Implement Stack Using Queue in DSA Python - Time & Space Complexity

Time Complexity: Implement Stack Using Queue

O(n)

Understanding Time Complexity

We want to understand how long it takes to perform stack operations when using a queue to build it.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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):
        return self.q.pop(0) if self.q else None

    def top(self):
        return self.q[0] if self.q else None

    def empty(self):
        return len(self.q) == 0
    

This code uses one queue to simulate a stack. The push operation rotates the queue to keep the newest element at the front.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop inside the push method that moves elements from front to back.
• How many times: It runs (n-1) times where n is the current number of elements in the queue.

How Execution Grows With Input

Each push moves almost all existing elements to the back of the queue to keep the newest element at front.

Input Size (n)	Approx. Operations in push
10	9 moves
100	99 moves
1000	999 moves

Pattern observation: The number of moves grows linearly as the stack grows.

Final Time Complexity

Time Complexity: O(n)

This means pushing an item takes longer as the stack gets bigger, roughly proportional to the number of items.

Common Mistake

[X] Wrong: "Push operation is always O(1) because we just add one item to the queue."

[OK] Correct: Because after adding, we rotate all other items to keep stack order, which takes O(n) time.

Interview Connect

Understanding how to build one data structure using another and analyzing its time helps you think flexibly and solve problems creatively.

Self-Check

"What if we used two queues instead of one? How would the time complexity of push and pop change?"

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