also be empty.
Test Case 2: Single Node without Neighbors (Edge Case)
 Input:
[[]]
 Output:
[[]]
 Analysis: The graph consists of one node with no neighbors. The clone should have one node with no neighbors.
Test Case 3: Single Node with a Selfloop (Not Allowed)
 Input: N/A
 Output: N/A
 Analysis: The problem specifies no selfloops, so this case is not allowed.
Test Case 4: Two Nodes (Boundary Case)
 Input:
[[2], [1]]
 Output:
[[2], [1]]
 Analysis: A simple graph of two nodes connected to each other. The clone should reflect this.
Test Case 5: Disconnected Nodes (Not Allowed)
 Input: N/A
 Output: N/A
 Analysis: The problem specifies the graph is connected. So, a graph with disconnected nodes is not allowed.
Test Case 6: Three Nodes, Full Mesh (Complex Case)
 Input:
[[2, 3], [1, 3], [1, 2]]
 Output:
[[2, 3], [1, 3], [1, 2]]
 Analysis: All nodes are connected to every other node. The cloned graph should also have this property.
Test Case 7: Three Nodes, Line (Complex Case)
 Input:
[[2], [1, 3], [2]]
 Output:
[[2], [1, 3], [2]]
 Analysis: Nodes are connected in a line. The clone should preserve this linear structure.
Test Case 8: Four Nodes, Square (Complex Case)
 Input:
[[2, 4], [1, 3], [2, 4], [1, 3]]
 Output:
[[2, 4], [1, 3], [2, 4], [1, 3]]
 Analysis: Nodes form a square. All sides and diagonals of the square in the cloned graph should be equivalent to the original.
Edge Cases
 Empty graph
 Single node without neighbors
Each of these test cases addresses different constraints and aspects of the problem. By considering these cases, we can better ensure that the solution is robust and accounts for edge and boundary conditions.
Visualizing these test cases can offer valuable insights into the problem. Here’s how you can picture each:
Test Case 1: Empty Graph
Visualize an empty space; there’s nothing to represent.
Test Case 2: Single Node without Neighbors
Picture a single dot with no lines connected to it.
Test Case 3: Single Node with a Selfloop (Not Allowed)
Not applicable as per the problem constraints.
Test Case 4: Two Nodes
Imagine two dots connected by a line, forming a simple graph.
Test Case 5: Disconnected Nodes (Not Allowed)
Not applicable as per the problem constraints.
Test Case 6: Three Nodes, Full Mesh
Visualize a triangle where each vertex is a node and each side is an edge.
Test Case 7: Three Nodes, Line
Think of three dots connected in a line, one after the other.
Test Case 8: Four Nodes, Square
Picture a square where each corner is a node and each side is an edge.
By visualizing these test cases, you make the abstract more concrete, making it easier to understand the underlying structure and behavior of the graph. This can be particularly helpful in grasping how the deep copy should reflect the original graph.
Empty Graph: An edge case that tests the ability to handle null or void conditions. Key insight: The algorithm should return an empty graph or a null value appropriately.
Single Node without Neighbors: Another edge case. It tests whether the algorithm can handle the minimal nonempty input. Key insight: The algorithm should create a clone node with the same value but no neighbors.
Two Nodes: This tests the basic case of multiple nodes and an edge. Key insight: The algorithm should correctly map nodes and their connections while creating a clone.
Three Nodes, Full Mesh: Tests handling of multiple edges per node. Key insight: Each node’s neighbor list should be accurately recreated in the clone.
Three Nodes, Line: Checks the algorithm’s ability to maintain sequential dependencies between nodes. Key insight: The algorithm should keep the order of neighbors consistent in the clone.
Four Nodes, Square: Tests more complex relationships. Key insight: The algorithm should preserve the structure and connections in a multiplenode graph.
By analyzing these cases, it’s evident that the algorithm must handle a range of scenarios, from empty or minimal graphs to more complex, interconnected ones. Each case reveals the need for robustness in the cloning process, from preserving node values to maintaining the graph’s overall structure.
Identification of Applicable Theoretical Concepts
Here are some concepts that could be useful in simplifying the problem:
Graph Theory: Understanding basic graph algorithms can help, such as DepthFirst Search (DFS) and BreadthFirst Search (BFS) for traversing the graph.
Hashing: A hash map can be used to keep track of visited nodes and their clones, reducing the lookup time to O(1).
Queue/Stack: These data structures are essential for BFS and DFS traversals, respectively, to keep track of which nodes to visit next.
Recursion: Recursive DFS can be employed for a more elegant solution, although it requires understanding of recursion mechanics and stack overflow conditions.
Adjacency List: The problem uses an adjacency list to represent the graph. Understanding its structure is critical for effectively solving the problem.
SpaceTime Tradeoff: Using additional memory can help speed up the cloning process by keeping track of already cloned nodes.
Complexity Analysis: Understanding the time and space complexity of your algorithm will help you finetune it for efficiency.
By employing these concepts, you can come up with an algorithm that is both efficient and easy to understand, making the problem more manageable.
Simple Explanation
Imagine you have a bunch of toy houses, and each house has strings connecting it to its nearby houses. Now, you want to create an exact duplicate set of these toy houses and their connecting strings. The challenge is to make sure that each new toy house is connected to the new neighbors in the same way as in the original set.
In simpler terms, you’re trying to make a copy of a network of connected houses, making sure that all connections in the copy match the original exactly.
Problem Breakdown and Solution Methodology
Approach to Solving the Problem
Mapping Original to Copy: The first step is to create a way to map each original toy house (node) to its copy. This ensures that you’re not creating multiple copies of the same house.
 Analogy: Think of it as labeling each original toy house and its copy with a unique sticker so that you can identify them later.
Start at the First House: You start by copying the first toy house, as that is the one specifically mentioned to be the starting point.
 Analogy: You have to start from your own home to recreate the entire neighborhood.
Exploring Neighbors: For each house, look at its connected neighbors. If a neighbor hasn’t been copied yet, create a copy and connect it to the copy of the current house.
 Analogy: You walk to your neighbor’s house and then to their neighbors, making sure to note each path you take.
Avoid Duplicate Work: As you explore and make copies, always check whether you’ve already made a copy of the house you’re currently at. If so, skip making a new copy, but establish the connection if it doesn’t exist.
 Analogy: Before building a copy of any house, you check if it already has a sticker. If it does, you know it’s been copied already.
Track Progress: Keep track of which houses you’ve visited and copied to ensure you’re not going in circles.
 Analogy: You mark houses you’ve visited with a different color sticker to avoid visiting them again unnecessarily.
Return the Copy: Once you’ve visited all connected houses, you’ll have your copied neighborhood. The function should return the copy of the initial house as the starting point.
 Analogy: You’ll know you’re done when every house in the neighborhood has a sticker and each house’s new copy is connected the same way as the original.
Parameters and Their Effects
Number of Nodes: The more houses (nodes), the more time it will take to create a copy. The algorithm is linear with respect to the number of houses and their connections.
Node Connections: The more neighbors a house has, the more connections you need to check and copy.
 Example: If a house is connected to every other house, it will take longer to process that house.
Example Case
Let’s assume a small neighborhood with three houses: House 1, House 2, and House 3.
 House 1 is connected to House 2 and House 3.
 House 2 is connected to House 1.
 House 3 is connected to House 1.
You start by copying House 1, then move to its neighbors (House 2 and House 3) and copy them. You also establish that the copy of House 1 is connected to the copies of House 2 and House 3. After visiting all the houses, you return the copy of House 1 as the new starting point for the copied neighborhood.
Inference of ProblemSolving Approach from the Problem Statement
Undirected Graph: The problem involves a graph where each edge is bidirectional. This informs us that we need to make sure both connected nodes are updated when we copy a single edge.
Deep Copy: The requirement for a deep copy suggests that we can’t just copy the pointers; we have to create entirely new nodes. This necessitates a mapping between the original and copied nodes to maintain connections.
Neighbors (List[Node]): Each node has a list of its neighbors. This indicates that we’ll have to perform recursive or iterative copying as we traverse from one node to its neighbors.
Connected: The graph is connected, meaning there is a path between every pair of nodes. This simplifies our traversal because we don’t have to worry about disjoint subgraphs.
Unique Node Values: Each node has a unique value, which can be exploited to create a map between the original and copied nodes, ensuring that each node is copied only once.
Constraints (Number of Nodes): The number of nodes is limited to 100. This tells us that computational complexity is probably not a big issue, allowing the use of simple algorithms.
Starting Node: The given node is always the first node with value 1. This is our entry point for both copying and traversal, simplifying where we begin our operation.
Each keyword or concept leads to specific strategies:
 “Undirected Graph” and “Connected” make BreadthFirst Search (BFS) or DepthFirst Search (DFS) suitable for traversal.
 “Deep Copy” and “Unique Node Values” make a HashMap ideal for mapping original nodes to their copies.
 “Neighbors” informs that we need to loop through these to complete the copying process.
 “Constraints” tell us that computational complexity is a secondary concern, allowing flexibility in choosing algorithms.
Undirected Graph: You can draw a graph where each edge has no direction, or arrows point both ways. If you are using a table, a nonzero entry at (i, j) should match with the entry at (j, i) to represent the undirected edge.
Deep Copy: This could be visualized by drawing a completely separate graph that has the same structure as the original but is clearly marked as a different entity. In a table, you’d create an entirely new table that has the same values but is labeled differently.
Neighbors (List[Node]): On your graph, you can circle each node and then draw lines to its immediate neighbors. In a table, the neighbors could be listed in the same row or column as the node.
Connected: You can ensure that all nodes are part of a single graph component, meaning you can reach any node from any other node. In a table, there should be no row or column with all zeros except the diagonal.
Unique Node Values: Each node can be labeled with a unique value. In a table, these unique values could be the row and column headings.
Constraints (Number of Nodes): You can label each node with numbers up to the constraint (e.g., 1100). In a table, limit the dimensions accordingly.
Starting Node: In your diagram, you could highlight or color the starting node to make it stand out. In a table, the starting node could be the first row/column.
Here’s how you could draw these:
For the graph, use circles for nodes and lines for edges. Use colors or labels to mark special properties like the “Starting Node”.
For the table, each row and column represents a node. The intersection between row i and column j can have a value (e.g., 1 or 0) to indicate the presence of an edge between nodes i and j.
By drawing these out, you can quickly recognize the properties and constraints mentioned in the problem, which can then inform your approach to solving it.
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
Stepwise Refinement for Solving the Problem:
Initialize a Hash Map: Create a hash map to keep track of visited nodes. This will help us avoid infinite loops and also store the new clones of nodes.
Define the Clone Function: Develop a function that takes a node as an argument and returns a deep clone of the node. This function will handle the recursion.
Base Case: Inside the clone function, if a node is already visited (exists in the hash map), return its clone.
Clone Current Node: Clone the node that is passed into the function and store it in the hash map as a visited node.
Iterate through Neighbors: For each neighbor of the current node, call the clone function. Add the clone of the neighbor to the neighbor list of the cloned node.
Return the Clone: After iterating through all neighbors, return the clone of the node back up the recursive chain.
Distill Into Granular, Actionable Steps:
Initialize a Hash Map: Just a single line using a hash map data structure.
Define the Clone Function:
 Accept a
Node
as a parameter.  Check if the node is in the hash map. If yes, return the corresponding value.
 If not, create a new node with the same value.
 Store this new node in the hash map.
 Accept a
Base Case:
if node in visited: return visited[node]
Clone Current Node:
new_node = Node(node.val, [])
visited[node] = new_node
Iterate through Neighbors:
for neighbor in node.neighbors: new_node.neighbors.append(clone(neighbor))
Return the Clone:
return new_node
Parts That Can Be Solved Independently:
 The cloning of each neighbor can be treated as an independent subproblem. However, due to the connected nature of the graph, these operations are not entirely independent of each other.
Repeatable Patterns Within Solution:
 The act of checking if a node is visited and then either returning the visited node or creating a new one is a pattern that repeats for each node in the graph.
 The process of iterating through neighbors and cloning them is also repetitive.
By understanding these steps, you can create a robust solution that adheres to the problem’s constraints and requirements.
Solution Approach and Analysis
Approach to Solving the Problem:
Initialize a “Visited” Hash Map:
 This map stores the original node as the key and the new cloned node as the value.
 Think of this map as a “passport check” for nodes. When a node is revisited, it shows its “passport,” and the map tells you if it already has a clone.
Start the Cloning Process:
 Call the cloning function and pass the initial node to it. This function is recursive and does the actual cloning.
Base Case in Cloning Function:
 Check if the node is already in the “Visited” map.
 If yes, return its clone immediately.
 This is like asking, “Have I seen you before?” If yes, you can skip the cloning process for this node.
Clone the Node:
 If the node is new, clone it. Add the clone to the “Visited” map.
 This is like giving a new “passport” to a firsttime traveler.
Clone the Neighbors:
 Go through each neighbor of the original node.
 Call the cloning function recursively for each neighbor.
 Add these clones to the neighbor list of the new cloned node.
 This is like traveling to each neighboring country and giving “passports” to all unknown nodes you meet.
Return the Cloned Node:
 Once all neighbors have been visited and cloned, return the cloned node to the previous recursive call.
Return the Clone of the Initial Node:
 Once the recursive calls are over, you’ll have a new cloned graph. Return the clone of the initial node as the starting point of this new graph.
How Parameters Affect the Solution:
Number of Nodes:
 More nodes mean more recursive calls, affecting time complexity.
Number of Edges:
 More edges mean each node will have more neighbors to clone, increasing the time spent in the function.
Example Case:
Let’s say you have 3 interconnected nodes:
1  2
\ /
\ /
3
Initialize “Visited” Hash Map: Empty initially.
Start Cloning: Pass Node 1 to the function.
Base Case: Node 1 is new, so continue.
Clone Node 1: Create a new node with value 1. Add to “Visited” map:
{1: 1'}
(1’ is the clone of 1).Clone the Neighbors: Node 1 has neighbors 2 and 3.
For Node 2:
 It’s new, so clone it.
 Add to “Visited”:
{1: 1', 2: 2'}
 Its only neighbor is 1, which already has a clone. So, add 1’ to 2’s neighbors.
For Node 3:
 It’s new, so clone it.
 Add to “Visited”:
{1: 1', 2: 2', 3: 3'}
 Its neighbors are 1 and 2, both have clones. Add 1’ and 2’ to 3’s neighbors.
Return Cloned Node: Return Node 1’ as the starting point of the new graph.
Your new cloned graph will look exactly like the original.
Identify Invariant
The invariant in this problem is the structure of the graph. Regardless of how you traverse the original graph or in what order you clone the nodes and their neighbors, the structure of the cloned graph should remain identical to the original. This means that if node A is connected to node B in the original graph, then the clone of node A should be connected to the clone of node B in the new graph.
The “Visited” hash map also acts as an invariant, ensuring that each node in the original graph maps to exactly one corresponding node in the cloned graph. This ensures the uniqueness of each node in the new graph and helps avoid cycles or inconsistencies during the cloning process.
Identify Loop Invariant
The loop invariant in this problem, particularly if you’re using DepthFirst Search (DFS) or BreadthFirst Search (BFS) to clone the graph, is that at any point during the traversal, each node that has been visited in the original graph should have an exact corresponding node in the cloned graph. Additionally, all connections (edges) of visited nodes should also be accurately represented in the cloned graph.
If you’re using a queue for BFS, for example, the loop invariant ensures that all nodes in the queue have already had their corresponding nodes created in the cloned graph. Before and after each iteration, this condition holds true. This helps maintain the integrity of the cloning process, ensuring that the cloned graph will ultimately be a correct deep copy of the original graph.
In the context of this graph cloning problem, the terms “invariant” and “loop invariant” can indeed point to the same underlying concept, especially if your algorithm uses loops for traversal (such as a while loop for BFS or DFS). The invariant, in a general sense, is a condition that remains true throughout the execution of the algorithm.
For this specific problem, the invariant would be that every node visited in the original graph has a corresponding node in the cloned graph with the same neighbors. If your algorithm employs loops for traversal, then this invariant becomes your loop invariant.
In summary, for this problem, the invariant and loop invariant could effectively be the same, as long as your algorithm employs looping structures for the traversal and cloning process.
For the graph cloning problem, if you’re using a loop to traverse the graph (DFS or BFS), here’s how you can establish the loop invariant. Let’s assume you’re using a BFS algorithm and a queue to handle nodes yet to be visited. The loop invariant is: “At the beginning of each iteration, each node in the ‘visited’ set must have a corresponding clone in the cloned graph, and each clone must have its neighbors set correctly.”
Initialization: Prior to the first iteration of the loop, we have an empty ‘visited’ set and an empty queue. We place the starting node in the queue and create a clone of it. We can claim the loop invariant holds as both the original graph and the cloned graph have just one node: the starting node, and its neighbors are set correctly (to empty).
Maintenance: To see that each iteration maintains the loop invariant, let’s look at the core logic of the BFS loop. In each iteration, we dequeue a node from the front of the queue and enqueue its neighbors that haven’t been visited yet. For each unvisited neighbor, we create a clone and set the neighbors of that clone correctly. Since we only enqueue neighbors that haven’t been visited and since we correctly set the neighbors for each cloned node, the loop invariant is maintained.
Termination: At termination, the queue is empty, meaning we have visited all nodes. According to our loop invariant, each visited node has a corresponding clone with the neighbors set correctly. Thus, we have a cloned graph that is a deep copy of the original graph, confirming that our algorithm works as intended.
Thought Process
The basic thought process and steps involved in solving this graph cloning problem can be broken down as follows:
Identify the Graph Type: The problem explicitly mentions that the graph is undirected. This means that if node
a
is connected to nodeb
, thenb
is also connected toa
.Choose a Traversal Method: Since we need to visit each node exactly once to clone it, we can use DepthFirst Search (DFS) or BreadthFirst Search (BFS).
Data Structures: We’ll need a data structure to keep track of visited nodes to avoid cycles. A dictionary works well for this; keys can be the original nodes and values their corresponding clones.
Clone Nodes and Edges: While traversing, each visited node is cloned, and its neighbors are set. If a neighbor is already visited (exists in the dictionary), we just link it; otherwise, we clone and then link it.
Entry Point: The entry point is the node given in the problem statement, which will always be the first node with val = 1.
Code:


Cues and Insights:
Graph Type: Knowing the graph is undirected tells us that backlinks are possible, crucial for setting neighbors correctly.
Traversal Method: Both BFS and DFS can work, so it gives us flexibility in approach.
Entry Point: Knowing that the entry point is the first node simplifies where to start the traversal and cloning.
Constraints: The number of nodes is limited to 100, indicating that O(N) or O(N+M) solutions will be acceptable. This constraint subtly tells us that using extra storage like a dictionary for tracking visited nodes will not exceed the space limitations.
By addressing these cues and constraints, the problem guides us towards a straightforward graph traversal solution.
Establishing Preconditions and Postconditions
Parameters:
 Inputs: A single input,
node
, of typeNode
.  Types:
Node
is a custom object type that has a value (val
) and a list of neighbors (neighbors
).  Context: The
node
parameter represents the first node (entry point) of an undirected graph to be cloned.
 Inputs: A single input,
Preconditions:
 State: None required; the method is selfcontained.
 Constraints: The number of nodes in the graph is between 1 and 100. Node values are unique. The
node
parameter can also beNone
.  Specific State: The program should have defined the custom
Node
object type.
Method Functionality:
 Expected to do: The method clones the entire graph starting from the input node and returns the clone of the input node.
 Interaction: The method uses DepthFirst Search (DFS) for traversal, keeps track of visited nodes using a dictionary, and clones nodes and their edges.
Postconditions:
 State: A cloned graph is created, rooted at the return value.
 Return Value: Represents the entry point of the cloned graph.
 Side Effects: The method does not modify the input graph but generates a new one.
Error Handling:
 Response: If the
node
parameter isNone
, the method returnsNone
.  Special Value: It returns
None
in case of invalid or null input, which is aligned with the expected behavior for such cases. No exceptions are thrown.
 Response: If the
Problem Decomposition
Problem Understanding:
 The problem asks us to clone an undirected graph. We’re given an entry node, and we need to make a deep copy of all the nodes and edges in the graph. The key requirement is that the cloned graph should have the same structure as the original.
Initial Breakdown:
 Broadly, the problem can be divided into three parts: graph traversal, node cloning, and edge cloning.
Subproblem Refinement:
 For graph traversal, we need to decide between DFS or BFS.
 Node cloning involves creating a new node for each original node.
 Edge cloning involves linking the cloned nodes in the same way as the original nodes are linked.
Task Identification:
 Traversing a node and cloning its edges is a repeatable task.
Task Abstraction:
 The task of traversing and cloning is abstract enough to be reusable but specific enough to make sense within this problem. A function named
clone_node
could handle this.
 The task of traversing and cloning is abstract enough to be reusable but specific enough to make sense within this problem. A function named
Method Naming:
traverse_and_clone
: Function to traverse the graph and initiate cloning.clone_node
: Function to clone individual nodes.clone_edges
: Function to clone edges between nodes.
Subproblem Interactions:
traverse_and_clone
initiates the process and calls the other functions.clone_node
andclone_edges
interact closely; after cloning a node, its edges need to be cloned. The order is traversal > node cloning > edge cloning for each node.
 Dependencies: Each node needs to be cloned before its edges can be.
From Brute Force to Optimal Solution
Brute Force Solution
A bruteforce way to solve this problem would be to perform multiple traversals of the original graph. On the first traversal, you clone all the nodes. On the second traversal, you’d copy all the edges.
Steps:
 Traverse the original graph using DFS or BFS and create a new node for every unique node encountered. Store the mapping of original node to the new node.
 Perform another traversal to clone the edges. While visiting each node, use the stored mapping to create edges in the cloned graph.
Code in Python3:


Inefficiencies:
 Multiple traversals: This causes extra time complexity.
 Repeated work: Some edges might be revisited, leading to extra computational effort.
Optimized Solution
We can avoid the multiple traversals and repeated work by combining the tasks of node cloning and edge cloning into a single traversal. This is possible because we can clone nodes and their edges “lazily,” i.e., as we encounter them.
Steps Towards Optimization:
 Merge the two separate traversals into one.
 As you visit a node, clone it and immediately clone its edges as well.
Code in Python3:


Improvements:
Time Complexity: Both the brute force and optimized solutions have a time complexity of O(N + E) where N is the number of nodes and E is the number of edges. However, the optimized version has a lower constant factor due to the reduced number of operations.
Space Complexity: The space complexity remains O(N) for both, as we need to store the mapping from original nodes to new nodes. However, the optimized solution uses slightly less auxiliary space as it eliminates the need for separate traversals.
This optimized solution is efficient and tackles the problem in a single traversal, reducing both time and space overheads.
Code Explanation and Design Decisions
Initial Parameters:
node
: This is the starting node of the original graph. It helps us navigate the rest of the graph through its neighbors.original_to_copy
: A dictionary that maps each original node to its corresponding cloned node. It’s crucial for keeping track of what has already been cloned and for connecting the edges in the cloned graph.
Primary Loop:
 The primary loop is the Depth First Search (DFS) traversal. Each iteration of DFS represents a visit to a unique node in the original graph. The visit clones the node and its edges. This contributes to building the cloned graph incrementally.
Conditions within the Loop:
if neighbor not in original_to_copy
: This condition checks if a neighboring node has already been cloned. If it hasn’t, it’s cloned and added to the DFS stack. This ensures no node is cloned more than once.
Updates within the Loop:
original_to_copy[neighbor] = Node(neighbor.val)
: Here, a new node is created for the neighbor if it hasn’t been cloned yet. This is necessary to advance the state of our cloned graph.original_to_copy[root].neighbors.append(original_to_copy[neighbor])
: This line connects the current node to its neighbor in the cloned graph. It mirrors the edge in the original graph, thereby preserving its structure.
Invariant:
 The invariant is that if a node is in
original_to_copy
, then it has been fully processed: it’s cloned and all its edges to previously visited nodes have also been cloned. This helps us ensure that the final cloned graph will be a correct replication of the original graph.
 The invariant is that if a node is in
Final Output:
 The final output is the starting node of the cloned graph. It represents an entirely separate graph that is a clone of the original graph, fulfilling the problem’s requirements of having the same structure as the original but being a distinct entity.
By understanding these facets of the algorithm, you grasp not just the “what” but the “why” of the solution. It addresses the complexities of graph cloning in an efficient manner, meeting all the problem’s constraints.
Coding Constructs
HighLevel Strategies:
 The code employs Depth First Search (DFS) to traverse the original graph nodebynode.
 It uses a dictionary (or a hashmap) to store the mapping of original nodes to their corresponding cloned nodes.
NonProgrammer Explanation:
 Imagine you have a maze made of rooms, and each room has doors to other rooms. This code makes an exact copy of that maze, room by room, door by door.
Logical Elements:
 Conditional Statements: To check if a room (node) is already copied or not.
 Loops: To visit each room and its neighboring rooms.
 Data Mapping: A storage that links each room in the original maze to its copy in the new maze.
Algorithmic Approach in Plain English:
 Start at the first room of the original maze.
 Make a copy of that room and remember that you’ve copied it.
 Look at all the rooms connected to it.
 For each connected room, if it’s not copied yet, copy it. Then, connect it back to the copy of the room you started with.
 Repeat these steps for every new room you enter until all rooms are copied and connected like the original.
Key Operations:
 Checking if a node is already cloned: To avoid cloning it again.
 Cloning a node: To extend the cloned graph.
 Adding edges between cloned nodes: To maintain the structure of the original graph.
Algorithmic Patterns:
 Graph Traversal: DFS is used for going through each node of the original graph.
 Memoization: Using a dictionary to store cloned nodes avoids redundant work and cycles.
 Recursion: The DFS algorithm inherently uses recursion to explore nodes until there are no new nodes left to explore.
Language Agnostic Coding Drills
Distinct Concepts in the Code:
 Variable Declaration: Declaring variables to hold temporary data.
 Data Structures: Using arrays, lists, or dictionaries to hold information.
 Conditional Statements:
ifelse
checks for making decisions.  Loops: Iterating through nodes or elements.
 Functions: Creating and calling functions.
 Recursion: Function calls itself for deeper exploration.
 Graph Traversal: Moving through nodes via edges.
 Memoization: Storing alreadycomputed values for reuse.
Concepts Ordered by Increasing Difficulty:
 Variable Declaration: Simplest form of storing data. No logic or algorithms involved.
 Conditional Statements: Basic decisionmaking. Only slight increase in complexity.
 Loops: Requires a deeper understanding of flow control but still straightforward.
 Data Structures: Arrays and dictionaries require understanding of data organization.
 Functions: A leap in complexity; requires understanding of scope, parameters, and return values.
 Graph Traversal: Understand the graph as a data structure and how to move through it.
 Recursion: Mentally challenging, as you need to trust the function to solve smaller versions of the problem.
 Memoization: Requires understanding of dynamic programming and optimization.
ProblemSolving Approach:
 Start Small: Begin by declaring necessary variables and initializing data structures (Variable Declaration, Data Structures).
 Condition Check: Implement basic conditions to check if a node is visited or not (Conditional Statements).
 Iterate: Write loops to traverse through each node and its neighbors (Loops).
 Functionalize: Move reusable code parts into separate functions (Functions).
 Explore: Use graph traversal algorithms like Depth First Search to explore nodes and edges (Graph Traversal).
 Go Deep: Implement recursion in your traversal function to explore each path deeply (Recursion).
 Optimize: Use memoization to avoid recomputing clones for nodes already visited (Memoization).
By progressively adding these components, you build up to the final, optimized solution. Each concept plays a critical role in constructing the solution, from the groundlevel building blocks to the more complex, problemspecific strategies.
Targeted Drills in Python
1. Coding Drills for General Concepts
Variable Declaration
1 2
a = 10 b = "Hello"
Why: Understand how to store and name data.
Conditional Statements
1 2 3 4
if a > 5: print("Greater") else: print("Smaller")
Why: Learn to make decisions in code.
Loops
1 2
for i in range(5): print(i)
Why: Iterate over a range of numbers, understanding flow control.
Data Structures
1 2
my_list = [1, 2, 3] my_dict = {"key": "value"}
Why: Learn how to organize and store multiple data elements.
Functions
1 2
def say_hello(name): return f"Hello, {name}"
Why: Understand code reuse, parameter passing, and return statements.
Graph Traversal
1 2 3 4 5 6 7 8
graph = {1: [2, 3], 2: [4], 3: [], 4: []} visited = [] def traverse(node): if node not in visited: print(node) visited.append(node) for neighbor in graph[node]: traverse(neighbor)
Why: Understand how to navigate through connected nodes.
Recursion
1 2 3 4
def factorial(n): if n == 0: return 1 return n * factorial(n1)
Why: Learn to break a problem into smaller similar problems.
Memoization
1 2 3 4 5 6 7 8 9 10
cache = {} def fib(n): if n in cache: return cache[n] if n == 0: return 0 if n == 1: return 1 cache[n] = fib(n1) + fib(n2) return cache[n]
Why: Store and reuse already computed values for optimization.
2. ProblemSpecific Drills
 None identified: For now, the problem does not require any additional problemspecific concepts.
3. Assembling the Drills into a Solution
 Start: Use the Variable Declaration drill to declare variables for storing your graph and visited nodes.
 Organization: Use the Data Structures drill to set up your graph and other necessary lists or dictionaries.
 Initial Check: Use Conditional Statements within your traversal function to check if a node has been visited.
 Loop Through: Employ Loops to iterate over your data structure and apply the traversal algorithm.
 Functions: Move your traversal code into a function, using the Functions drill.
 Depth: Incorporate Recursion into your traversal function to explore the graph deeply.
 Optimization: Implement Memoization to store and reuse results to avoid redundant calculations.
Each of these drills teaches a fundamental skill. As you go through the drills, you build up your coding toolbox, enabling you to assemble these into more complex programs.
Q&A
Similar Problems
Here are 10 distinct problems that use similar underlying concepts:
Two Sum
 Why: It involves iterating through a list and using a hash map, similar to the traversal and storage components we’ve discussed.
Longest Substring Without Repeating Characters
 Why: Uses a sliding window technique that could be likened to our graph traversal where we look ahead before making moves.
3Sum
 Why: Requires sorting and then traversing the array, which can be seen as a simplified version of graph traversal.
Palindrome Partitioning
 Why: This involves recursion to solve subproblems, akin to our graph traversal approach.
Climbing Stairs
 Why: This problem involves simple recursion and memoization, similar to our optimized graph traversal method.
Valid Parentheses
 Why: This problem involves using a stack for matching brackets, which is another form of traversal and state checking.
Search in Rotated Sorted Array
 Why: It involves a binary search, a form of traversal that efficiently navigates through a sorted list.
Subsets
 Why: Involves generating all possible subsets which can be related to visiting all nodes in a graph traversal.
Course Schedule
 Why: Directly involves graph traversal to check for cycles in course prerequisites.
Fibonacci Number
 Why: A straightforward example of recursion and memoization, akin to our optimized graph traversal algorithm.
Each of these problems has a component that could be considered parallel to a part of our original problem, whether it be traversal, storage, recursion, or optimization.
The problem requires creating a deep copy of a given graph. A deep copy means that not only the nodes themselves are duplicated, but also the connections between them. Here’s an implementation for this:


Explanation
Special Case: If the input node is null, return null since there’s no graph to clone.
Create Visited Dictionary: A dictionary
visited
is used to keep track of the nodes that have been cloned. The keys are the original nodes, and the values are the corresponding cloned nodes.BreadthFirst Search (BFS): Using a queue, we perform a BFS starting from the given node. For each node, we clone it and its neighbors, updating the neighbors list of the clone.
Key Takeaway
The algorithm performs a deep copy of a graph using BFS, ensuring that both the nodes and the connections between them are duplicated. It uses a dictionary to map original nodes to cloned nodes, allowing efficient access and avoiding duplication of work. The complexity of the algorithm is O(n + m), where n is the number of nodes, and m is the number of edges in the graph, satisfying the constraints provided.
Define the Interface
Input: Node reference (adjacency list?) Output: Node reference (new adjacency list)
Define the Terms
 Deep copy (we need to copy such that a cloned node has the same number of neighbors with the same values as the copied node values)
 Undirected graph 0 is connected to 1, this also means 1 is connected to 0
Simplifying Assumptions
 The value of the node is the same as the node’s index
Graph Problems
 edges and the number of nodes were given We created the adjacency list
 adjacency list
Graph Traversal
 DFS
 BFS
Identify the Invariants
 You cannot visit the node more than once
 visited array to keep track of which nodes we have already visited
Recursive vs Iterative
 DFS
Unit of Work
What is our goal? Clone the entire graph
Break down the goal into smaller tasks. Create a new node Copy the neighbors of that node Base Case When should we stop the recursion? Do we have multiple base cases?
We should store in a hashtable, the orinal graph’s node as key and value is the cloned node.
Identify the constraints
Search the Problem Space
Classify the Problem
Analyze the Examples
Solve the Problem by Hand
Describe the Pattern
Brute Force Approach
Analyze Time and Space Complexity
Outline the Solution
Key Takeaways
At index 0, the node has value 1, but since the list of neighbors is empty.
The indices of the array represents the node, the position at which an array is found represents the neighbors for that node.
Rearrange the examples, so it goes from the small test cases to the biggest test cases.
Cases:
No node with no neighbors []
One node with no neighbors [[]] 0
Two nodes with only one neighbor
[[2],[1]] 0 1
Outline the Solution
 We need to create all the node objects first
 Start traversing from the first node and copy its neighbors
 Repeat for all the remaining nodes
 Return the reference to the new cloned graph’s first node
We start from node 1 and go depth first manner We keep track of the visited nodes When we reach the node where we started, we stop Do we now copy all the neighbors for every node?
Create a node object node.neighbors = [] node.neighbors « node2

