Clone Linked List with Random Pointer in DSA Java - 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 connections match the original without mixing 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, so the list looks like: original1 -> clone1 -> original2 -> clone2 -> ...

advanced

How do we set the random pointers for cloned nodes in the interleaved list?

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

advanced

How do we separate the cloned list from the original after setting next and random pointers?

We restore the original list by skipping cloned nodes and build the cloned list by linking cloned nodes together, effectively splitting the interleaved list into two.

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

A The head of the list

B Only the next node

C Only the previous node

D Any node in the list or null

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

In the cloning process, where is the cloned node inserted initially?

A Before the original node

B After the original node

C At the end of the list

D At the head of the list

The cloned node is inserted immediately after its original node.

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

A Clone's random = original's random

B Clone's random = original's next

C Clone's random = original's random's next

D Clone's random = null

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

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

A Separate the cloned nodes to form the new list

B Reverse the cloned list

C Delete the original list

D Set all random pointers to null

We separate the interleaved list into original and cloned lists.

Why can't we just copy nodes one by one without interleaving?

A Because random pointers need to be linked correctly

B Because next pointers are missing

C Because the list is circular

D Because nodes have no data

Random pointers can point anywhere, so interleaving helps link clones correctly.

Explain the three main steps to clone a linked list with random pointers.

Think about how to keep track of original and cloned nodes together.

Why is it important to set the cloned node's random pointer using original's random's next node?

Remember the interleaved structure of original and cloned nodes.

Practice

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

easy

A. Copying both next and random pointers correctly

B. Only copying the next pointers

C. Ignoring the random pointers

D. Reversing the linked list during cloning

Solution

1. Step 1: Understand linked list structure

   A linked list with random pointers has two references per node: next and random.

2. Step 2: Identify cloning challenge

   Cloning requires copying both next and random pointers so the new list matches the original structure.

3. Final Answer:

   Copying both next and random pointers correctly -> Option A

4. Quick Check:

   Both pointers must be copied = B [OK]

Hint: Remember to copy both next and random pointers [OK]

Common Mistakes:

• Copying only next pointers
• Ignoring random pointers
• Assuming random pointers are null

2. Which Java code snippet correctly declares a node class for a linked list with a random pointer?

easy

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

B. class Node { int val; Node next; Node random; Node() { val = 0; } }

C. class Node { int val; Node next; Node random; void Node(int val) { this.val = val; } }

D. class Node { int val; Node next; Node random; Node(int val) { this.val = val; } }

Solution

1. Step 1: Check constructor syntax

   Constructor must have the same name as class and assign value with this.val = val.

2. Step 2: Validate field declarations

   Fields val, next, and random must be declared correctly as int and Node types.

3. Final Answer:

   class Node { int val; Node next; Node random; Node(int val) { this.val = val; } } -> Option D

4. Quick Check:

   Constructor with this.val = val is correct = A [OK]

Hint: Constructor name must match class and use this keyword [OK]

Common Mistakes:

• Using void return type for constructor
• Not using this keyword for assignment
• Incorrect constructor name

3. What will be the output of the following code snippet that clones a linked list with random pointers?

class Node {
  int val;
  Node next, random;
  Node(int val) { this.val = val; }
}

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

Node head = new Node(1);
head.next = new Node(2);
head.random = head.next;
head.next.random = head;

Node clonedHead = cloneList(head);
System.out.println(clonedHead.val + " -> " + clonedHead.next.val + " -> null");
System.out.println(clonedHead.random.val + " -> " + clonedHead.next.random.val + " -> null");

// Assume cloneList correctly clones the list with random pointers

medium

A. 2 -> 1 -> null
1 -> 2 -> null

B. 1 -> 2 -> null
2 -> 1 -> null

C. 1 -> 2 -> null
1 -> 2 -> null

D. 1 -> 1 -> null
2 -> 2 -> null

Solution

1. Step 1: Understand original list and random pointers

   Original list nodes: 1 -> 2 -> null; random: 1.random = 2, 2.random = 1.

2. Step 2: Cloned list should replicate values and random links

   Cloned list nodes: 1 -> 2 -> null; cloned random pointers: cloned 1.random = cloned 2, cloned 2.random = cloned 1.

3. Final Answer:

   1 -> 2 -> null 2 -> 1 -> null -> Option B

4. Quick Check:

   Random pointers cycle = C [OK]

Hint: Random pointers in clone match original's random targets [OK]

Common Mistakes:

• Printing original random values instead of cloned
• Mixing up random pointer values
• Assuming random pointers are null

4. Identify the bug in this cloning function for a linked list with random pointers:

Node cloneList(Node head) {
  if (head == null) return null;
  Node curr = head;
  while (curr != null) {
    Node copy = new Node(curr.val);
    copy.next = curr.next;
    curr.next = copy;
    curr = curr.next.next;
  }
  // Set random pointers
  curr = head;
  while (curr != null) {
    curr.next.random = curr.random != null ? curr.random.next : null;
    curr = curr.next.next;
  }
  // Separate lists
  curr = head;
  Node copyHead = head.next;
  while (curr != null) {
    Node copy = curr.next;
    curr.next = copy.next;
    copy.next = (copy.next != null) ? copy.next.next : null;
    curr = curr.next;
  }
  return copyHead;
}

medium

A. In the last while loop, copy.next = copy.next.next causes error when copy.next is null

B. In the second while loop, curr.next.random = curr.random.next may cause NullPointerException

C. In the first while loop, curr = curr.next should be curr = curr.next.next

D. The function does not clone the random pointers at all

Solution

1. Step 1: Analyze last while loop for separating lists

   copy.next = copy.next.next assumes copy.next is not null, but at list end copy.next is null causing error.

2. Step 2: Identify fix for null check

   Must check if copy.next is null before accessing copy.next.next to avoid NullPointerException.

3. Final Answer:

   In the last while loop, copy.next = copy.next.next causes error when copy.next is null -> Option A

4. Quick Check:

   Null check needed before copy.next.next = A [OK]

Hint: Check for null before accessing next of next pointers [OK]

Common Mistakes:

• Not checking null before copy.next.next
• Incorrect iteration step in first loop
• Ignoring random pointer cloning

5. Given a linked list with random pointers, how can you clone it in O(n) time and O(1) extra space? Choose the correct approach.

hard

A. Use a hash map to store original to clone node mapping

B. Copy only next pointers and ignore random pointers

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

D. Create new nodes and assign random pointers by traversing multiple times

Solution

1. Step 1: Understand space constraints

   Using a hash map uses O(n) extra space, so not allowed for O(1) space.

2. Step 2: Use interleaving method

   Interleave cloned nodes between original nodes, assign random pointers using original nodes, then separate lists to get cloned list.

3. Final Answer:

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

4. Quick Check:

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

Hint: Interleave clones to avoid extra space for mapping [OK]

Common Mistakes:

• Using hash map increases space complexity
• Multiple traversals without interleaving
• Ignoring random pointers