Clone Linked List with Random Pointer in DSA Python - 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

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Return cloned list head

The process clones nodes by inserting them next to originals, sets random pointers using original's random, then detaches the clone list.

Execution Sample

DSA Python

class Node:
    def __init__(self, val, next=None, random=None):
        self.val = val
        self.next = next
        self.random = random

def clone(head):
    # cloning logic here

This code clones a linked list where each node has a next and a random pointer.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Insert cloned nodes after each original node	1->2->3->null	Node1.next = Clone1, Clone1.next = Node2; Node2.next = Clone2, Clone2.next = Node3; Node3.next = Clone3, Clone3.next = null	1->1'->2->2'->3->3'->null
2	Assign random pointers for cloned nodes	1->1'->2->2'->3->3'->null	Clone1.random = Node1.random.next; Clone2.random = Node2.random.next; Clone3.random = Node3.random.next	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	1->2->3->null and 1'->2'->3'->null	Restore original next pointers; set clone next pointers	Original: 1->2->3->null; Clone: 1'->2'->3'->null
4	Return cloned list head	Clone list head at 1'	No pointer changes	Cloned list: 1'->2'->3'->null
5	End	Original and clone separated	No changes	Process complete

💡 All nodes cloned and separated; original list restored; cloned list returned

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	1->1'->2->2'->3->3'->null	1->2->3->null	1->2->3->null
clone_head	null	null	null	1'	1'->2'->3'->null
Node1.next	2	1'	1'	2	2
Clone1.next	null	2	2	2'	2'
Node1.random	3	3	3	3	3
Clone1.random	null	null	3'	3'	3'

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 nodes' random pointers to set clone random pointers correctly, as shown in Step 2 of the execution_table.

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

We assign clone.random = original.random.next because the clone of any node is right after the original, so original.random.next points to the clone, as shown in Step 2.

Why do we need to separate the cloned list from the original after setting random pointers?

Because after Step 1, original and cloned nodes are interleaved, we restore original next pointers and extract cloned nodes to form a separate list, as shown in Step 3.

Visual Quiz - 3 Questions

Look at the execution_table at Step 1, what is the visual state of the list after inserting cloned nodes?

A 1->2->3->null

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

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

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

💡 Hint

Check 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

Look at the Operation column in execution_table where random pointers are assigned

If we skip Step 3 (separating the cloned list), what would be the state of the list?

A Only cloned nodes remain

B Only original nodes remain

C Original and cloned nodes interleaved

D List is empty

💡 Hint

Refer to the Visual State after Step 1 and Step 2 in execution_table

Concept Snapshot

Clone Linked List with Random Pointer:
1. Insert cloned nodes after originals.
2. Set clone.random = original.random.next.
3. Separate cloned list from original.
4. Return cloned list head.
This preserves original list and clones random pointers correctly.

Full Transcript

To clone a linked list with random pointers, first insert cloned nodes immediately after each original node. This interleaves original and cloned nodes. Next, assign the random pointers of cloned nodes by using the original node's random pointer's next node, which is the clone. Finally, separate the cloned nodes to form the cloned list and restore the original list's next pointers. Return the head of the cloned list. This method ensures the cloned list has correct next and random pointers without extra space for mapping.

Practice

1. What is the main challenge when cloning a linked list that has a random pointer in addition to the usual next pointer?

easy

A. Reversing the linked list before cloning

B. Copying the next pointers correctly

C. Ensuring the random pointers in the cloned list point to the correct cloned nodes

D. Sorting the linked list nodes before 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 in sequence, while the random pointer can point to any node or null.

2. Step 2: Identify the cloning challenge

   When cloning, copying next pointers is straightforward, but random pointers must point to the corresponding cloned nodes, not the original nodes.

3. Final Answer:

   Ensuring the random pointers in the cloned list point to the correct cloned nodes -> Option C

4. Quick Check:

   Random pointer cloning challenge = Ensuring the random pointers in the cloned list point to the correct cloned nodes [OK]

Hint: Focus on mapping original to cloned nodes for random pointers [OK]

Common Mistakes:

• Ignoring random pointers and copying only next pointers
• Setting random pointers to original nodes instead of cloned nodes
• Assuming random pointers are always null or sequential

2. Which of the following Python code snippets correctly creates a new node with the same value as an existing node curr in a linked list with random pointers?

easy

A. new_node = Node(curr)

B. new_node = Node(curr.val)

C. new_node = Node(curr.next)

D. new_node = Node(curr.random)

Solution

1. Step 1: Understand the Node constructor

   The Node constructor takes a value to create a new node. To clone a node, we pass the value of the current node.

2. Step 2: Analyze each option

   new_node = Node(curr.val) passes curr.val, which is the node's value. Options A, C, and D pass node objects or pointers, which is incorrect for creating a new node with the same value.

3. Final Answer:

   new_node = Node(curr.val) -> Option B

4. Quick Check:

   Node creation with value = new_node = Node(curr.val) [OK]

Hint: Use node.val to copy value, not the node itself [OK]

Common Mistakes:

• Passing the node object instead of its value
• Passing next or random pointers instead of value
• Confusing node creation syntax

3. Given the following code snippet to clone a linked list with random pointers, what will be the output of the print statement?

class Node:
    def __init__(self, val, next=None, random=None):
        self.val = val
        self.next = next
        self.random = random

def print_list(head):
    curr = head
    result = []
    while curr:
        rand_val = curr.random.val if curr.random else None
        result.append((curr.val, rand_val))
        curr = curr.next
    print(result)

# Original list
node1 = Node(1)
node2 = Node(2)
node3 = Node(3)
node1.next = node2
node2.next = node3
node1.random = node3
node2.random = node1
node3.random = node2

# Cloned list (shallow copy for example)
clone1 = Node(1)
clone2 = Node(2)
clone3 = Node(3)
clone1.next = clone2
clone2.next = clone3
clone1.random = node3
clone2.random = node1
clone3.random = node2

print_list(clone1)

medium

A. [(1, 3), (2, 1), (3, 2)]

B. [(1, None), (2, None), (3, None)]

C. [(1, 1), (2, 2), (3, 3)]

D. [(1, 2), (2, 3), (3, 1)]

Solution

1. Step 1: Understand the random pointers in the cloned list

   The cloned nodes' random pointers point to the original nodes (node3, node1, node2), not the cloned nodes.

2. Step 2: Trace the print_list output

   For clone1: val=1, random points to node3 with val=3.
   For clone2: val=2, random points to node1 with val=1.
   For clone3: val=3, random points to node2 with val=2.
   So the printed list is [(1, 3), (2, 1), (3, 2)].

3. Final Answer:

   [(1, 3), (2, 1), (3, 2)] -> Option A

4. Quick Check:

   Random pointers show original node values = [(1, 3), (2, 1), (3, 2)] [OK]

Hint: Check if random points to original or cloned nodes [OK]

Common Mistakes:

• Assuming random points to cloned nodes
• Printing None for random pointers incorrectly
• Mixing up node values in output

4. Consider the following code snippet intended to clone a linked list with random pointers. What is the main bug that will cause incorrect cloning?

def clone_list(head):
    if not head:
        return None
    curr = head
    # Step 1: Create new nodes and insert them after original nodes
    while curr:
        new_node = Node(curr.val)
        new_node.next = curr.next
        curr.next = new_node
        curr = new_node.next
    # Step 2: Assign random pointers for new nodes
    curr = head
    while curr:
        if curr.random:
            curr.next.random = curr.random
        curr = curr.next.next
    # Step 3: Separate the cloned list
    curr = head
    clone_head = head.next
    while curr:
        clone = curr.next
        curr.next = clone.next
        if clone.next:
            clone.next = clone.next.next
        curr = curr.next
    return clone_head

medium

A. Random pointers are assigned incorrectly by missing .next on curr.random

B. Random pointers are assigned to original nodes instead of cloned nodes

C. New nodes are not inserted after original nodes

D. The cloned list is not separated from the original list

Solution

1. Step 1: Analyze random pointer assignment

   In Step 2, the code assigns curr.next.random = curr.random. But curr.random points to an original node, so the cloned node's random pointer points to an original node, not the cloned one.

2. Step 2: Correct assignment

   The correct assignment should be curr.next.random = curr.random.next to point to the cloned node corresponding to curr.random.

3. Final Answer:

   Random pointers are assigned incorrectly by missing .next on curr.random -> Option A

4. Quick Check:

   Random pointer bug = missing .next on curr.random [OK]

Hint: Assign random as curr.next.random = curr.random.next [OK]

Common Mistakes:

• Assigning random pointer directly to original node
• Forgetting to separate cloned list
• Not inserting cloned nodes after originals

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? A) Use a hash map to map original nodes to cloned nodes. B) Reverse the original list, clone nodes, then reverse back. C) Create a new list by copying values and random pointers by traversing twice. D) Insert cloned nodes after original nodes, assign random pointers using next pointers, then separate lists.

hard

A. Use a hash map to map original nodes to cloned nodes.

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

C. Create a new list by copying values and random pointers by traversing twice.

D. Insert cloned nodes after original nodes, assign random pointers using next pointers, then separate lists.

Solution

1. Step 1: Understand space and time constraints

   O(n) time means one or two passes over the list. O(1) extra space means no extra data structures like hash maps.

2. Step 2: Analyze each approach

   Use a hash map to map original nodes to cloned nodes. uses extra space (hash map), so not O(1) space.
   Insert cloned nodes after original nodes, assign random pointers using next pointers, then separate lists. inserts cloned nodes between original nodes, uses next pointers to assign random pointers correctly, then separates the lists. This uses no extra space and runs in O(n).
   Create a new list by copying values and random pointers by traversing twice. requires extra space or multiple traversals without the in-place trick.
   Reverse the original list, clone nodes, then reverse back. is unrelated and does not solve the problem efficiently.

3. Final Answer:

   Insert cloned nodes after original nodes, assign random pointers using next pointers, then separate lists. -> Option D

4. Quick Check:

   In-place cloning with next pointers = Insert cloned nodes after original nodes, assign random pointers using next pointers, then separate lists. [OK]

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

Common Mistakes:

• Using hash maps increases space complexity
• Trying to clone without linking cloned nodes properly
• Ignoring the need to separate cloned list at the end