Implement Stack Using Queue in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Push and Pop Operations on Stack Implemented Using One Queue

What is the output of the following code that uses one queue to implement a stack?

DSA Python

from collections import deque

class StackUsingQueue:
    def __init__(self):
        self.q = deque()

    def push(self, x):
        self.q.append(x)
        for _ in range(len(self.q) - 1):
            self.q.append(self.q.popleft())

    def pop(self):
        if self.q:
            return self.q.popleft()
        return None

stack = StackUsingQueue()
stack.push(10)
stack.push(20)
stack.push(30)
print(stack.pop())
print(stack.pop())
print(stack.pop())

A 30\n10\n20

B 10\n20\n30

C 30\n20\n10

D 20\n30\n10

💡 Hint

Think about how the queue is rotated after each push to simulate stack behavior.

The push method rotates the queue so the last pushed element is always at the front, making pop remove the last pushed element first, mimicking stack LIFO behavior.

🧠 Conceptual

intermediate

Understanding Time Complexity of Stack Operations Using Two Queues

If a stack is implemented using two queues where push operation is costly and pop operation is cheap, what is the time complexity of push and pop respectively?

A Push: O(n), Pop: O(1)

B Push: O(n), Pop: O(n)

C Push: O(1), Pop: O(n)

D Push: O(1), Pop: O(1)

💡 Hint

Consider which operation involves moving all elements between queues.

In the costly push approach, push moves all elements to the other queue to maintain order, taking O(n). Pop just dequeues from the front, O(1).

🔧 Debug

advanced

Identify the Error in Stack Using Queue Implementation

What error will the following code produce when calling pop on an empty stack implemented using one queue?

DSA Python

from collections import deque

class StackUsingQueue:
    def __init__(self):
        self.q = deque()

    def push(self, x):
        self.q.append(x)
        for _ in range(len(self.q) - 1):
            self.q.append(self.q.popleft())

    def pop(self):
        return self.q.popleft()

stack = StackUsingQueue()
print(stack.pop())

A AttributeError

B IndexError: pop from an empty deque

C None

D IndexError

💡 Hint

What happens when you pop from an empty deque?

Calling popleft() on an empty deque raises IndexError with message 'pop from an empty deque'.

❓ Predict Output

advanced

Output of Mixed Push and Pop Operations Using Two Queues

What is the output of the following code implementing stack with two queues where pop is costly?

DSA Python

from collections import deque

class StackUsingTwoQueues:
    def __init__(self):
        self.q1 = deque()
        self.q2 = deque()

    def push(self, x):
        self.q1.append(x)

    def pop(self):
        if not self.q1:
            return None
        while len(self.q1) > 1:
            self.q2.append(self.q1.popleft())
        res = self.q1.popleft()
        self.q1, self.q2 = self.q2, self.q1
        return res

stack = StackUsingTwoQueues()
stack.push(5)
stack.push(10)
print(stack.pop())
stack.push(15)
print(stack.pop())
print(stack.pop())

A 10\n15\n5

B 5\n10\n15

C 15\n10\n5

D 10\n5\n15

💡 Hint

Pop removes the last pushed element by moving all but last to the other queue.

Pop removes 10 first, then after pushing 15, pop removes 15, then finally 5 remains.

🧠 Conceptual

expert

Minimum Number of Queues to Implement a Stack with O(1) Amortized Push and Pop

What is the minimum number of standard FIFO queues required to implement a stack with both push and pop operations having O(1) amortized time complexity?

A Two queues

B One queue

C Three queues

D Impossible with any finite number of queues

💡 Hint

Consider the fundamental difference between FIFO and LIFO behaviors and the cost of rearranging elements.

It is proven that implementing a stack with both push and pop in O(1) amortized time using only FIFO queues is impossible due to their ordering constraints.

Practice

1. What is the main idea behind implementing a stack using a queue?

easy

A. Use a stack to implement queue operations

B. Use a queue to sort elements in ascending order

C. Use one or two queues to simulate stack operations like push and pop

D. Use two stacks to implement a queue

Solution

1. Step 1: Understand stack and queue differences

   A stack follows Last-In-First-Out (LIFO), while a queue follows First-In-First-Out (FIFO).

2. Step 2: Simulate stack using queue(s)

   By using one or two queues cleverly, we can reorder elements to behave like a stack for push and pop operations.

3. Final Answer:

   Use one or two queues to simulate stack operations like push and pop -> Option C

4. Quick Check:

   Stack via queue = simulate push/pop [OK]

Hint: Remember stack is LIFO, queue is FIFO; simulate LIFO with queue [OK]

Common Mistakes:

• Confusing stack and queue order
• Trying to sort queue instead of simulating stack
• Using stack to implement queue instead

2. Which of the following is the correct syntax to initialize an empty queue in Python using collections.deque?

easy

A. queue = deque()

B. queue = []

C. queue = Queue()

D. queue = queue()

Solution

1. Step 1: Recall Python queue initialization

   collections.deque() is the correct way to create a double-ended queue in Python.

2. Step 2: Check options

   Only queue = deque() uses deque() correctly; others are either list or undefined.

3. Final Answer:

   queue = deque() -> Option A

4. Quick Check:

   collections.deque() creates queue [OK]

Hint: Use deque() from collections for queue in Python [OK]

Common Mistakes:

• Using list [] instead of deque
• Calling undefined Queue() class
• Using lowercase queue() which is not defined

3. What will be the output of the following code implementing push and pop using one queue?

from collections import deque

class Stack:
    def __init__(self):
        self.q = deque()

    def push(self, x):
        self.q.append(x)
        for _ in range(len(self.q) - 1):
            self.q.append(self.q.popleft())

    def pop(self):
        return self.q.popleft() if self.q else None

s = Stack()
s.push(10)
s.push(20)
s.push(30)
print(s.pop())
print(s.pop())
print(s.pop())

medium

A. 20\n30\n10

B. 10\n20\n30

C. 30\n10\n20

D. 30\n20\n10

Solution

1. Step 1: Understand push operation

   Push appends element then rotates queue so new element is at front.

2. Step 2: Trace push and pop calls

   Push 10: queue=[10]
   Push 20: queue=[20,10]
   Push 30: queue=[30,20,10]
   Pop removes front: 30, then 20, then 10.

3. Final Answer:

   30 20 10 -> Option D

4. Quick Check:

   Stack pop order = 30,20,10 [OK]

Hint: Push rotates queue to front; pop removes front [OK]

Common Mistakes:

• Ignoring rotation in push
• Assuming pop removes last element instead of front
• Confusing queue order with stack order

4. Find the error in this stack implementation using two queues:

from collections import deque

class Stack:
    def __init__(self):
        self.q1 = deque()
        self.q2 = deque()

    def push(self, x):
        self.q1.append(x)

    def pop(self):
        while len(self.q1) > 1:
            self.q2.append(self.q1.popleft())
        return self.q1.popleft()
        self.q1, self.q2 = self.q2, self.q1

medium

A. The queues are not swapped after pop, so q1 becomes empty

B. The pop method does not remove the last element

C. The push method should append to q2 instead of q1

D. The pop method should return from q2 instead of q1

Solution

1. Step 1: Analyze pop method flow

   Pop moves all but last from q1 to q2, then pops last from q1.

2. Step 2: Identify missing swap

   After popping, q1 is empty but queues are not swapped, so next push/pop fail.

3. Final Answer:

   The queues are not swapped after pop, so q1 becomes empty -> Option A

4. Quick Check:

   Missing swap causes empty q1 [OK]

Hint: Swap queues after pop to keep q1 as main queue [OK]

Common Mistakes:

• Not swapping queues after pop
• Returning before swapping queues
• Appending to wrong queue in push

5. You want to implement a stack using a single queue with minimal operations. Which approach correctly achieves this for the push operation to maintain stack order?

hard

A. Use two queues to enqueue the new element in the empty queue

B. Append the new element, then rotate the queue by dequeuing and enqueuing all previous elements

C. Dequeue all elements, enqueue the new element, then enqueue all dequeued elements

D. Append the new element and do nothing else

Solution

1. Step 1: Understand stack order with one queue

   To simulate LIFO, after adding new element, rotate queue so new element is front.

2. Step 2: Compare approaches

   Append the new element, then rotate the queue by dequeuing and enqueuing all previous elements rotates queue after append, maintaining stack order. Append the new element and do nothing else does not reorder. Dequeue all elements, enqueue the new element, then enqueue all dequeued elements is inefficient but correct logically. Use two queues to enqueue the new element in the empty queue uses two queues, not single.

3. Final Answer:

   Append the new element, then rotate the queue by dequeuing and enqueuing all previous elements -> Option B

4. Quick Check:

   Rotate after push keeps stack order [OK]

Hint: Rotate queue after push to keep new element on top [OK]

Common Mistakes:

• Not rotating queue after push
• Using two queues when only one is allowed
• Appending without reordering elements