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What is the time complexity of implementing a queue using two stacks?

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Question: What is the time complexity of implementing a queue using two stacks?

Options:

  1. O(1) for enqueue, O(n) for dequeue
  2. O(n) for enqueue, O(1) for dequeue
  3. O(n) for both
  4. O(1) for both

Correct Answer: O(1) for enqueue, O(n) for dequeue

Solution: 

Implementing a queue using two stacks allows O(1) time for enqueue and O(n) time for dequeue, as elements may need to be transferred between stacks.

What is the time complexity of implementing a queue using two stacks?

Practice Questions

Q1
What is the time complexity of implementing a queue using two stacks?
  1. O(1) for enqueue, O(n) for dequeue
  2. O(n) for enqueue, O(1) for dequeue
  3. O(n) for both
  4. O(1) for both

Questions & Step-by-Step Solutions

What is the time complexity of implementing a queue using two stacks?
  • Step 1: Understand what a queue is. A queue is a data structure that follows the First In First Out (FIFO) principle, meaning the first element added is the first one to be removed.
  • Step 2: Understand what a stack is. A stack is a data structure that follows the Last In First Out (LIFO) principle, meaning the last element added is the first one to be removed.
  • Step 3: Learn how to implement a queue using two stacks. We will use one stack for enqueueing (adding elements) and another stack for dequeueing (removing elements).
  • Step 4: When we enqueue (add) an element, we simply push it onto the first stack. This operation takes O(1) time because it is a simple push operation.
  • Step 5: When we dequeue (remove) an element, we need to check if the second stack is empty. If it is empty, we pop all elements from the first stack and push them onto the second stack. This reverses the order of elements, making the oldest element on top of the second stack.
  • Step 6: If the second stack is not empty, we can simply pop the top element from it. The time taken to pop from a stack is O(1).
  • Step 7: The transfer of elements from the first stack to the second stack takes O(n) time, where n is the number of elements in the first stack. This is the worst-case scenario for dequeueing.
  • Step 8: Therefore, the time complexity for enqueueing is O(1) and for dequeueing is O(n) when using two stacks to implement a queue.
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