Clone Linked List with Random Pointer in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Cloning a Simple Linked List with Random Pointers

What is the printed output of the cloned linked list's nodes' values and their random pointers after running the given cloning code?

DSA C#

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

public 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 = copy.next;
    }
    curr = head;
    while (curr != null) {
        if (curr.random != null) {
            curr.next.random = curr.random.next;
        }
        curr = curr.next.next;
    }
    curr = head;
    Node dummy = new Node(0);
    Node copyCurr = dummy;
    while (curr != null) {
        copyCurr.next = curr.next;
        curr.next = curr.next.next;
        curr = curr.next;
        copyCurr = copyCurr.next;
    }
    return dummy.next;
}

// Setup linked list:
// Node1(val=1) -> Node2(val=2) -> null
// Node1.random = Node2
// Node2.random = Node2

Node node1 = new Node(1);
Node node2 = new Node(2);
node1.next = node2;
node1.random = node2;
node2.random = node2;

Node clonedHead = CloneList(node1);
Node temp = clonedHead;
while (temp != null) {
    Console.WriteLine($"Node val: {temp.val}, Random val: {(temp.random != null ? temp.random.val.ToString() : "null")}");
    temp = temp.next;
}

A

Node val: 1, Random val: 2
Node val: 2, Random val: 2

B

Node val: 1, Random val: 1
Node val: 2, Random val: 1

C

Node val: 1, Random val: null
Node val: 2, Random val: null

D

Node val: 1, Random val: 2
Node val: 2, Random val: 1

💡 Hint

Focus on how the random pointers are assigned in the cloned list after interleaving nodes.

The cloning process creates new nodes interleaved with original nodes. The random pointer of each cloned node is set to the next node of the original node's random pointer. Since node1.random points to node2, cloned node1.random points to cloned node2. Similarly, node2.random points to itself, so cloned node2.random points to cloned node2.

❓ Predict Output

intermediate

Output of Cloning a Linked List with Null Random Pointers

What is the output when cloning a linked list where all random pointers are null?

DSA C#

Node node1 = new Node(10);
Node node2 = new Node(20);
Node node3 = new Node(30);
node1.next = node2;
node2.next = node3;
// All random pointers are null
Node clonedHead = CloneList(node1);
Node temp = clonedHead;
while (temp != null) {
    Console.WriteLine($"Node val: {temp.val}, Random val: {(temp.random != null ? temp.random.val.ToString() : "null")}");
    temp = temp.next;
}

A

Node val: 10, Random val: null
Node val: 20, Random val: null
Node val: 30, Random val: null

B

Node val: 10, Random val: 10
Node val: 20, Random val: 20
Node val: 30, Random val: 30

C

Node val: 10, Random val: 20
Node val: 20, Random val: 30
Node val: 30, Random val: null

D

Node val: 10, Random val: 30
Node val: 20, Random val: 10
Node val: 30, Random val: 20

💡 Hint

If original random pointers are null, cloned random pointers should also be null.

Since all original nodes have random pointers set to null, the cloned nodes' random pointers remain null after cloning.

🔧 Debug

advanced

Identify the Bug in Cloning Random Pointer Assignment

Given the cloning code below, what error will occur when running it on a linked list with random pointers?

DSA C#

Node curr = head;
while (curr != null) {
    curr.next.random = curr.random.next;
    curr = curr.next.next;
}

A No error, code runs correctly

B NullReferenceException when curr.random is null

C IndexOutOfRangeException due to accessing next of last node

D StackOverflowException due to infinite loop

💡 Hint

Consider what happens if curr.random is null before accessing curr.random.next.

If curr.random is null, accessing curr.random.next causes a NullReferenceException. The code should check if curr.random is not null before accessing curr.random.next.

🚀 Application

advanced

Number of Nodes in Cloned List After Cloning

If the original linked list has 5 nodes, how many nodes will the cloned linked list have after running the cloning algorithm?

A 0 nodes

B 10 nodes

C 5 nodes

D 6 nodes

💡 Hint

The cloned list is a separate copy with the same number of nodes as the original.

The cloning algorithm creates exactly one new node for each original node, so the cloned list has the same number of nodes as the original.

🧠 Conceptual

expert

Why Interleaving Nodes Helps in Cloning Random Pointers

Why does the cloning algorithm insert cloned nodes between original nodes before assigning random pointers?

A It sorts the nodes by their values

B It merges the original and cloned lists permanently

C It reduces the total number of nodes in the list

D It allows easy access to the cloned node corresponding to any original node's random pointer without extra space

💡 Hint

Think about how to find the clone of a node's random pointer without using a dictionary.

By placing cloned nodes right after their originals, the clone of any node's random pointer is simply the next node of that random pointer. This avoids using extra memory for mapping.

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