Queue Using Two Stacks in DSA Python - Time & Space Complexity

Time Complexity: Queue Using Two Stacks

O(1) amortized

Understanding Time Complexity

We want to understand how fast operations run when we build a queue using two stacks.

Specifically, how does the time to add or remove items grow as we add more items?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class QueueUsingStacks:
    def __init__(self):
        self.stack_in = []
        self.stack_out = []

    def enqueue(self, item):
        self.stack_in.append(item)

    def dequeue(self):
        if not self.stack_out:
            while self.stack_in:
                self.stack_out.append(self.stack_in.pop())
        return self.stack_out.pop() if self.stack_out else None

This code adds items to one stack and removes items from another, moving items only when needed.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving all items from stack_in to stack_out when stack_out is empty.
• How many times: This move happens only when stack_out is empty, not on every dequeue.

How Execution Grows With Input

Adding items is always quick, but removing items sometimes requires moving many items.

Input Size (n)	Approx. Operations
10	About 10 moves when stack_out empties
100	About 100 moves occasionally, but mostly 1 per dequeue
1000	About 1000 moves occasionally, but mostly 1 per dequeue

Pattern observation: Most dequeues are fast, but occasionally a slow step moves many items all at once.

Final Time Complexity

Time Complexity: O(1) amortized per operation

This means on average, each add or remove takes a constant amount of time, even if some removals take longer.

Common Mistake

[X] Wrong: "Each dequeue always takes time proportional to the number of items."

[OK] Correct: Because items are moved only once from stack_in to stack_out, so the cost spreads out over many dequeues.

Interview Connect

Understanding this helps you explain how clever data structures can make slow steps rare, keeping average speed fast.

Self-Check

What if we moved items from stack_in to stack_out on every dequeue, even if stack_out is not empty? How would the time complexity change?

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