Clone Linked List with Random Pointer in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a 'random pointer' in a linked list?

A random pointer is an extra pointer in each node of a linked list that can point to any node in the list or be null, not necessarily the next node.

intermediate

Why do we need to clone a linked list with random pointers carefully?

Because each node has two pointers (next and random), we must copy both correctly so the new list's structure and random links match the original without sharing nodes.

intermediate

What is the first step in the common approach to clone a linked list with random pointers?

Insert a new cloned node right after each original node in the list, so the list alternates between original and cloned nodes.

advanced

How do you set the random pointers of the cloned nodes?

For each original node, set the cloned node's random pointer to the original node's random pointer's next node (which is the clone of the random target).

advanced

How do you separate the cloned list from the original after cloning?

Traverse the combined list and restore the original list's next pointers, while extracting the cloned nodes into a separate list by fixing their next pointers.

What does the random pointer in a linked list node point to?

A Only the next node

B Any node in the list or null

C Only the previous node

D The head node

The random pointer can point to any node in the list or be null.

What is the purpose of inserting cloned nodes between original nodes in the cloning algorithm?

A To reverse the list

B To sort the list

C To delete original nodes

D To easily assign random pointers for cloned nodes

Inserting cloned nodes between original nodes helps assign random pointers by using the original nodes' random pointers.

How do you assign the random pointer of a cloned node?

A Clone's random = original's random's next

B Clone's random = original's next

C Clone's random = original's random

D Clone's random = null

The cloned node's random pointer is set to the next node of the original node's random pointer, which is the clone of the random target.

What is the final step in cloning a linked list with random pointers?

A Sort the cloned list

B Reverse the original list

C Separate the cloned nodes to form the new list

D Delete all original nodes

After setting next and random pointers, we separate the cloned nodes to form the new cloned list.

What is the time complexity of cloning a linked list with random pointers using the common approach?

A O(n)

B O(n^2)

C O(log n)

D O(1)

The cloning algorithm runs in linear time O(n) because it traverses the list a constant number of times.

Explain step-by-step how to clone a linked list with random pointers without using extra space.

Think about weaving cloned nodes into the original list first.

Why can't we simply copy nodes and random pointers by traversing the original list once?

Consider what happens if random points forward or backward.

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