Stack Using Two Queues in Java

The stack works as a linear data structure that implements the Last In First Out (LIFO) method, so the last added element gets removed first. Two FIFO queues need to be used to implement a LIFO stack since they operate according to First In, First Out principles. LIFO (Last In, First Out) functions will be replicated using FIFO procedures.

This problem is an excellent exercise in algorithm design, requiring efficient enqueue and dequeue operations to ensure proper functionality. The challenge lies in deciding whether to make a push or pop costly, optimizing performance based on the expected usage patterns.

There are the following two approaches:

1. Making push (insertion) Costly: It ensures that the top element remains at the front by strategically transferring elements between queues. This allows efficient access to the most recently inserted element, maintaining stack behavior using queues
2. Making pop (removal) Costly: It transfers elements between two queues to mimic stack behavior, ensuring Last In, First Out (LIFO) order. This movement helps maintain correct push and pop operations using only queue operations.

This implementation is useful in situations where we are limited to using only queue operations and need stack functionality.

Algorithm

We can implement a stack using two queues (q1 and q2) by making either push or pop operations costly. Below is an approach where push operation is costly while making pop and top operations efficient.

1. Initialize Two Queues
   
   • Maintain two queues: q1 (primary queue) and q2 (temporary queue).
2. Push Operation (Costly)
   
   • Insert the new element into q2.
   • Move all elements from q1 to q2.
   • Swap q1 and q2 to maintain the order.
3. Pop Operation (Efficient)
   
   • Remove the front element from q1.
4. Top Operation (Efficient)
   
   • Return the front element of q1.
5. IsEmpty Operation
   
   • Return true if q1 is empty, otherwise false.

Output:

 Top element: 30 Popped element: 30 Popped element: 20 Stack empty: false Popped element: 10 Stack empty: true 

Explanation

This Java program implements a stack using two queues (q1 and q2). The push operation is costly, as the new element is first added to q2, and all elements from q1 are moved to q2 before swapping the queues. The pop and top operations are efficient since elements are dequeued directly from q1. This approach ensures LIFO behavior while maintaining efficient retrieval operations.

Key Observations

Push is Costly: The new element is added and the order is rearranged, making this O(n) time complexity.

Pop and Top are Efficient: Since the last inserted element is always at the front, pop and top operations take O(1) time.

Two Queues Are Used: The swap of queues helps maintain the order without additional storage.

Alternative Approach: If pop were costly, elements would be moved only when popping, making push O(1) and pop O(n).

Brute Force Vs. Optimal Solution

Brute Force Approach

1. Using an Array:

• Insert elements at the end (O(1)).
• Remove elements from the end (O(1)).
• The problem is maintaining LIFO with FIFO operations.

Optimal Approach (Using Two Queues)

• Uses queue operations efficiently.
• Ensures LIFO behavior with O(n) for push and O(1) for pop/top.
• Uses queue swapping to maintain order efficiently.

Conclusion

Implementing a stack using two queues is a fascinating problem in data structure transformation. By cleverly utilizing enqueue and dequeue operations, we can mimic stack behavior while only using queues. This approach is useful in scenarios with hardware limitations or specific programming constraints where only queue operations are available.

The efficiency of this method depends on whether the push or pop operation is made costly. If frequent pop operations are needed, making push expensive (as in our implementation) is beneficial. However, if insertions are more frequent, a costly pop approach might be preferred. The choice between these implementations depends on the specific requirements of the use case.