Queue Using Two Stacks in DSA Python - Execution Trace

Concept Flow - Queue Using Two Stacks

Enqueue Operation

↓

Push element onto Stack1

↓

Done

↓

Dequeue Operation

↓

Is Stack2 empty?

No→Pop element from Stack2

Yes↓

Transfer all elements from Stack1 to Stack2

↓

Pop element from Stack2

↓

Done

Enqueue adds elements to Stack1. Dequeue removes from Stack2. If Stack2 is empty, elements from Stack1 are moved to Stack2 to reverse order.

Execution Sample

DSA Python

stack1 = []
stack2 = []

# Enqueue 1
stack1.append(1)

# Enqueue 2
stack1.append(2)

# Dequeue
if not stack2:
    while stack1:
        stack2.append(stack1.pop())
stack2.pop()

This code enqueues 1 and 2, then dequeues one element using two stacks.

Execution Table

Step	Operation	Stack1 State	Stack2 State	Action Description	Visual State
1	Enqueue 1	[]	[]	Push 1 onto Stack1	Stack1: [1], Stack2: []
2	Enqueue 2	[1]	[]	Push 2 onto Stack1	Stack1: [1, 2], Stack2: []
3	Dequeue check Stack2 empty	[1, 2]	[]	Stack2 empty, transfer all from Stack1	Stack1: [], Stack2: [2, 1]
4	Dequeue pop from Stack2	[]	[2, 1]	Pop top element (1) from Stack2	Stack1: [], Stack2: [2]
5	End	[]	[2]	Dequeue complete, returned 1	Stack1: [], Stack2: [2]

💡 Dequeue stops after popping from Stack2 which now has elements reversed from Stack1

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
stack1	[]	[1]	[1, 2]	[]	[]	[]
stack2	[]	[]	[]	[2, 1]	[2]	[2]

Key Moments - 3 Insights

Why do we transfer all elements from stack1 to stack2 during dequeue?

Because stack1 stores elements in enqueue order (newest on top), transferring to stack2 reverses order so oldest element is on top for dequeue, as shown in step 3 of execution_table.

What happens if we try to dequeue when both stacks are empty?

There is no element to dequeue; the code would not pop from stack2 because it is empty. This is not shown in the table but is important to check before dequeue.

Why don't we pop directly from stack1 during dequeue?

Popping directly from stack1 would remove the newest element, not the oldest. Using stack2 after transfer ensures FIFO order, as seen in step 4 where stack2.pop() returns the oldest element.

Visual Quiz - 3 Questions

Look at the execution_table, what is the state of stack2 after step 3?

A [2, 1]

B [1, 2]

C []

D [1]

💡 Hint

Refer to the 'Stack2 State' column in row for step 3 in execution_table.

At which step does the oldest element get removed from the queue?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Check the 'Action Description' column for when pop from Stack2 happens.

If we enqueue another element 3 after step 5, what will be the state of stack1?

A [2, 3]

B [3]

C []

D [1, 3]

💡 Hint

Look at variable_tracker for stack1 final state and recall enqueue pushes onto stack1.

Concept Snapshot

Queue Using Two Stacks:
- Enqueue: push element onto stack1
- Dequeue: if stack2 empty, move all from stack1 to stack2
- Pop from stack2 to dequeue
- This reverses order to maintain FIFO
- Efficient amortized operations

Full Transcript

This visualization shows how a queue can be implemented using two stacks. Enqueue operations push elements onto stack1. Dequeue operations pop elements from stack2. If stack2 is empty during dequeue, all elements from stack1 are moved to stack2, reversing their order so the oldest element is on top. This ensures the queue behaves in FIFO order. The execution table traces enqueueing 1 and 2, then dequeuing one element, showing stack states and actions at each step. Variable tracker shows how stack1 and stack2 change after each operation. Key moments clarify why elements are transferred and why popping directly from stack1 would break queue order. The quiz tests understanding of stack states and operation steps. This method efficiently simulates a queue using two stacks.

Practice

1. What is the main idea behind implementing a Queue using two stacks?

easy

A. Use two stacks to reverse the order twice to maintain FIFO

B. Use two stacks to store elements in sorted order

C. Use one stack for enqueue and another for dequeue operations

D. Use one stack and a queue together

Solution

1. Step 1: Understand queue order

   A queue follows FIFO (First In First Out), meaning the first element added is the first removed.

2. Step 2: Use two stacks to reverse order twice

   One stack reverses the order when pushing elements, and the second stack reverses it back when popping, preserving FIFO.

3. Final Answer:

   Use two stacks to reverse the order twice to maintain FIFO -> Option A

4. Quick Check:

   Two stacks reverse order twice = FIFO queue [OK]

Hint: Two reversals with stacks simulate queue order [OK]

Common Mistakes:

• Thinking one stack alone can maintain FIFO
• Confusing stack sorting with queue behavior
• Using a queue inside the stack implementation

2. Which of the following is the correct syntax to push an element x onto a stack named stack1 in Python?

easy

A. stack1.push(x)

B. stack1.append(x)

C. stack1.insert(x)

D. stack1.add(x)

Solution

1. Step 1: Recall Python list methods for stack

   Python lists use append() to add elements at the end, simulating stack push.

2. Step 2: Check other options

   push() and add() are not list methods; insert() requires an index.

3. Final Answer:

   stack1.append(x) -> Option B

4. Quick Check:

   Python stack push = append() [OK]

Hint: Use append() to push on Python list stack [OK]

Common Mistakes:

• Using push() which is not a Python list method
• Using insert() without index argument
• Confusing add() with set method

3. What will be the output of the following code implementing dequeue operation in a queue using two stacks?

stack1 = [1, 2, 3]
stack2 = []

if not stack2:
    while stack1:
        stack2.append(stack1.pop())

print(stack2.pop())

medium

A. 1

B. 3

C. 2

D. IndexError

Solution

1. Step 1: Transfer elements from stack1 to stack2

   Elements pop from stack1 in order 3, 2, 1 and append to stack2, so stack2 becomes [3, 2, 1].

2. Step 2: Pop from stack2

   Popping from stack2 removes last element, which is 1, the first element enqueued.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   Two stacks reverse order, first out is 1 [OK]

Hint: Pop from second stack after transfer gives queue front [OK]

Common Mistakes:

• Assuming pop from stack1 directly returns front
• Confusing order after transfer
• Expecting 3 instead of 1

4. Identify the error in the following dequeue method of a queue implemented using two stacks:

def dequeue():
    if stack2.is_empty():
        while stack1:
            stack2.append(stack1.pop())
    return stack1.pop()

medium

A. Using stack2.is_empty() instead of checking if stack2 is empty

B. Using append() instead of push() on stack2

C. Popping from stack1 instead of stack2 after transfer

D. Not checking if both stacks are empty before popping

Solution

1. Step 1: Check stack transfer logic

   Elements are moved from stack1 to stack2 when stack2 is empty to reverse order.

2. Step 2: Identify pop operation error

   After transfer, popping should be from stack2 to get the correct queue front, but code pops from stack1 incorrectly.

3. Final Answer:

   Popping from stack1 instead of stack2 after transfer -> Option C

4. Quick Check:

   Pop from stack2 after transfer, not stack1 [OK]

Hint: Pop from stack2 after transfer, not stack1 [OK]

Common Mistakes:

• Popping from wrong stack after transfer
• Using non-existent is_empty() method on list
• Ignoring empty queue check

5. You want to implement a queue using two stacks where enqueue and dequeue operations are both efficient. Which approach below achieves amortized O(1) time per operation?

hard

A. Push elements to stack1; on dequeue, pop directly from stack1

B. Push elements alternately to stack1 and stack2; pop from the other stack

C. Push elements to stack2; on dequeue, pop from stack1

D. Push all elements to stack1 on enqueue; on dequeue, transfer all to stack2 if empty, then pop from stack2

Solution

1. Step 1: Understand enqueue operation

   All new elements are pushed onto stack1, which is fast (O(1)).

2. Step 2: Understand dequeue operation

   When stack2 is empty, transfer all elements from stack1 to stack2 (reversing order), then pop from stack2. This transfer happens only when needed, spreading cost.

3. Step 3: Analyze amortized time

   Each element moves at most twice (push to stack1, move to stack2), so average cost per operation is O(1).

4. Final Answer:

   Push all elements to stack1 on enqueue; on dequeue, transfer all to stack2 if empty, then pop from stack2 -> Option D

5. Quick Check:

   Amortized O(1) by lazy transfer from stack1 to stack2 [OK]

Hint: Transfer only when needed for amortized O(1) dequeue [OK]

Common Mistakes:

• Transferring elements on every dequeue causing O(n) each time
• Popping directly from stack1 breaking queue order
• Alternating pushes causing confusion