Clone Linked List with Random Pointer in DSA C# - Execution Trace

Concept Flow - Clone Linked List with Random Pointer

Start at head of original list

↓

Create new node copy and insert after original node

↓

Move to next original node

↓

Repeat until end of original list

↓

Assign random pointers for copied nodes

↓

Separate copied nodes to form new cloned list

↓

Return head of cloned list

We first insert copied nodes next to original nodes, then assign random pointers for copies, and finally separate the copied list from the original.

Execution Sample

DSA C#

Node CloneList(Node head) {
  // Step 1: Insert copies
  // Step 2: Assign random pointers
  // Step 3: Separate lists
  return clonedHead;
}

This code clones a linked list with random pointers by interleaving copied nodes, setting random pointers, and separating the clone.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Insert copy of node 1 after node 1	1 -> 1' -> 2 -> 3 -> null	1.next = 1', 1'.next = 2	1 -> 1' -> 2 -> 3 -> null
2	Insert copy of node 2 after node 2	1 -> 1' -> 2 -> 2' -> 3 -> null	2.next = 2', 2'.next = 3	1 -> 1' -> 2 -> 2' -> 3 -> null
3	Insert copy of node 3 after node 3	1 -> 1' -> 2 -> 2' -> 3 -> 3' -> null	3.next = 3', 3'.next = null	1 -> 1' -> 2 -> 2' -> 3 -> 3' -> null
4	Assign random pointer for 1'	No change in nodes	1'.random = 1.random.next (3')	1'.random -> 3'
5	Assign random pointer for 2'	No change in nodes	2'.random = 2.random.next (1')	2'.random -> 1'
6	Assign random pointer for 3'	No change in nodes	3'.random = 3.random.next (2')	3'.random -> 2'
7	Separate original and copied lists	Original: 1 -> 2 -> 3 -> null Copy: 1' -> 2' -> 3' -> null	Restore original next pointers, set copy next pointers	Two separate lists formed
8	Return head of copied list	Copy list head = 1'	No pointer changes	Cloned list ready

💡 All nodes copied, random pointers assigned, and lists separated successfully.

Variable Tracker

Variable	Start	After Step 3	After Step 6	After Step 7	Final
head	1 -> 2 -> 3 -> null	1 -> 1' -> 2 -> 2' -> 3 -> 3' -> null	Same	Restored to 1 -> 2 -> 3 -> null	Same
current	1	3'	3'	null	null
copyHead	null	null	null	1'	1' -> 2' -> 3' -> null
random pointers	1.random->3, 2.random->1, 3.random->2	Not set for copies	Set for copies	Same	Same

Key Moments - 3 Insights

Why do we insert copied nodes immediately after original nodes?

Inserting copies after originals lets us easily assign random pointers by using original.random.next (see steps 1-4 in execution_table).

How do we assign random pointers for copied nodes?

Copied node's random pointer is set to original node's random pointer's next node (the copy), as shown in steps 4-6.

How do we separate the copied list from the original without losing connections?

We restore original next pointers and set copied next pointers in step 7, carefully unlinking the interleaved nodes.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what does 1'.random point to?

A Node 1

B Node 3

C Node 3'

D Null

💡 Hint

Check the Pointer Changes column at step 4 where 1'.random = 1.random.next (3')

At which step do we separate the original and copied lists?

A Step 7

B Step 3

C Step 5

D Step 8

💡 Hint

Look at the Operation column for 'Separate original and copied lists' in execution_table

If we skip inserting copied nodes after original nodes, what problem occurs?

A List becomes circular

B Random pointers cannot be assigned easily

C Original list is lost

D No problem, cloning works fine

💡 Hint

Refer to key_moments about why copies are inserted after originals

Concept Snapshot

Clone Linked List with Random Pointer:
1. Insert copied nodes after originals.
2. Assign copied nodes' random pointers using original.random.next.
3. Separate copied list from original restoring next pointers.
This creates a deep copy preserving random links.

Full Transcript

To clone a linked list with random pointers, we first insert a copy of each node right after the original node. This interleaves the original and copied nodes. Then, for each copied node, we set its random pointer by looking at the original node's random pointer and moving to the next node, which is the copy. Finally, we separate the copied nodes into a new list and restore the original list's next pointers. This method ensures the random pointers are correctly assigned without extra space. The execution steps show how nodes are inserted, random pointers assigned, and lists separated step-by-step.

Practice

1. What is the main challenge when cloning a linked list with a random pointer?

easy

A. Reversing the linked list before cloning

B. Sorting the linked list before cloning

C. Copying the next pointers correctly while preserving the random pointers

D. Ignoring the random pointers during cloning

Solution

1. Step 1: Understand the structure of the linked list

   The linked list has two pointers: next and random. The next pointer points to the next node, and the random pointer can point to any node or null.

2. Step 2: Identify the cloning challenge

   While cloning, copying next pointers is straightforward, but random pointers can point anywhere, so they must be copied carefully to preserve the original structure.

3. Final Answer:

   Copying the next pointers correctly while preserving the random pointers -> Option C

4. Quick Check:

   Random pointer preservation = main cloning challenge [OK]

Hint: Remember random pointers can point anywhere, handle them carefully [OK]

Common Mistakes:

• Ignoring random pointers during cloning
• Assuming random pointers follow next pointers
• Trying to sort or reverse the list before cloning

2. Which of the following C# code snippets correctly declares a node class for a linked list with a random pointer?

easy

A. public class Node { public int val; public Node next; public Node random; public Node(int val) { this.val = val; next = null; random = null; } }

B. public class Node { int val; Node next; Node random; }

C. class Node { public int val; public Node next; }

D. public struct Node { public int val; public Node next; public Node random; }

Solution

1. Step 1: Check for correct access modifiers and constructor

   public class Node { public int val; public Node next; public Node random; public Node(int val) { this.val = val; next = null; random = null; } } declares a public class with public fields and a constructor initializing val, next, and random to null, which is correct.

2. Step 2: Identify issues in other options

   public class Node { int val; Node next; Node random; } lacks access modifiers and constructor. class Node { public int val; public Node next; } misses random pointer. public struct Node { public int val; public Node next; public Node random; } uses struct which is not ideal for linked list nodes.

3. Final Answer:

   Option A code snippet correctly declares the node class -> Option A

4. Quick Check:

   Public class with constructor = correct node declaration [OK]

Hint: Use public class with constructor initializing all pointers [OK]

Common Mistakes:

• Missing public access modifiers
• Not initializing pointers in constructor
• Using struct instead of class for nodes

3. Given the following code snippet for cloning a linked list with random pointers, what will be the output of the cloned list's node values in order?

Node head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.random = head.next.next; // 1's random points to 3
head.next.random = head; // 2's random points to 1
head.next.next.random = head.next; // 3's random points to 2

Node clonedHead = CloneLinkedList(head);
Node current = clonedHead;
while (current != null) {
    Console.Write(current.val + " ");
    current = current.next;
}

medium

A. 3 2 1

B. 1 2 3

C. 1 3 2

D. 2 1 3

Solution

1. Step 1: Understand the original list structure

   The original list nodes are 1 -> 2 -> 3 with next pointers in that order. Random pointers do not affect the order of next pointers.

2. Step 2: Cloning preserves next order

   The CloneLinkedList function clones nodes preserving the next pointer order, so the cloned list will have nodes 1 -> 2 -> 3 in sequence.

3. Final Answer:

   1 2 3 -> Option B

4. Quick Check:

   Next pointers order = 1 2 3 [OK]

Hint: Cloning preserves next order; random pointers don't change sequence [OK]

Common Mistakes:

• Confusing random pointer order with next pointer order
• Assuming cloned list reverses nodes
• Printing random pointer values instead of node values

4. Identify the bug in the following CloneLinkedList method that clones a linked list with random pointers:

public Node CloneLinkedList(Node head) {
    if (head == null) return null;
    Node current = head;
    while (current != null) {
        Node copy = new Node(current.val);
        copy.next = current.next;
        current.next = copy;
        current = current.next;
    }
    // Setting random pointers omitted for brevity
    return head.next;
}

medium

A. The method returns the original list head instead of cloned head

B. Random pointers are not set, so cloned list is incomplete

C. The copy node's next pointer is set incorrectly

D. The current pointer update causes an infinite loop

Solution

1. Step 1: Analyze the while loop pointer update

   Inside the loop, current is updated with current = current.next, but after insertion, current.next points to the copy node, so current moves to copy node again, causing an infinite loop.

2. Step 2: Correct pointer update to skip copy node

   To avoid infinite loop, current should be updated to current.next.next to move to the next original node.

3. Final Answer:

   The current pointer update causes an infinite loop -> Option D

4. Quick Check:

   Pointer update must skip copy node to avoid infinite loop [OK]

Hint: Update current to current.next.next after inserting copy [OK]

Common Mistakes:

• Updating current to current.next causes infinite loop
• Forgetting to set random pointers
• Returning original head instead of cloned head

5. You want to clone a linked list with random pointers in O(n) time and O(1) extra space. Which approach correctly achieves this?

hard

A. Interleave cloned nodes with original nodes, set random pointers, then separate lists

B. Reverse the original list, clone nodes, then reverse back

C. Create a new list by copying values only, ignoring random pointers

D. Use a hash map to store original to cloned node mappings

Solution

1. Step 1: Understand space and time constraints

   O(n) time and O(1) extra space means no extra data structures like hash maps can be used.

2. Step 2: Use interleaving method

   Interleave cloned nodes between original nodes, set random pointers using original nodes, then separate the two lists. This uses no extra space and runs in linear time.

3. Final Answer:

   Interleave cloned nodes with original nodes, set random pointers, then separate lists -> Option A

4. Quick Check:

   Interleaving nodes = O(n) time & O(1) space cloning [OK]

Hint: Interleave clones between originals to avoid extra space [OK]

Common Mistakes:

• Using hash map increases space to O(n)
• Ignoring random pointers during cloning
• Reversing list does not help cloning random pointers