Show (depict an algorithm or describe in words) how to implement a queue using two stacks. Analyze the running time of the basic queue operations.

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**Implementing a Queue using Two Stacks**

To implement a queue using two stacks, you can follow these steps:

1. **Initialize Two Stacks:**
   - Stack1 (inputStack): Used for enqueue operations.
   - Stack2 (outputStack): Used for dequeue operations.

2. **Enqueue Operation:**
   - Push the new element onto inputStack.

3. **Dequeue Operation:**
   - If outputStack is empty, transfer all elements from inputStack to outputStack by popping each item from inputStack and pushing it onto outputStack.
   - Pop the element from outputStack. This is the dequeued element.

**Running Time Analysis:**

- **Enqueue Operation:**
  - Time Complexity: O(1)
  - Explanation: Each enqueue operation involves a single push operation on inputStack.

- **Dequeue Operation:**
  - Average Time Complexity: O(1)
  - Worst-Case Time Complexity: O(n)
  - Explanation: In the worst case, all elements are transferred from inputStack to outputStack, which takes O(n). However, when averaged over a series of enqueue and dequeue operations, each element is moved at most twice, leading to an amortized O(1) complexity.

This approach efficiently uses the LIFO (Last-In-First-Out) nature of stacks to simulate FIFO (First-In-First-Out) queue behavior.
Transcribed Image Text:**Implementing a Queue using Two Stacks** To implement a queue using two stacks, you can follow these steps: 1. **Initialize Two Stacks:** - Stack1 (inputStack): Used for enqueue operations. - Stack2 (outputStack): Used for dequeue operations. 2. **Enqueue Operation:** - Push the new element onto inputStack. 3. **Dequeue Operation:** - If outputStack is empty, transfer all elements from inputStack to outputStack by popping each item from inputStack and pushing it onto outputStack. - Pop the element from outputStack. This is the dequeued element. **Running Time Analysis:** - **Enqueue Operation:** - Time Complexity: O(1) - Explanation: Each enqueue operation involves a single push operation on inputStack. - **Dequeue Operation:** - Average Time Complexity: O(1) - Worst-Case Time Complexity: O(n) - Explanation: In the worst case, all elements are transferred from inputStack to outputStack, which takes O(n). However, when averaged over a series of enqueue and dequeue operations, each element is moved at most twice, leading to an amortized O(1) complexity. This approach efficiently uses the LIFO (Last-In-First-Out) nature of stacks to simulate FIFO (First-In-First-Out) queue behavior.
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