Dequeue Operation in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Dequeue Pop Front Operation

What is the printed state of the deque after popping an element from the front?

DSA Rust

use std::collections::VecDeque;

fn main() {
    let mut dq: VecDeque<i32> = VecDeque::new();
    dq.push_back(10);
    dq.push_back(20);
    dq.push_back(30);
    dq.pop_front();
    for val in dq.iter() {
        print!("{} -> ", val);
    }
    print!("null");
}

A 30 -> null

B 10 -> 20 -> 30 -> null

C 20 -> 30 -> null

D 10 -> 30 -> null

💡 Hint

Remember that pop_front removes the first element inserted.

The deque initially has 10, 20, 30. Popping front removes 10, leaving 20 and 30.

❓ Predict Output

intermediate

Output after Multiple Dequeue Operations

What is the output after popping one element from the back and one from the front of the deque?

DSA Rust

use std::collections::VecDeque;

fn main() {
    let mut dq: VecDeque<i32> = VecDeque::new();
    dq.push_back(5);
    dq.push_back(15);
    dq.push_back(25);
    dq.pop_back();
    dq.pop_front();
    for val in dq.iter() {
        print!("{} -> ", val);
    }
    print!("null");
}

A 5 -> 15 -> null

B 15 -> null

C 5 -> null

D null

💡 Hint

Pop_back removes last, pop_front removes first.

Initial deque: 5, 15, 25. pop_back removes 25, pop_front removes 5, leaving 15.

🔧 Debug

advanced

Identify the Error in Dequeue Pop Operation

What error does this code produce when trying to pop from an empty deque?

DSA Rust

use std::collections::VecDeque;

fn main() {
    let mut dq: VecDeque<i32> = VecDeque::new();
    dq.pop_front().unwrap();
}

A SyntaxError: missing semicolon

B TypeError: cannot pop from VecDeque

C No error, prints null

D panic at 'called `Option::unwrap()` on a `None` value'

💡 Hint

unwrap on None causes panic.

pop_front returns None on empty deque. unwrap on None causes panic runtime error.

🧠 Conceptual

advanced

Result of Popping All Elements from Deque

If you pop all elements from a deque one by one from the front, what will be the state of the deque?

A Empty deque with length 0

B Deque with last element remaining

C Deque unchanged

D Deque with first element remaining

💡 Hint

Popping all elements removes everything.

Popping all elements from front removes every item, leaving an empty deque.

🚀 Application

expert

Final Deque State After Mixed Operations

Given the sequence of operations, what is the final printed state of the deque? Operations: 1. push_back(1) 2. push_back(2) 3. push_front(3) 4. pop_back() 5. push_front(4) 6. pop_front()

DSA Rust

use std::collections::VecDeque;

fn main() {
    let mut dq: VecDeque<i32> = VecDeque::new();
    dq.push_back(1);
    dq.push_back(2);
    dq.push_front(3);
    dq.pop_back();
    dq.push_front(4);
    dq.pop_front();
    for val in dq.iter() {
        print!("{} -> ", val);
    }
    print!("null");
}

A 3 -> 1 -> null

B 4 -> 3 -> 1 -> null

C 1 -> null

D 4 -> 1 -> null

💡 Hint

Track each operation step-by-step.

After all operations, deque contains 3 and 1 in order.

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