How would you implement a queue using two stacks?

Explanation:

Implementing a queue using two stacks leverages the Last In, First Out (LIFO) nature of stacks to mimic the First In, First Out (FIFO) behavior of a queue. The main idea is to use one stack for enqueue operations and the other for dequeue operations. When you need to dequeue an element and the dequeue stack is empty, you transfer all elements from the enqueue stack to the dequeue stack, reversing their order in the process.

Key Talking Points:

• Enqueue Operation: Push the element onto the first stack.
• Dequeue Operation: If the second stack is empty, pop all elements from the first stack and push them onto the second stack. Then, pop the element from the second stack.
• Amortized Time Complexity:
  • Enqueue: O(1)
  • Dequeue: O(1) (amortized, due to potential transfer of elements between stacks)
• Space Complexity: O(n), where n is the number of elements in the queue.

NOTES:

Reference Table:

Pseudocode:

class QueueUsingTwoStacks:
    def __init__(self):
        self.stack1 = []
        self.stack2 = []

    def enqueue(self, item):
        self.stack1.append(item)

    def dequeue(self):
        if not self.stack2:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
        if not self.stack2:
            raise IndexError("Dequeue from empty queue")
        return self.stack2.pop()

# Example usage:
queue = QueueUsingTwoStacks()
queue.enqueue(1)
queue.enqueue(2)
print(queue.dequeue())  # Outputs: 1
print(queue.dequeue())  # Outputs: 2

Follow-Up Questions and Answers:

1. What are the advantages of using two stacks over a single stack for implementing a queue?

   • Using two stacks allows you to reverse the order of elements, enabling you to mimic the FIFO order of a queue while using the LIFO nature of stacks. This separation of operations between two stacks allows for efficient enqueue and dequeue operations.

2. Can you optimize the space usage further than O(n)?

   • The space complexity is already optimal at O(n), as you need to store all elements in the queue. However, you can minimize overhead by using dynamic arrays or linked lists for stack implementation.

3. How does this implementation compare to using a doubly linked list?

   • A doubly linked list allows O(1) time complexity for both enqueue and dequeue operations without the need to transfer elements between structures, but it may have higher overhead due to additional pointers for each node.