Implement Stack Using Queue in DSA Python - Execution Trace

Concept Flow - Implement Stack Using Queue

Push operation

↓

Enqueue element in queue

↓

Rotate queue to put new element at front

↓

Top element is at front of queue

↓

Pop operation

↓

Dequeue front element from queue

↓

Stack size decreases

↓

Repeat for next operations

Push adds element by enqueue and rotates queue to keep stack order; pop removes front element simulating stack pop.

Execution Sample

DSA Python

class StackUsingQueue:
    def __init__(self):
        self.q = []
    def push(self, x):
        self.q.append(x)
        for _ in range(len(self.q)-1):
            self.q.append(self.q.pop(0))
    def pop(self):
        if self.q:
            return self.q.pop(0)
        else:
            return None

This code pushes elements onto a stack using a single queue by rotating elements.

Execution Table

Step	Operation	Queue State Before	Action	Queue State After	Visual State
1	push(1)	[]	Enqueue 1, rotate 0 times	[1]	1 -> null
2	push(2)	[1]	Enqueue 2, rotate 1 time	[2, 1]	2 -> 1 -> null
3	push(3)	[2, 1]	Enqueue 3, rotate 2 times	[3, 2, 1]	3 -> 2 -> 1 -> null
4	pop()	[3, 2, 1]	Dequeue front (3)	[2, 1]	2 -> 1 -> null
5	pop()	[2, 1]	Dequeue front (2)	[1]	1 -> null
6	push(4)	[1]	Enqueue 4, rotate 1 time	[4, 1]	4 -> 1 -> null
7	pop()	[4, 1]	Dequeue front (4)	[1]	1 -> null
8	pop()	[1]	Dequeue front (1)	[]	empty
9	pop()	[]	Queue empty, cannot pop	[]	empty

💡 Execution stops after last pop attempt on empty queue.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	After Step 9
q (queue)	[]	[1]	[2, 1]	[3, 2, 1]	[2, 1]	[1]	[4, 1]	[1]	[]	[]
stack size	0	1	2	3	2	1	2	1	0	0

Key Moments - 3 Insights

Why do we rotate the queue after enqueue during push?

Rotating moves the newly added element to the front, so pop can remove it first, simulating stack's LIFO order (see steps 2 and 3 in execution_table).

What happens if we pop when the queue is empty?

Pop operation cannot remove any element and the queue remains empty, as shown in step 9 of execution_table.

Why does the queue state after push show the newest element at front?

Because after enqueue, we rotate the queue so the newest element moves to front, making it the next to pop (see steps 2, 3, and 6).

Visual Quiz - 3 Questions

Look at the execution_table, what is the queue state after step 3 (push(3))?

A [3, 2, 1]

B [2, 1, 3]

C [1, 2, 3]

D [3, 1, 2]

💡 Hint

Check the 'Queue State After' column for step 3 in execution_table.

At which step does the stack size become 0 according to variable_tracker?

A After Step 7

B After Step 8

C After Step 9

D After Step 6

💡 Hint

Look at the 'stack size' row in variable_tracker for when it reaches 0.

If we skip rotating the queue after enqueue in push, what would happen to the pop operation?

A Pop would remove the newest element correctly

B Queue would become empty immediately

C Pop would remove the oldest element, not the newest

D Push would fail to add elements

💡 Hint

Refer to key_moments about why rotation is needed after enqueue.

Concept Snapshot

Implement Stack Using Queue:
- Use one queue to simulate stack
- push(x): enqueue x, then rotate queue to move x to front
- pop(): dequeue front element
- Top element is always at front of queue
- Maintains LIFO order using queue operations

Full Transcript

This concept shows how to build a stack using a single queue. When pushing, we add the element to the queue and then rotate the queue so the new element moves to the front. This way, when we pop, we remove the front element, which is the last pushed element, simulating stack behavior. The execution table tracks each step showing queue states and visual linked list style representation. The variable tracker shows how the queue and stack size change after each operation. Key moments clarify why rotation is necessary and what happens when popping from an empty queue. The visual quiz tests understanding of queue states and stack size changes during execution.

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