Delete by Value in Doubly Linked List in DSA Rust - Build from Scratch

Delete by Value in Doubly Linked List

📖 Scenario: Imagine you have a playlist of songs where you can move forward and backward easily. This playlist is like a doubly linked list where each song knows the next and previous song. Sometimes, you want to remove a song by its name from the playlist.

🎯 Goal: You will build a doubly linked list of songs and write code to delete a song by its name. Finally, you will print the playlist to see the remaining songs.

📋 What You'll Learn

Create a doubly linked list with exact songs: "SongA", "SongB", "SongC", "SongD", "SongE"

Create a variable called song_to_delete with the value "SongC"

Write a function called delete_by_value that deletes the first node with the given song name

Print the playlist after deletion in the format: SongA <-> SongB <-> SongD <-> SongE <-> null

💡 Why This Matters

🌍 Real World

Doubly linked lists are used in music players, web browsers, and other applications where you need to move forward and backward through items efficiently.

💼 Career

Understanding linked lists and how to manipulate them is fundamental for software engineers working with low-level data structures, memory management, and performance-critical applications.

1

Create the doubly linked list with songs

Create a struct called Node with fields data (String), prev (Option>), and next (Option>). Then create a struct called DoublyLinkedList with a field head (Option>). Finally, create a DoublyLinkedList called playlist and add nodes with these exact songs in order: "SongA", "SongB", "SongC", "SongD", "SongE".

DSA Rust

# Define Node and DoublyLinkedList structs
# Create playlist with nodes SongA, SongB, SongC, SongD, SongE
# Your code here

Hint

Use Rc and RefCell to allow multiple ownership and mutability of nodes. Append nodes one by one to the list.

2

Create the song_to_delete variable

Create a variable called song_to_delete and set it to the string "SongC".

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

#[derive(Debug)]
struct Node {
    data: String,
    prev: Option>>,
    next: Option>>,
}

#[derive(Debug)]
struct DoublyLinkedList {
    head: Option>>,
    tail: Option>>,
}

impl DoublyLinkedList {
    fn new() -> Self {
        DoublyLinkedList { head: None, tail: None }
    }

    fn append(&mut self, data: String) {
        let new_node = Rc::new(RefCell::new(Node { data, prev: None, next: None }));
        match self.tail.take() {
            Some(old_tail) => {
                old_tail.borrow_mut().next = Some(Rc::clone(&new_node));
                new_node.borrow_mut().prev = Some(old_tail);
                self.tail = Some(new_node);
            }
            None => {
                self.head = Some(Rc::clone(&new_node));
                self.tail = Some(new_node);
            }
        }
    }
}

fn main() {
    let mut playlist = DoublyLinkedList::new();
    playlist.append("SongA".to_string());
    playlist.append("SongB".to_string());
    playlist.append("SongC".to_string());
    playlist.append("SongD".to_string());
    playlist.append("SongE".to_string());
    
    // Create song_to_delete variable
    # Your code here
}

Hint

Use let song_to_delete = "SongC".to_string(); to create the variable.

3

Write the delete_by_value function

Inside the DoublyLinkedList impl block, write a function called delete_by_value that takes &mut self and a value: &str. It should find the first node with data equal to value and remove it from the list by updating the prev and next pointers correctly. If the node is the head or tail, update self.head or self.tail accordingly. Stop after deleting the first matching node.

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

#[derive(Debug)]
struct Node {
    data: String,
    prev: Option>>,
    next: Option>>,
}

#[derive(Debug)]
struct DoublyLinkedList {
    head: Option>>,
    tail: Option>>,
}

impl DoublyLinkedList {
    fn new() -> Self {
        DoublyLinkedList { head: None, tail: None }
    }

    fn append(&mut self, data: String) {
        let new_node = Rc::new(RefCell::new(Node { data, prev: None, next: None }));
        match self.tail.take() {
            Some(old_tail) => {
                old_tail.borrow_mut().next = Some(Rc::clone(&new_node));
                new_node.borrow_mut().prev = Some(old_tail);
                self.tail = Some(new_node);
            }
            None => {
                self.head = Some(Rc::clone(&new_node));
                self.tail = Some(new_node);
            }
        }
    }

    // Write delete_by_value function here
    # Your code here
}

fn main() {
    let mut playlist = DoublyLinkedList::new();
    playlist.append("SongA".to_string());
    playlist.append("SongB".to_string());
    playlist.append("SongC".to_string());
    playlist.append("SongD".to_string());
    playlist.append("SongE".to_string());
    
    let song_to_delete = "SongC".to_string();
}

Hint

Traverse the list from head. When you find the node with matching data, update the prev node's next and the next node's prev. Handle the case when the node is head or tail.

4

Print the playlist after deletion

Call playlist.delete_by_value(&song_to_delete) to delete the song. Then print the playlist from head to tail in this format: each song's name followed by <-> , ending with null. For example: SongA <-> SongB <-> SongD <-> SongE <-> null.

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

#[derive(Debug)]
struct Node {
    data: String,
    prev: Option>>,
    next: Option>>,
}

#[derive(Debug)]
struct DoublyLinkedList {
    head: Option>>,
    tail: Option>>,
}

impl DoublyLinkedList {
    fn new() -> Self {
        DoublyLinkedList { head: None, tail: None }
    }

    fn append(&mut self, data: String) {
        let new_node = Rc::new(RefCell::new(Node { data, prev: None, next: None }));
        match self.tail.take() {
            Some(old_tail) => {
                old_tail.borrow_mut().next = Some(Rc::clone(&new_node));
                new_node.borrow_mut().prev = Some(old_tail);
                self.tail = Some(new_node);
            }
            None => {
                self.head = Some(Rc::clone(&new_node));
                self.tail = Some(new_node);
            }
        }
    }

    fn delete_by_value(&mut self, value: &str) {
        let mut current = self.head.clone();
        while let Some(node_rc) = current {
            if node_rc.borrow().data == value {
                let prev = node_rc.borrow().prev.clone();
                let next = node_rc.borrow().next.clone();

                if let Some(prev_node) = prev.clone() {
                    prev_node.borrow_mut().next = next.clone();
                } else {
                    self.head = next.clone();
                }

                if let Some(next_node) = next.clone() {
                    next_node.borrow_mut().prev = prev.clone();
                } else {
                    self.tail = prev.clone();
                }

                break;
            }
            current = node_rc.borrow().next.clone();
        }
    }
}

fn main() {
    let mut playlist = DoublyLinkedList::new();
    playlist.append("SongA".to_string());
    playlist.append("SongB".to_string());
    playlist.append("SongC".to_string());
    playlist.append("SongD".to_string());
    playlist.append("SongE".to_string());
    
    let song_to_delete = "SongC".to_string();

    // Delete the song and print the playlist
    # Your code here
}

Hint

After deleting, start from head and print each node's data followed by <-> . End with null.

Practice

1. What is the main purpose of the delete_by_value function in a doubly linked list?

easy

A. To remove the first node that contains the given value

B. To add a new node with the given value at the end

C. To reverse the entire doubly linked list

D. To count how many nodes have the given value

Solution

1. Step 1: Understand the function's goal

   The delete_by_value function searches for a node with the specified value.

2. Step 2: Identify the action performed

   Once found, it removes that node from the doubly linked list, adjusting pointers accordingly.

3. Final Answer:

   To remove the first node that contains the given value -> Option A

4. Quick Check:

   Delete by value means remove matching node [OK]

Hint: Delete by value removes first matching node [OK]

Common Mistakes:

• Confusing delete with insert
• Thinking it reverses the list
• Assuming it counts nodes

2. Which Rust syntax correctly declares a mutable doubly linked list node struct with prev and next pointers?

easy

A. struct Node { value: i32, prev: Option>, next: Option> }

B. struct Node { value: i32, prev: Option<*mut Node>, next: Option<*mut Node> }

C. struct Node { value: i32, prev: Option, next: Option }

D. struct Node { value: i32, prev: Option<Node>, next: Option<Node> }

Solution

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

   Rust needs shared ownership and mutability for prev and next pointers, so Rc is used.

2. Step 2: Check options for correct pointer usage

   struct Node { value: i32, prev: Option>, next: Option> } uses Option> which allows multiple owners and interior mutability, suitable for doubly linked lists.

3. Final Answer:

   struct Node { value: i32, prev: Option>, next: Option> } -> Option A

4. Quick Check:

   Use Rc and RefCell for mutable shared nodes [OK]

Hint: Use Rc for mutable shared pointers [OK]

Common Mistakes:

• Using Box for both prev and next causes ownership issues
• Using raw pointers without unsafe code
• Using direct Node type causing infinite size

3. Given the following Rust code snippet for deleting a value from a doubly linked list, what will be the printed list after deleting value 3?

let mut list = DoublyLinkedList::new();
list.push_back(1);
list.push_back(3);
list.push_back(5);
list.delete_by_value(3);
list.print();

medium

A. 1 -> 3 -> 5 -> null

B. 1 -> 5 -> null

C. 3 -> 5 -> null

D. 1 -> null

Solution

1. Step 1: Understand initial list state

   After pushing 1, 3, 5, the list is 1 -> 3 -> 5 -> null.

2. Step 2: Delete node with value 3

   The delete_by_value(3) removes the node containing 3, linking 1 directly to 5.

3. Final Answer:

   1 -> 5 -> null -> Option B

4. Quick Check:

   Deleted 3, list is 1 -> 5 [OK]

Hint: Deleting value removes that node, links neighbors [OK]

Common Mistakes:

• Not updating prev and next pointers correctly
• Deleting wrong node or none
• Printing old list without deletion

4. Consider this Rust function snippet for deleting a node by value in a doubly linked list:

fn delete_by_value(&mut self, val: i32) {
    let mut current = self.head.clone();
    while let Some(node) = current {
        if node.borrow().value == val {
            // deletion logic here
            break;
        }
        current = node.borrow().next.clone();
    }
}

What is a common bug if the deletion logic forgets to update the prev pointer of the next node?

medium

A. The list will reverse automatically

B. The head pointer will always point to null

C. The deleted node will remain in memory forever

D. The list may become corrupted with dangling backward links

Solution

1. Step 1: Identify pointer updates needed on deletion

   When deleting a node, the previous node's next and the next node's prev pointers must be updated to bypass the deleted node.

2. Step 2: Consequence of missing prev update

   If the next node's prev pointer is not updated, it still points to the deleted node, causing backward links to be incorrect and list corruption.

3. Final Answer:

   The list may become corrupted with dangling backward links -> Option D

4. Quick Check:

   Missing prev update breaks backward links [OK]

Hint: Always update both prev and next pointers when deleting [OK]

Common Mistakes:

• Updating only next pointer
• Not breaking loop after deletion
• Forgetting to handle head or tail cases

5. You have a doubly linked list with nodes: 2 -> 4 -> 6 -> 4 -> 8 -> null. You want to delete all nodes with value 4 using a delete_by_value function that removes only the first matching node per call. How many times must you call delete_by_value(4) to remove all 4s, and what will the final list look like?

hard

A. Call once; final list: 2 -> 4 -> 6 -> 8 -> null

B. Call once; final list: 2 -> 6 -> 4 -> 8 -> null

C. Call twice; final list: 2 -> 6 -> 8 -> null

D. Call twice; final list: 2 -> 4 -> 6 -> 8 -> null

Solution

1. Step 1: Understand deletion per call

   Each call to delete_by_value(4) removes only the first node with value 4.

2. Step 2: Count how many 4s and calls needed

   There are two nodes with value 4. So, two calls are needed to remove both.

3. Step 3: Final list after deletions

   After two deletions, the list becomes 2 -> 6 -> 8 -> null.

4. Final Answer:

   Call twice; final list: 2 -> 6 -> 8 -> null -> Option C

5. Quick Check:

   Two 4s, two deletes, final list without 4s [OK]

Hint: Delete one node per call; repeat to remove all matches [OK]

Common Mistakes:

• Assuming one call removes all matching nodes
• Not updating pointers after each deletion
• Confusing node values or order