A random pointer is an extra pointer in each node of a linked list that can point to any node in the list or be null, not necessarily the next node.
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intermediate
Why do we need to clone a linked list with random pointers carefully?
Because each node has two pointers (next and random), we must copy both correctly so the new list's structure and random links match the original without sharing nodes.
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intermediate
What is the main idea behind the three-step approach to clone a linked list with random pointers?
1. Insert cloned nodes after original nodes. 2. Set random pointers of cloned nodes using original nodes. 3. Separate the cloned list from the original list.
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advanced
In Rust, why is careful ownership management important when cloning a linked list with random pointers?
Rust enforces strict ownership and borrowing rules, so we must use smart pointers like Rc and RefCell to allow multiple ownership and mutable access to nodes safely.
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beginner
What does the output look like after cloning a linked list with random pointers?
The output is a new linked list where each node has the same value, next pointer, and random pointer structure as the original, but all nodes are distinct copies.
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What does the 'random' pointer in a linked list node point to?
AThe head of the list
BOnly the next node
COnly the previous node
DAny node in the list or null
✗ Incorrect
The random pointer can point to any node in the list or be null.
Which step is NOT part of the common cloning algorithm for linked lists with random pointers?
AReverse the original list
BSet random pointers of cloned nodes
CInsert cloned nodes after original nodes
DSeparate cloned list from original
✗ Incorrect
Reversing the original list is not part of the cloning process.
In Rust, which smart pointer is commonly used to allow multiple ownership of linked list nodes?
ABox
BRefCell
CRc
DArc
✗ Incorrect
Rc (Reference Counted) pointer allows multiple ownership in Rust.
What is the main challenge when cloning a linked list with random pointers?
ACopying both next and random pointers correctly
BCopying next pointers only
CSorting the list
DCopying node values only
✗ Incorrect
Both next and random pointers must be copied correctly to preserve structure.
After cloning, how do the nodes in the new list relate to the original nodes?
AThey are the same nodes
BThey are new nodes with the same values and pointers
CThey only share random pointers
DThey only share next pointers
✗ Incorrect
Cloned nodes are new copies with the same values and pointer structure.
Explain the three main steps to clone a linked list with random pointers.
Think about how to link cloned nodes without losing original structure.
You got /3 concepts.
Describe how Rust's ownership rules affect cloning a linked list with random pointers.
Consider how Rust prevents multiple mutable references.
You got /3 concepts.
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
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.
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.
Final Answer:
Copying the random pointers correctly to the new nodes -> Option B
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?
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
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.
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.
Final Answer:
1 -> Random: 2
2 -> Random: 1 -> Option C
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
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.
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.
Final Answer:
Inserting Box<Node> into map moves ownership of cloned nodes, preventing proper linking -> Option D
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
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.
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.
Final Answer:
Interleave cloned nodes between original nodes, then separate lists -> Option A
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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