Double Ended Queue Deque in DSA Rust - Build from Scratch

Double Ended Queue (Deque) in Rust

📖 Scenario: Imagine you are managing a line of customers waiting for service. Sometimes customers join the line at the front, sometimes at the back. You also serve customers from both ends depending on priority. This is like a double ended queue, called a deque.

🎯 Goal: You will build a simple deque using Rust's VecDeque. You will add customers to the front and back, remove from front and back, and then print the final state of the deque.

📋 What You'll Learn

Create a VecDeque named customers with initial customers

Add a customer to the front and another to the back

Remove one customer from the front and one from the back

Print the final state of customers showing all remaining customers

💡 Why This Matters

🌍 Real World

Deques are useful in real-world scenarios like managing lines where people can join or leave from both ends, such as ticket counters or call centers.

💼 Career

Understanding deques helps in jobs involving system design, real-time processing, and implementing efficient data structures for queues and stacks.

1

Create the initial deque

Create a VecDeque called customers with these exact strings in order: "Alice", "Bob", "Charlie"

DSA Rust

use std::collections::VecDeque;

// Create a VecDeque named customers with "Alice", "Bob", "Charlie"
// Your code here

Hint

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

2

Add customers to front and back

Add "Diana" to the front of customers using push_front and add "Evan" to the back using push_back

DSA Rust

use std::collections::VecDeque;

fn main() {
    let mut customers: VecDeque = VecDeque::from(vec!["Alice", "Bob", "Charlie"]);
    
    // Add "Diana" to front and "Evan" to back
    // Your code here
}

Hint

Use push_front and push_back methods on customers.

3

Remove customers from front and back

Remove one customer from the front of customers using pop_front and one from the back using pop_back. Store the removed values in variables front_removed and back_removed respectively

DSA Rust

use std::collections::VecDeque;

fn main() {
    let mut customers: VecDeque = VecDeque::from(vec!["Alice", "Bob", "Charlie"]);
    customers.push_front("Diana");
    customers.push_back("Evan");

    // Remove one from front and one from back
    // Your code here
}

Hint

Use pop_front and pop_back and assign results to front_removed and back_removed.

4

Print the final deque state

Print the final state of customers using println! with the exact format: "Final customers: {:?}"

DSA Rust

use std::collections::VecDeque;

fn main() {
    let mut customers: VecDeque = VecDeque::from(vec!["Alice", "Bob", "Charlie"]);
    customers.push_front("Diana");
    customers.push_back("Evan");

    let front_removed = customers.pop_front();
    let back_removed = customers.pop_back();

    // Print final customers
    // Your code here
}

Hint

Use println!("Final customers: {:?}", customers); to print the deque.

Practice

1. What is a deque in data structures?

easy

A. A double-ended queue allowing insertion and removal from both ends

B. A queue that only allows insertion at the front

C. A stack that only allows removal from the top

D. A list that only supports random access

Solution

1. Step 1: Understand the deque concept

   A deque is a data structure where elements can be added or removed from both the front and the back.

2. Step 2: Compare options with deque definition

   Only double-ended queue allowing insertion and removal from both ends correctly describes this behavior; others describe different structures.

3. Final Answer:

   A double-ended queue allowing insertion and removal from both ends -> Option A

4. Quick Check:

   Deque = double-ended queue [OK]

Hint: Deque means double-ended queue, both ends usable [OK]

Common Mistakes:

• Confusing deque with stack or regular queue
• Thinking deque only allows insertion at one end
• Assuming deque is a fixed-size array

2. Which of the following is the correct way to create an empty VecDeque in Rust?

easy

A. let dq = VecDeque::empty();

B. let dq = VecDeque();

C. let dq = new VecDeque();

D. let dq = VecDeque::new();

Solution

1. Step 1: Recall Rust VecDeque creation syntax

   In Rust, the standard way to create an empty VecDeque is by calling VecDeque::new().

2. Step 2: Check each option for Rust syntax correctness

   let dq = VecDeque::new(); uses correct Rust syntax. The other options are invalid Rust syntax for VecDeque creation.

3. Final Answer:

   let dq = VecDeque::new(); -> Option D

4. Quick Check:

   VecDeque::new() creates empty deque [OK]

Hint: Use VecDeque::new() to create empty deque in Rust [OK]

Common Mistakes:

• Using constructor syntax like VecDeque() which is invalid in Rust
• Trying to call new as a method without ::
• Using non-existent empty() method

3. What will be the output of the following Rust code?

use std::collections::VecDeque;

fn main() {
    let mut dq = VecDeque::new();
    dq.push_back(10);
    dq.push_front(20);
    dq.push_back(30);
    dq.pop_front();
    println!("{:?}", dq);
}

medium

A. [10, 30]

B. [20, 10, 30]

C. [30, 10]

D. [10, 20, 30]

Solution

1. Step 1: Trace deque operations step-by-step

   Initially empty deque: [] - push_back(10): [10] - push_front(20): [20, 10] - push_back(30): [20, 10, 30] - pop_front(): removes 20, resulting in [10, 30]

2. Step 2: Confirm final deque state

   After all operations, the deque contains [10, 30].

3. Final Answer:

   [10, 30] -> Option A

4. Quick Check:

   After pop_front, deque = [10, 30] [OK]

Hint: Track push/pop order carefully front and back [OK]

Common Mistakes:

• Forgetting pop_front removes front element
• Mixing order of push_front and push_back
• Assuming pop_front removes back element

4. Identify the error in the following Rust code using VecDeque:

use std::collections::VecDeque;

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

medium

A. pop_back() does not return a value

B. Cannot print Option<i32> directly with {}

C. VecDeque does not have push_front method

D. dq must be immutable to pop elements

Solution

1. Step 1: Understand pop_back return type

   pop_back() returns an Option containing the popped value or None if empty.

2. Step 2: Check println! usage

   Using {} tries to print the Option<i32> directly, which causes a type mismatch error. {:?} or unwrap() is needed.

3. Final Answer:

   Cannot print Option<i32> directly with {} -> Option B

4. Quick Check:

   Option<i32> needs {:?} or unwrap() to print [OK]

Hint: Use {:?} or unwrap() to print Option values [OK]

Common Mistakes:

• Trying to print Option with {} instead of {:?}
• Assuming pop_back returns value directly, not Option
• Thinking VecDeque lacks push_front method

5. You want to implement a sliding window maximum using a VecDeque in Rust. Which approach correctly maintains the deque to always have the maximum at the front while sliding the window?

hard

A. Remove elements from front smaller than new element before pushing it to front

B. Always push new element to front without removing any elements

C. Remove elements from back smaller than new element before pushing it to back

D. Remove elements from back larger than new element before pushing it to back

Solution

1. Step 1: Understand sliding window max logic with deque

   To keep max at front, remove all smaller elements from back before adding new element at back.

2. Step 2: Analyze options for correct deque maintenance

   Remove elements from back smaller than new element before pushing it to back correctly describes removing smaller elements from back before pushing new element to back. Others do not maintain max at front properly.

3. Final Answer:

   Remove elements from back smaller than new element before pushing it to back -> Option C

4. Quick Check:

   Sliding window max uses back removals of smaller elements [OK]

Hint: Pop smaller from back before push_back new max candidate [OK]

Common Mistakes:

• Removing from front instead of back
• Pushing new elements to front incorrectly
• Removing larger elements instead of smaller