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

Concept Flow - Clone Linked List with Random Pointer

Start with original list

↓

Step 1: Insert cloned nodes after originals

↓

Step 2: Assign random pointers for cloned nodes

↓

Step 3: Separate cloned list from original

↓

Result: Two separate lists, original and cloned

The cloning process inserts new nodes next to originals, copies random pointers, then detaches the clone list.

Execution Sample

DSA C++

Node* cloneList(Node* head) {
  // Step 1: Insert clones
  // Step 2: Copy random pointers
  // Step 3: Separate lists
}

This code clones a linked list with random pointers by interleaving cloned nodes, copying random links, and separating the lists.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Insert clone after each original node	Original: 1->2->3->null After insertion: 1->1'->2->2'->3->3'->null	For each node, new node inserted as next	1 -> 1' -> 2 -> 2' -> 3 -> 3' -> null
2	Assign random pointers for clones	List unchanged in nodes count	clone.random = original.random.next if original.random exists	1(r->3) -> 1'(r->3') -> 2(r->1) -> 2'(r->1') -> 3(r->2) -> 3'(r->2') -> null
3	Separate cloned list from original	Two lists: Original: 1->2->3->null Clone: 1'->2'->3'->null	Restore original next pointers, set clone next pointers	Original: 1 -> 2 -> 3 -> null Clone: 1' -> 2' -> 3' -> null
4	End	Lists separated, cloning done	No pointer changes	Original and clone lists fully independent

💡 All nodes cloned, random pointers copied, lists separated

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
head	1->2->3->null	1->1'->2->2'->3->3'->null	Same as after Step 1	1->2->3->null (restored original list)	1->2->3->null (original list head)
clone_head	null	null	null	1'->2'->3'->null (start of cloned list)	1'->2'->3'->null (cloned list head)
current	head (1)	head (1)	head (1)	head (1)	null (end of traversal)

Key Moments - 3 Insights

Why do we insert cloned nodes immediately after original nodes instead of creating a separate list first?

Inserting clones next to originals allows easy access to original's random pointer's clone by original.random.next, as shown in Step 2 of execution_table.

How do we correctly assign the random pointer of the cloned nodes?

We use original.random.next because the clone of any node is right after it, so clone.random = original.random.next, as detailed in Step 2.

Why do we need to separate the two lists at the end?

Because after insertion, original and cloned nodes are interleaved, so we restore original next pointers and extract cloned nodes to form an independent cloned list, as in Step 3.

Visual Quiz - 3 Questions

Look at the execution_table at Step 1, what is the visual state of the list after insertion?

A 1 -> 2 -> 3 -> null

B 1 -> 1' -> 2 -> 2' -> 3 -> 3' -> null

C 1' -> 2' -> 3' -> null

D 1 -> 2 -> 3 -> 1' -> 2' -> 3' -> null

💡 Hint

Refer to the 'Visual State' column in Step 1 of execution_table.

At which step do we assign the random pointers for the cloned nodes?

A Step 2

B Step 1

C Step 3

D Step 4

💡 Hint

Check the 'Operation' column in execution_table for Step 2.

If we skip Step 3, what would be the state of the list?

A Only original list remains

B Two separate lists

C Original and cloned nodes interleaved in one list

D Only cloned list remains

💡 Hint

Look at the 'Nodes in List' and 'Visual State' columns before Step 3.

Concept Snapshot

Clone Linked List with Random Pointer:
1. Insert cloned node after each original node.
2. Set clone.random = original.random.next.
3. Separate cloned list from original.
Result: Two independent lists with correct random pointers.

Full Transcript

To clone a linked list with random pointers, first insert cloned nodes immediately after each original node. This interleaving helps assign random pointers easily by pointing clone.random to original.random.next. Finally, separate the interleaved list into two independent lists: the original and the cloned. This method ensures the cloned list has the same structure and random links as the original.

Practice

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

easy

A. Maintaining the correct random pointer connections in the cloned list

B. Copying the data values of each node

C. Reversing the linked list before cloning

D. Sorting the linked list nodes after 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 copying nodes is straightforward, correctly assigning the random pointers in the cloned list is tricky because they can point anywhere in the original list.

3. Final Answer:

   Maintaining the correct random pointer connections in the cloned list -> Option A

4. Quick Check:

   Random pointer cloning = main challenge [OK]

Hint: Focus on copying random pointers correctly [OK]

Common Mistakes:

• Ignoring random pointers during cloning
• Only copying next pointers
• Assuming random pointers follow next pointers

2. Which of the following is the correct way to declare a node structure for a linked list with a random pointer in C++?

easy

A. struct Node { int data; Node* next; Node* random; };

B. struct Node { int data; Node next; Node random; };

C. struct Node { int data; Node* next; int random; };

D. struct Node { int data; Node next; int random; };

Solution

1. Step 1: Understand pointer usage in C++ structs

   To link nodes, pointers to Node type are needed for next and random pointers.

2. Step 2: Check each option's correctness

   struct Node { int data; Node* next; Node* random; }; uses Node* for next and random, which is correct. Those using Node without pointer are invalid. struct Node { int data; Node* next; int random; }; uses int for random, which is incorrect.

3. Final Answer:

   struct Node { int data; Node* next; Node* random; }; -> Option A

4. Quick Check:

   Use pointers for next and random [OK]

Hint: Use Node* for next and random pointers [OK]

Common Mistakes:

• Using Node instead of Node* for pointers
• Using int type for random pointer
• Missing semicolon after struct

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

struct Node {
  int data;
  Node* next;
  Node* random;
  Node(int x) : data(x), next(nullptr), random(nullptr) {}
};

// Original list: 1 -> 2 -> 3 -> null
// random pointers: 1.random = 3, 2.random = 1, 3.random = 2

// After cloning, print cloned list data:
Node* cloneHead = cloneLinkedList(head); // Assume correct cloning function
Node* temp = cloneHead;
while(temp) {
  std::cout << temp->data << " ";
  temp = temp->next;
}

medium

A. 1 3 2

B. 3 2 1

C. 1 2 3

D. 2 1 3

Solution

1. Step 1: Understand the cloning process

   The cloneLinkedList function creates a new list with the same data and next pointer order as the original list.

2. Step 2: Trace the output loop

   The loop prints data of cloned nodes in order: first node data 1, then 2, then 3.

3. Final Answer:

   1 2 3 -> Option C

4. Quick Check:

   Cloned list data order = original order [OK]

Hint: Cloned list preserves original node order [OK]

Common Mistakes:

• Confusing random pointer order with next pointer order
• Printing random pointer data instead of next
• Assuming reversed list after cloning

4. Identify the error in the following code snippet for cloning a linked list with random pointers:

Node* cloneLinkedList(Node* head) {
  if (!head) return nullptr;
  Node* curr = head;
  while (curr) {
    Node* newNode = new Node(curr->data);
    newNode->next = curr->next;
    curr->next = newNode;
    curr = newNode->next;
  }
  curr = head;
  while (curr) {
    if (curr->random) {
      curr->next->random = curr->random->next;
    }
    curr = curr->next->next;
  }
  Node* original = head, *copy = head->next, *temp = copy;
  while (original && copy) {
    original->next = original->next->next;
    copy->next = copy->next ? copy->next->next : nullptr;
    original = original->next;
    copy = copy->next;
  }
  return temp;
}

medium

A. The random pointer assignment is incorrect

B. No error; the code correctly clones the list

C. The final while loop incorrectly restores original and cloned lists

D. The first while loop incorrectly inserts cloned nodes

Solution

1. Step 1: Analyze node insertion loop

   The first loop inserts cloned nodes after each original node correctly.

2. Step 2: Analyze random pointer assignment and list separation

   The second loop assigns random pointers correctly. The final loop restores original and cloned lists properly.

3. Final Answer:

   No error; the code correctly clones the list -> Option B

4. Quick Check:

   Code follows standard cloning steps [OK]

Hint: Check each step matches cloning algorithm phases [OK]

Common Mistakes:

• Misreading pointer assignments
• Assuming random pointer assignment is wrong without checking
• Confusing original and cloned list restoration

5. You want to clone a linked list with random pointers but without using extra space for a hash map. Which approach below correctly achieves this with O(1) extra space?

hard

A. Reverse the original list, clone it, then reverse both lists back

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

C. Traverse the list twice, first to copy data, second to copy random pointers using a vector

D. Create cloned nodes interleaved with original nodes, assign random pointers, then separate lists

Solution

1. Step 1: Understand space constraints

   Using extra data structures like hash maps or vectors uses extra space.

2. Step 2: Identify O(1) space cloning method

   Interleaving cloned nodes with original nodes allows assigning random pointers without extra space, then separating lists restores original and cloned lists.

3. Final Answer:

   Create cloned nodes interleaved with original nodes, assign random pointers, then separate lists -> Option D

4. Quick Check:

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

Hint: Interleave cloned nodes to avoid extra space [OK]

Common Mistakes:

• Using hash maps when O(1) space is required
• Trying to clone by reversing list
• Using extra arrays or vectors unnecessarily