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DSA Rustprogramming~30 mins

Clone Linked List with Random Pointer in DSA Rust - Build from Scratch

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Clone Linked List with Random Pointer
📖 Scenario: Imagine you have a special linked list where each node has two pointers: one points to the next node, and another points randomly to any node in the list or null. This structure is used in complex applications like social networks to represent friends and random connections.
🎯 Goal: You will build a Rust program to create such a linked list, then clone it exactly, including the random pointers. This means the new list should be a deep copy with the same values and random connections, but completely independent from the original.
📋 What You'll Learn
Create a linked list node struct with val, next, and random fields
Build an example linked list with 3 nodes and specific random pointers
Write a function to clone the linked list with random pointers
Print the cloned list showing each node's value and the value of its random pointer
💡 Why This Matters
🌍 Real World
Cloning complex linked structures with random pointers is useful in applications like social networks, compilers, and games where objects have arbitrary references.
💼 Career
Understanding how to clone data structures with complex pointers is important for software engineers working on memory management, data serialization, and system design.
Progress0 / 4 steps
1
Create the linked list node struct and build the list
Define a struct called Node with fields val: i32, next: Option>, and random: Option<*const Node>. Then create three nodes with values 1, 2, and 3. Link them so that node1.next = Some(node2) and node2.next = Some(node3). Set node1.random to Some(&node3 as *const Node), node2.random to Some(&node1 as *const Node), and node3.random to None.
DSA Rust
Hint

Use Option> for next to own the next node. Use raw pointers *const Node for random because it can point anywhere or be None.

2
Add a helper function to clone the linked list
Create a function called clone_list that takes head: &Option> and returns Option>. Inside, create a HashMap to map original node pointers to cloned nodes. This will help set random pointers correctly.
DSA Rust
Hint

Use two passes: first clone nodes and next pointers while storing mapping from original to clone. Then set random pointers using the map.

3
Use the clone function and prepare to print the cloned list
Call clone_list with the original list head and store the result in a variable called cloned_head. Then write a function called print_list that takes head: &Option> and prints each node's val and the val of the node pointed to by random (or None if no random).
DSA Rust
Hint

Use a loop to go through the list. For each node, print its value and the value of the node its random pointer points to. Use unsafe to dereference raw pointers.

4
Run the program and verify the cloned list output
Run the program. The output should print each node's value and the value of its random pointer in the cloned list. The expected output is:
Node val: 1, Random points to: 3
Node val: 2, Random points to: 1
Node val: 3, Random points to: None
DSA Rust
Hint

Check that the printed values match the expected output exactly, showing the cloned list nodes and their random pointers.

Practice

(1/5)
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
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