Queue Using Two Stacks in DSA Python - Build from Scratch

Queue Using Two Stacks

📖 Scenario: Imagine you are managing a line of customers waiting to buy tickets. You want to keep track of the order they arrive and serve them in the same order. But you only have two stacks (like two piles of plates) to help you manage this line.

🎯 Goal: Build a queue using two stacks. You will add customers to the line and serve them in the order they arrived, using only two stacks.

📋 What You'll Learn

Create two empty stacks named stack_in and stack_out

Create a variable queue_size to keep track of the number of customers

Write a function enqueue to add a customer to stack_in and update queue_size

Write a function dequeue to remove a customer from stack_out, transferring from stack_in if needed, and update queue_size

Print the state of stack_in and stack_out after operations

💡 Why This Matters

🌍 Real World

Queues are used in many places like customer service lines, printer job management, and task scheduling. Using two stacks to build a queue is a common programming technique to understand data structure transformations.

💼 Career

Understanding how to implement one data structure using others shows problem-solving skills and knowledge of data structures, which is important for software development and technical interviews.

1

Create two empty stacks

Create two empty lists called stack_in and stack_out to represent the two stacks.

DSA Python

# Your code here

Hint

Use empty lists to represent stacks in Python.

2

Create a variable to track queue size

Create a variable called queue_size and set it to 0 to track how many customers are in the queue.

DSA Python

stack_in = []
stack_out = []
# Your code here

Hint

Start with zero customers in the queue.

3

Write enqueue and dequeue functions

Write a function called enqueue that takes a parameter customer, adds it to stack_in, and increases queue_size by 1. Then write a function called dequeue that removes and returns the oldest customer from the queue by using stack_out. If stack_out is empty, move all items from stack_in to stack_out first. Decrease queue_size by 1 when a customer is removed.

DSA Python

stack_in = []
stack_out = []
queue_size = 0
# Your code here

Hint

Remember to move all items from stack_in to stack_out only when stack_out is empty before dequeuing.

4

Print the stacks after enqueue and dequeue

Call enqueue three times with customers 'Alice', 'Bob', and 'Charlie'. Then call dequeue once. Finally, print the contents of stack_in and stack_out to show the current state of the queue.

DSA Python

stack_in = []
stack_out = []
queue_size = 0

def enqueue(customer):
    global queue_size
    stack_in.append(customer)
    queue_size += 1

def dequeue():
    global queue_size
    if not stack_out:
        while stack_in:
            stack_out.append(stack_in.pop())
    if stack_out:
        queue_size -= 1
        return stack_out.pop()
    else:
        return None

# Your code here

Hint

After dequeuing once, stack_out should have the remaining customers in reverse order, and stack_in should be empty.

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