Clone Linked List with Random Pointer in DSA Python - Time & Space Complexity

Time Complexity: Clone Linked List with Random Pointer

O(n)

Understanding Time Complexity

When cloning a linked list with random pointers, we want to know how the time needed grows as the list gets longer.

We ask: How many steps does it take to copy all nodes and their random links?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

def clone_linked_list(head):
    if not head:
        return None
    
    # Step 1: Insert cloned nodes
    curr = head
    while curr:
        new_node = Node(curr.val, curr.next)
        curr.next = new_node
        curr = new_node.next
    
    # Step 2: Assign random pointers
    curr = head
    while curr:
        if curr.random:
            curr.next.random = curr.random.next
        curr = curr.next.next
    
    # Step 3: Separate the lists
    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
    

This code clones a linked list where each node has a next and a random pointer, creating a new list with the same structure.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Three separate while loops each traverse the entire list once.
• How many times: Each loop runs approximately n times, where n is the number of nodes.

How Execution Grows With Input

Each loop goes through all nodes once, so the total steps add up but grow linearly with the list size.

Input Size (n)	Approx. Operations
10	About 30 steps (3 loops x 10 nodes)
100	About 300 steps
1000	About 3000 steps

Pattern observation: The total work grows directly in proportion to the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to clone the list grows in a straight line as the list gets longer.

Common Mistake

[X] Wrong: "Because there are three loops, the time complexity is O(n³)."

[OK] Correct: The loops run one after another, not nested inside each other, so their times add up, not multiply.

Interview Connect

Understanding this linear time cloning method shows you can handle complex pointer structures efficiently, a skill valued in many coding challenges.

Self-Check

"What if we used a hash map to store old-to-new node mappings instead of inserting cloned nodes inline? How would the time complexity change?"

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