Peek Front Element of Queue in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of peeking the front element after enqueue operations?

Consider a queue where you enqueue the elements 10, 20, and 30 in that order. What will be the front element after these operations?

DSA Python

from collections import deque

queue = deque()
queue.append(10)
queue.append(20)
queue.append(30)
front = queue[0]
print(front)

A 10

B 20

C 30

D IndexError

💡 Hint

Remember, the front element is the first one added that has not been removed.

In a queue, elements are added at the back and removed from the front. After adding 10, 20, and 30, the front is still 10 because it was the first added and not removed.

❓ Predict Output

intermediate

What happens when peeking an empty queue?

What will happen if you try to peek the front element of an empty queue implemented using deque?

DSA Python

from collections import deque

queue = deque()
front = queue[0]
print(front)

A 0

B None

C IndexError

D AttributeError

💡 Hint

Think about what happens when you access an index that does not exist in a list or deque.

Trying to access queue[0] when the queue is empty raises an IndexError because there is no element at index 0.

🔧 Debug

advanced

Identify the error when peeking front element in this queue implementation

This code tries to peek the front element of a queue implemented as a list. What error will it produce?

DSA Python

queue = []
queue.append(5)
queue.append(10)
front = queue.pop(0)
print(front)
print(queue[0])

A No error, prints 5 then 5

B IndexError on print(queue[0])

C TypeError on queue.pop(0)

D No error, prints 5 then 10

💡 Hint

Check what happens to the queue after popping the front element and then accessing queue[0].

The first pop(0) removes 5, so front is 5. Then print(queue[0]) tries to print the new front, which is 10. But if the queue had only one element, accessing queue[0] after pop(0) would cause IndexError. Here, since queue has 10 after pop, no error occurs. So the correct answer is no error and prints 5 then 10.

❓ Predict Output

advanced

What is the output after multiple enqueue and dequeue operations?

Given the following operations on a queue, what will be printed as the front element?

DSA Python

from collections import deque

queue = deque()
queue.append(1)
queue.append(2)
queue.append(3)
queue.popleft()
queue.append(4)
front = queue[0]
print(front)

A 2

B 1

C 3

D 4

💡 Hint

Remember that popleft removes the front element.

Initially queue is [1,2,3]. popleft removes 1, so queue becomes [2,3]. Then 4 is appended, queue is [2,3,4]. The front element is queue[0] which is 2.

🧠 Conceptual

expert

Which data structure property ensures peek front is O(1)?

Why does a queue implemented with a linked list or deque allow peeking the front element in constant time?

A Because the queue uses hashing for quick access

B Because the front element is directly referenced without traversal

C Because the queue elements are stored in contiguous memory

D Because the queue stores elements in sorted order

💡 Hint

Think about how the front element is accessed in these data structures.

In linked lists and deques, the front element is directly accessible via a pointer or index, so peeking does not require traversing the structure, making it O(1).

Practice

1. What does the peek operation do in a queue?

easy

A. Returns the front element without removing it

B. Removes the front element from the queue

C. Adds an element to the back of the queue

D. Checks if the queue is empty

Solution

1. Step 1: Understand the peek operation

   The peek operation looks at the front element of the queue without changing the queue.

2. Step 2: Compare with other operations

   Unlike dequeue, peek does not remove the element; it only returns it.

3. Final Answer:

   Returns the front element without removing it -> Option A

4. Quick Check:

   Peek = front element without removal [OK]

Hint: Peek shows front without removing it [OK]

Common Mistakes:

• Confusing peek with dequeue
• Thinking peek adds elements
• Assuming peek checks emptiness

2. Which of the following is the correct syntax to peek the front element of a queue named q in Python using collections.deque?

easy

A. front = q[0]

B. front = q.peek()

C. front = q.front()

D. front = q.get()

Solution

1. Step 1: Recall deque indexing

   In Python's deque, the front element is accessed by index 0, like q[0].

2. Step 2: Check other options

   Methods peek(), front(), and get() do not exist for deque.

3. Final Answer:

   front = q[0] -> Option A

4. Quick Check:

   deque front element = q[0] [OK]

Hint: Use q[0] to peek front in deque [OK]

Common Mistakes:

• Using non-existent peek() method
• Trying q.front() which is invalid
• Using q.get() which is for other structures

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

from collections import deque
q = deque([10, 20, 30])
print(q[0])
q.popleft()
print(q[0])

medium

A. 30\n20

B. 10\n10

C. 20\n30

D. 10\n20

Solution

1. Step 1: Peek front before removal

   Initially, q = [10, 20, 30], so q[0] is 10.

2. Step 2: Remove front and peek again

   After q.popleft(), q becomes [20, 30], so q[0] is now 20.

3. Final Answer:

   10\n20 -> Option D

4. Quick Check:

   First peek 10, after removal peek 20 [OK]

Hint: Peek before and after popleft to see front change [OK]

Common Mistakes:

• Assuming front stays same after popleft
• Confusing popleft with pop
• Printing wrong elements

4. Identify the error in the following code that tries to peek the front element of a queue:

from queue import Queue
q = Queue()
q.put(5)
q.put(10)
print(q[0])

medium

A. print statement syntax error

B. Queue is empty, cannot peek

C. Queue object does not support indexing

D. put() method is incorrect

Solution

1. Step 1: Understand Queue class behavior

   Python's queue.Queue does not support direct indexing like q[0].

2. Step 2: Identify correct peek method

   To peek, one must use q.queue[0] or other methods, but direct q[0] causes error.

3. Final Answer:

   Queue object does not support indexing -> Option C

4. Quick Check:

   queue.Queue no indexing [OK]

Hint: queue.Queue objects can't use q[0] indexing [OK]

Common Mistakes:

• Trying to index queue.Queue directly
• Assuming put() is wrong
• Thinking queue is empty

5. You have a queue implemented using a list in Python: q = [5, 10, 15]. You want to peek the front element safely without causing an error if the queue is empty. Which code snippet correctly does this?

hard

A. front = q.pop(0)

B. front = q[0] if len(q) > 0 else None

C. front = q[-1]

D. front = q.get()

Solution

1. Step 1: Check for empty queue before peeking

   Accessing q[0] when q is empty causes error, so check length first.

2. Step 2: Understand other options

   pop(0) removes element, not peek; q[-1] accesses last element; q.get() is invalid for list.

3. Final Answer:

   front = q[0] if len(q) > 0 else None -> Option B

4. Quick Check:

   Safe peek with length check [OK]

Hint: Check length before accessing q[0] to avoid errors [OK]

Common Mistakes:

• Using pop(0) which removes element
• Accessing q[0] without checking empty
• Using q.get() on list