Dequeue Operation in DSA Rust - Build from Scratch

Dequeue Operation

📖 Scenario: You are managing a line of customers waiting for service. Sometimes customers leave from the front or the back of the line. You want to keep track of the line using a double-ended queue (deque) so you can add or remove customers from both ends easily.

🎯 Goal: Build a Rust program that creates a deque with some customers, removes one customer from the front, removes one customer from the back, and then shows the remaining customers in the line.

📋 What You'll Learn

Create a deque with the exact customers: "Alice", "Bob", "Charlie", "Diana", "Ethan"

Create a variable called line to hold the deque

Remove one customer from the front of the deque using pop_front()

Remove one customer from the back of the deque using pop_back()

Print the remaining customers in the deque separated by ->

💡 Why This Matters

🌍 Real World

Deques are useful in real-world situations like managing lines, undo-redo stacks, or sliding window problems where you need to add or remove items from both ends efficiently.

💼 Career

Understanding deque operations is important for software developers working on performance-critical applications, game development, or systems programming where efficient data structure manipulation is required.

1

Create the initial deque

Create a variable called line that is a VecDeque containing these exact customers in order: "Alice", "Bob", "Charlie", "Diana", "Ethan".

DSA Rust

use std::collections::VecDeque;

// Create the deque called line with the customers
// Your code here

Hint

Use VecDeque::from with a vec! of the exact customer names.

2

Remove one customer from the front

Use pop_front() on the line deque to remove one customer from the front. Store the removed customer in a variable called front_customer.

DSA Rust

use std::collections::VecDeque;

let mut line: VecDeque = VecDeque::from(vec!["Alice", "Bob", "Charlie", "Diana", "Ethan"]);

// Remove one customer from the front and store in front_customer
// Your code here

Hint

Call pop_front() on line and assign it to front_customer.

3

Remove one customer from the back

Use pop_back() on the line deque to remove one customer from the back. Store the removed customer in a variable called back_customer.

DSA Rust

use std::collections::VecDeque;

let mut line: VecDeque = VecDeque::from(vec!["Alice", "Bob", "Charlie", "Diana", "Ethan"]);

let front_customer = line.pop_front();

// Remove one customer from the back and store in back_customer
// Your code here

Hint

Call pop_back() on line and assign it to back_customer.

4

Print the remaining customers

Print the remaining customers in line separated by -> . Use println! with join on an iterator of line.

DSA Rust

use std::collections::VecDeque;

let mut line: VecDeque = VecDeque::from(vec!["Alice", "Bob", "Charlie", "Diana", "Ethan"]);

let front_customer = line.pop_front();
let back_customer = line.pop_back();

// Print the remaining customers separated by ' -> '
// Your code here

Hint

Use line.iter() to get an iterator, then cloned() to copy the &str, collect into a Vec<&str>, then join(" -> ") to make a string.

Practice

1. What does the pop_front() operation do in a dequeue?

easy

A. Removes and returns the element at the front of the dequeue

B. Removes and returns the element at the back of the dequeue

C. Adds an element to the front of the dequeue

D. Adds an element to the back of the dequeue

Solution

1. Step 1: Understand pop_front() operation

   The pop_front() removes the first element from the front of the dequeue.

2. Step 2: Compare with other operations

   Unlike pop_back() which removes from the back, pop_front() targets the front element.

3. Final Answer:

   Removes and returns the element at the front of the dequeue -> Option A

4. Quick Check:

   pop_front() = remove front element [OK]

Hint: pop_front() always removes from the front [OK]

Common Mistakes:

• Confusing pop_front() with pop_back()
• Thinking pop_front() adds elements
• Mixing push and pop operations

2. Which of the following is the correct Rust syntax to remove an element from the back of a VecDeque named dq?

easy

A. dq.pop_front();

B. dq.pop_back();

C. dq.remove_back();

D. dq.pop();

Solution

1. Step 1: Identify VecDeque method for back removal

   The correct method to remove from the back is pop_back().

2. Step 2: Check other options

   pop_front() removes from front, remove_back() and pop() are invalid methods for VecDeque.

3. Final Answer:

   dq.pop_back(); -> Option B

4. Quick Check:

   pop_back() removes back element [OK]

Hint: Use pop_back() to remove from back in VecDeque [OK]

Common Mistakes:

• Using pop() which is for Vec, not VecDeque
• Using remove_back() which does not exist
• Confusing pop_front() with pop_back()

3. What is the output of the following Rust code?

use std::collections::VecDeque;

fn main() {
    let mut dq = VecDeque::new();
    dq.push_back(10);
    dq.push_back(20);
    dq.push_front(5);
    dq.pop_back();
    dq.push_front(1);
    println!("{:?}", dq);
}

medium

A. [5, 10, 20]

B. [1, 10, 20]

C. [10, 20, 1]

D. [1, 5, 10]

Solution

1. Step 1: Trace push_back and push_front calls

   After push_back(10) and push_back(20), dq = [10, 20]. Then push_front(5) makes dq = [5, 10, 20].

2. Step 2: Apply pop_back and push_front

   pop_back() removes 20, dq = [5, 10]. Then push_front(1) makes dq = [1, 5, 10].

3. Final Answer:

   [1, 5, 10] -> Option D

4. Quick Check:

   Operations result in [1, 5, 10] [OK]

Hint: Track each operation step-by-step on the deque [OK]

Common Mistakes:

• Forgetting pop_back removes last element
• Mixing order of push_front and push_back
• Misreading the final print format

4. Identify the error in this Rust code that tries to remove an element from the front of a VecDeque named dq:

use std::collections::VecDeque;

fn main() {
    let mut dq = VecDeque::new();
    dq.push_back(1);
    dq.push_back(2);
    let front = dq.pop();
    println!("{:?}", front);
}

medium

A. VecDeque has no method named pop()

B. pop() removes from back, not front

C. dq is not mutable

D. push_back() is used incorrectly

Solution

1. Step 1: Check VecDeque methods

   VecDeque does not have a method named pop(). It uses pop_front() or pop_back().

2. Step 2: Identify the error cause

   Calling dq.pop() causes a compile-time error because the method does not exist.

3. Final Answer:

   VecDeque has no method named pop() -> Option A

4. Quick Check:

   pop() method missing in VecDeque [OK]

Hint: Use pop_front() or pop_back() for VecDeque removals [OK]

Common Mistakes:

• Using pop() from Vec on VecDeque
• Ignoring compiler error messages
• Assuming pop() removes front element

5. Given a VecDeque with elements [3, 6, 9, 12], which sequence of operations will result in the dequeue [1, 3, 6]?

hard

A. dq.pop_back(); dq.pop_front(); dq.push_front(1);

B. dq.pop_front(); dq.pop_back(); dq.push_front(1);

C. dq.pop_back(); dq.pop_back(); dq.push_front(1);

D. dq.pop_front(); dq.pop_front(); dq.push_back(1);

Solution

1. Step 1: Start with dq = [3, 6, 9, 12]

   Initial dequeue has elements 3, 6, 9, 12 in order.

2. Step 2: Apply operations in dq.pop_back(); dq.pop_back(); dq.push_front(1);

   First pop_back() removes 12, dq = [3, 6, 9]. Second pop_back() removes 9, dq = [3, 6]. Then push_front(1) adds 1 at front, dq = [1, 3, 6].

3. Final Answer:

   dq.pop_back(); dq.pop_back(); dq.push_front(1); -> Option C

4. Quick Check:

   Operations produce [1, 3, 6] [OK]

Hint: Remove from back twice, then add 1 at front [OK]

Common Mistakes:

• Removing from front instead of back
• Adding 1 at back instead of front
• Wrong operation order