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

Time Complexity: Clone Linked List with Random Pointer

O(n)

Understanding Time Complexity

We want to understand how long it takes to copy a linked list where each node has a random pointer.

How does the time needed grow as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct Node {
    val: i32,
    next: Option>,
    random: Option<*mut Node>,
}

fn clone_list(head: &Option>) -> Option> {
    // Step 1: Insert cloned nodes
    // Step 2: Assign random pointers
    // Step 3: Separate cloned list
    None // placeholder
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Three passes over the linked list nodes.
• How many times: Each pass visits every node once.

How Execution Grows With Input

Each node is visited three times, so the total work grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	~30 (3 x 10)
100	~300 (3 x 100)
1000	~3000 (3 x 1000)

Pattern observation: The operations increase linearly as the list size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to clone the list grows in a straight line with the number of nodes.

Common Mistake

[X] Wrong: "Cloning the list takes more than linear time because of the random pointers."

[OK] Correct: Each node's random pointer is handled in the same pass without extra nested loops, so it does not multiply the work.

Interview Connect

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

Self-Check

"What if we used a hash map to store cloned nodes instead of modifying the original list? How would the time complexity change?"

Practice

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

easy

A. Copying the data values of each node

B. Copying the random pointers correctly to the new nodes

C. Reversing the linked list before cloning

D. Sorting the linked list nodes

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:

   Copying the random pointers correctly to the new nodes -> Option B

4. Quick Check:

   Random pointer cloning = main challenge [OK]

Hint: Focus on random pointer mapping for correct cloning [OK]

Common Mistakes:

• Ignoring random pointers during cloning
• Assuming random pointers follow next pointers
• Copying only data without pointers

2. Which Rust syntax correctly defines a node struct for a linked list with a random pointer?

easy

A. struct Node { val: i32, next: Option, random: Option<*mut Node> }

B. struct Node { val: i32, next: Option, random: Option }

C. struct Node { val: i32, next: Option<Node>, random: Option<Node> }

D. struct Node { val: i32, next: Box<Node>, random: Box<Node> }

Solution

1. Step 1: Understand Rust pointer types for linked lists

   Rust uses Option for owning next nodes. For random pointers, raw pointers (*mut Node) are used to avoid ownership conflicts.

2. Step 2: Analyze each struct definition

   struct Node { val: i32, next: Option, random: Option<*mut Node> } uses Option for next and Option<*mut Node> for random, which is correct. struct Node { val: i32, next: Option, random: Option } wrongly uses Box<Node> for random, causing ownership issues. struct Node { val: i32, next: Option<Node>, random: Option<Node> } uses Node directly, which is invalid. struct Node { val: i32, next: Box<Node>, random: Box<Node> } uses Box<Node> without Option, disallowing null.

3. Final Answer:

   struct Node { val: i32, next: Option, random: Option<*mut Node> } -> Option A

4. Quick Check:

   Correct pointer types = struct Node { val: i32, next: Option, random: Option<*mut Node> } [OK]

Hint: Use Option for next, raw pointer for random [OK]

Common Mistakes:

• Using Box<Node> without Option for nullable pointers
• Using Box<Node> for random causing ownership errors
• Using Node directly instead of Box or pointer

3. Given the following Rust code snippet for cloning a linked list with random pointers, what will be the output printed?

fn print_list(head: &Option) {
    let mut current = head;
    while let Some(node) = current {
        let random_val = match node.random {
            Some(ptr) => unsafe { (*ptr).val },
            None => -1,
        };
        println!("{} -> Random: {}", node.val, random_val);
        current = &node.next;
    }
}

// Original list: 1 -> 2 -> null
// Random pointers: 1.random -> 2, 2.random -> 1
// After cloning, print cloned list using print_list(cloned_head)

medium

A. 1 -> Random: 1
2 -> Random: 2

B. 1 -> Random: -1
2 -> Random: -1

C. 1 -> Random: 2
2 -> Random: 1

D. Compilation error due to unsafe block

Solution

1. Step 1: Understand the print_list function

   The function prints each node's value and the value of the node pointed by its random pointer. If random is None, it prints -1.

2. Step 2: Analyze the original and cloned list random pointers

   Original list has 1.random -> 2 and 2.random -> 1. The cloned list should preserve these random pointer relations, so printing cloned list shows the same pattern.

3. Final Answer:

   1 -> Random: 2 2 -> Random: 1 -> Option C

4. Quick Check:

   Random pointers preserved = output B [OK]

Hint: Random pointers keep same target values after cloning [OK]

Common Mistakes:

• Assuming random pointers reset to self or None
• Confusing node values with random pointer values
• Thinking unsafe block causes compile error

4. In the Rust cloning function below, what is the bug causing incorrect random pointers in the cloned list?

fn clone_list(head: &Option) -> Option {
    let mut map = std::collections::HashMap::new();
    let mut current = head;
    while let Some(node) = current {
        let new_node = Box::new(Node { val: node.val, next: None, random: None });
        map.insert(node as *const _, new_node);
        current = &node.next;
    }
    // Second pass to assign next and random pointers
    // ...
    None
}

medium

A. Not cloning node.val causes wrong values

B. Using raw pointer as key causes runtime error

C. Missing unsafe block around map insertion

D. Inserting Box<Node> into map moves ownership of cloned nodes, preventing proper linking

Solution

1. Step 1: Analyze map insertion of new_node

   Inserting new_node (Box<Node>) into map moves ownership of the cloned node into the HashMap, preventing proper linking of next and random pointers.

2. Step 2: Understand ownership rules in Rust cloning

   To keep original nodes intact, cloned nodes should be stored separately without moving original nodes. The map should store references or pointers, not move ownership.

3. Final Answer:

   Inserting Box<Node> into map moves ownership of cloned nodes, preventing proper linking -> Option D

4. Quick Check:

   Ownership moved incorrectly = bug C [OK]

Hint: Avoid moving ownership when mapping original to cloned nodes [OK]

Common Mistakes:

• Assuming raw pointers cause runtime errors
• Ignoring ownership move semantics
• Forgetting to clone node values

5. You want to clone a linked list with random pointers in Rust without using extra space (no HashMap). Which approach correctly achieves this?

hard

A. Interleave cloned nodes between original nodes, then separate lists

B. Use recursion to clone nodes and random pointers

C. Copy nodes into a vector, then assign random pointers by index

D. Clone nodes and random pointers in a single pass using a stack

Solution

1. Step 1: Understand the interleaving technique

   Insert cloned nodes immediately after original nodes, so each original node points to its clone. This allows setting random pointers by referencing original nodes' random pointers' next nodes.

2. Step 2: Separate the interleaved list into original and cloned lists

   After setting random pointers, detach cloned nodes to form the cloned list, restoring the original list.

3. Final Answer:

   Interleave cloned nodes between original nodes, then separate lists -> Option A

4. Quick Check:

   Interleaving nodes = no extra space cloning [OK]

Hint: Interleave clones to link random pointers without extra space [OK]

Common Mistakes:

• Using recursion causes stack overflow on large lists
• Using vector wastes extra space
• Stack approach is not standard for this problem