What does 'Delete Node at End' mean in a linked list?
It means removing the last node from the linked list, so the second last node becomes the new end.
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beginner
In a singly linked list, why do we need to traverse the list to delete the last node?
Because we need to find the second last node to update its next pointer to None, as singly linked lists don't have backward links.
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beginner
What happens if we try to delete the last node in an empty linked list?
Nothing happens because the list is empty; there is no node to delete.
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intermediate
In Rust, what is important to consider when deleting a node from a linked list?
We must manage ownership and borrowing rules carefully to avoid dangling pointers or memory leaks.
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beginner
What is the time complexity of deleting the last node in a singly linked list?
O(n), because we need to traverse the list to find the second last node.
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What pointer do we update when deleting the last node in a singly linked list?
AThe head pointer
BThe next pointer of the second last node
CThe previous pointer of the last node
DThe tail pointer
✗ Incorrect
We update the next pointer of the second last node to None to remove the last node.
What happens if the linked list has only one node and we delete the last node?
AThe list becomes empty
BThe head points to itself
CThe list remains unchanged
DThe last node is duplicated
✗ Incorrect
Deleting the only node makes the list empty, so head becomes None.
What is the time complexity to delete the last node in a singly linked list?
AO(1)
BO(log n)
CO(n)
DO(n^2)
✗ Incorrect
We must traverse the list to find the second last node, so it takes O(n) time.
In Rust, which concept helps prevent memory errors when deleting nodes?
AOwnership and borrowing
BGarbage collection
CManual memory management
DPointers without safety
✗ Incorrect
Rust uses ownership and borrowing rules to ensure safe memory management.
If a linked list is empty, what should happen when deleting the last node?
ADelete the head node
BThrow an error
CCreate a new node
DReturn immediately without changes
✗ Incorrect
Since the list is empty, no deletion is needed.
Explain step-by-step how to delete the last node in a singly linked list.
Think about what pointers you need to update and how to handle edge cases.
You got /4 concepts.
Describe how Rust's ownership model affects deleting nodes in a linked list.
Consider how Rust manages memory differently from other languages.
You got /4 concepts.
Practice
(1/5)
1. What does the operation Delete Node at End do in a linked list?
easy
A. Adds a new node at the end of the linked list
B. Removes the last node from the linked list
C. Removes the first node from the linked list
D. Searches for a node at the end of the linked list
Solution
Step 1: Understand the linked list structure
A linked list is a chain of nodes where each node points to the next one.
Step 2: Identify the effect of deleting the last node
Deleting the last node means removing the node that has no next node, effectively shortening the list by one node at the end.
Final Answer:
Removes the last node from the linked list -> Option B
Quick Check:
Delete Node at End = Remove last node [OK]
Hint: Deleting at end always removes the last node [OK]
Common Mistakes:
Confusing delete at end with delete at start
Thinking it adds a node instead of removing
Assuming it searches without deleting
2. Which of the following Rust code snippets correctly deletes the last node of a singly linked list?
easy
A. fn delete_end(head: &mut Option) { if head.is_none() { return; } let mut current = head.as_mut().unwrap(); while current.next.as_ref().unwrap().next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; }
B. fn delete_end(head: &mut Option) { *head = None; }
C. fn delete_end(head: &mut Option) { let mut current = head.as_mut().unwrap(); current.next = None; }
D. fn delete_end(head: &mut Option) { while head.is_some() { head = head.unwrap().next; } }
Solution
Step 1: Analyze the code logic for deleting last node
fn delete_end(head: &mut Option) { if head.is_none() { return; } let mut current = head.as_mut().unwrap(); while current.next.as_ref().unwrap().next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; } traverses the list until the second last node and sets its next to None, removing the last node.
Step 2: Check other options for correctness
fn delete_end(head: &mut Option) { *head = None; } deletes the entire list, not just last node. fn delete_end(head: &mut Option) { let mut current = head.as_mut().unwrap(); current.next = None; } incorrectly sets next of head only. fn delete_end(head: &mut Option) { while head.is_some() { head = head.unwrap().next; } } tries to reassign head incorrectly causing error.
Final Answer:
Option A code correctly deletes the last node -> Option A
Quick Check:
Traverse to second last node and set next to None [OK]
Hint: Traverse to second last node, then set next to None [OK]
Common Mistakes:
Setting head to None deletes whole list
Not checking for empty or single node list
Incorrectly reassigning head pointer
3. Given the Rust linked list code below, what will be the printed output after deleting the last node?
struct Node { val: i32, next: Option } fn print_list(head: &Option) { let mut current = head; while let Some(node) = current { print!("{} -> ", node.val); current = &node.next; } println!("null"); } fn main() { let mut head = Some(Box::new(Node { val: 1, next: Some(Box::new(Node { val: 2, next: Some(Box::new(Node { val: 3, next: None })) })) })); delete_end(&mut head); print_list(&head); } fn delete_end(head: &mut Option) { if head.is_none() { return; } if head.as_ref().unwrap().next.is_none() { *head = None; return; } let mut current = head.as_mut().unwrap(); while current.next.as_ref().unwrap().next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; }
medium
A. null
B. 1 -> 2 -> 3 -> null
C. 1 -> 2 -> null
D. 1 -> null
Solution
Step 1: Understand initial list and delete_end effect
The list starts as 1 -> 2 -> 3 -> null. delete_end removes last node (3).
Step 2: Trace print_list after deletion
After deletion, list is 1 -> 2 -> null, so print_list outputs "1 -> 2 -> null".
Final Answer:
1 -> 2 -> null -> Option C
Quick Check:
Last node removed, list ends at 2 [OK]
Hint: Deleting last node removes final value from print output [OK]
Common Mistakes:
Expecting original list unchanged
Confusing print output format
Not handling single node list case
4. The following Rust function to delete the last node of a linked list causes a runtime panic. What is the error?
fn delete_end(head: &mut Option) { let mut current = head.as_mut().unwrap(); while current.next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; }
medium
A. It panics because it sets current to None instead of current.next
B. It panics because it modifies head directly without checking
C. It panics because it never stops the loop
D. It panics because it tries to unwrap None when list is empty
Solution
Step 1: Locate the unwrap without check
let mut current = head.as_mut().unwrap(); executes without prior if head.is_none() { return; }, causing panic on empty list when as_mut() returns None and unwrap panics.
Step 2: Verify other options
The code does not modify head directly. The while loop terminates when current.next.is_none(). Setting current.next = None after reaching the last node causes no panic.
Final Answer:
It panics because it tries to unwrap None when list is empty -> Option D
Quick Check:
head.as_mut().unwrap() without empty check [OK]
Hint: Check head.is_none() before any unwrap on head [OK]
Common Mistakes:
No check for empty list before unwrap
Not checking for empty or single node list
Looping until current.next is None instead of second last
5. You have a singly linked list with nodes containing values [10, 20, 30, 40]. Write a Rust function to delete the last node and return the updated list head. Which of the following implementations correctly handles the edge case when the list has only one node?
hard
A. fn delete_end(head: &mut Option) { if head.is_none() { return; } if head.as_ref().unwrap().next.is_none() { *head = None; return; } let mut current = head.as_mut().unwrap(); while current.next.as_ref().unwrap().next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; }
B. fn delete_end(head: &mut Option) { let mut current = head.as_mut().unwrap(); while current.next.is_some() { current = current.next.as_mut().unwrap(); } current = None; }
C. fn delete_end(head: &mut Option) { if head.is_none() { return; } let mut current = head.as_mut().unwrap(); while current.next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; }
D. fn delete_end(head: &mut Option) { *head = None; }
Solution
Step 1: Check for empty and single node list
fn delete_end(head: &mut Option) { if head.is_none() { return; } if head.as_ref().unwrap().next.is_none() { *head = None; return; } let mut current = head.as_mut().unwrap(); while current.next.as_ref().unwrap().next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; } correctly checks if list is empty or has only one node, setting head to None in that case.
Step 2: Traverse to second last node and delete last
fn delete_end(head: &mut Option) { if head.is_none() { return; } if head.as_ref().unwrap().next.is_none() { *head = None; return; } let mut current = head.as_mut().unwrap(); while current.next.as_ref().unwrap().next.is_some() { current = current.next.as_mut().unwrap(); } current.next = None; } then traverses to second last node and sets its next to None, deleting last node properly.
Final Answer:
Option A correctly handles all cases including single node list -> Option A
Quick Check:
Check empty/single node, then delete last node [OK]
Hint: Always check if list has one node before deleting last [OK]
Common Mistakes:
Not handling single node list causing panic
Setting current = None instead of current.next = None
Deleting entire list instead of last node
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