Implement a Stack using a Queue

This problem focuses on implementing a Stack data structure using one or more Queues. The challenge lies in the fact that stacks are LIFO (Last-In, First-Out) while queues are FIFO (First-In, First-Out). We need to cleverly use queue operations to mimic stack behavior.

Approach 1: Using Two Queues

This approach uses two queues to ensure that the most recently inserted element is always at the front of one of the queues, ready to be returned by the top and pop operations. Let's call the queues q1 and q2.

push(x): Add element x to q1. Then, transfer all elements from q2 to q1 so that x becomes the first element.
pop(): Dequeue the first element from q1. This element will be the most recently pushed element, which adheres to the stack's LIFO principle.
top(): Return the first element in q1.
empty(): Return true if both q1 and q2 are empty, false otherwise.

Code (Python)

class Stack:
    def __init__(self):
        self.q1 = []
        self.q2 = []

    def push(self, x):
        self.q1.append(x)
        while self.q2:
            self.q1.append(self.q2.pop(0))
        self.q1, self.q2 = self.q2, self.q1

    def pop(self):
        if not self.empty():
            return self.q1.pop(0)
        return None

    def top(self):
        if not self.empty():
            return self.q1[0]
        return None

    def empty(self):
        return not self.q1 and not self.q2

# Example Usage
stack = Stack()
stack.push(1)
stack.push(2)
print(stack.top())   # Output: 2
print(stack.pop())   # Output: 2
print(stack.empty()) # Output: False

Code (Java)

import java.util.LinkedList;
import java.util.Queue;

class Stack {
    private Queue<Integer> q1 = new LinkedList<>();
    private Queue<Integer> q2 = new LinkedList<>();

    public void push(int x) {
        q1.offer(x);
        while (!q2.isEmpty()) {
            q1.offer(q2.poll());
        }
        Queue<Integer> temp = q1;
        q1 = q2;
        q2 = temp;
    }

    public int pop() {
        if (!empty()) {
            return q2.poll();
        }
        return -1; // Or throw an exception
    }

    public int top() {
        if (!empty()) {
            return q2.peek();
        }
        return -1; // Or throw an exception
    }

    public boolean empty() {
        return q2.isEmpty();
    }

    public static void main(String[] args) {
        Stack stack = new Stack();
        stack.push(1);
        stack.push(2);
        System.out.println(stack.top());   // Output: 2
        System.out.println(stack.pop());   // Output: 2
        System.out.println(stack.empty()); // Output: false
    }
}

Time Complexity Analysis

push(x): O(n), where n is the number of elements already in the stack. This is because we need to move all elements from q2 to q1.
pop(): O(1), as we simply dequeue from q1.
top(): O(1), as we simply peek at the front of q1.
empty(): O(1), as we check if both queues are empty.

Space Complexity Analysis

The space complexity is O(n), where n is the maximum number of elements stored in the stack, as we use two queues to store the elements.

Approach 2: Using One Queue

This approach uses only one queue. When a new element is pushed onto the stack, we enqueue it to the queue. Then, we dequeue all the existing elements and enqueue them back to the queue. This operation will make sure the new element will be at the front of the queue, which simulates the stack's LIFO property.

push(x): Enqueue x to the queue. Then, move all the other elements to the end of the queue.
pop(): Dequeue the first element from the queue.
top(): Return the first element in the queue.
empty(): Return true if the queue is empty, false otherwise.

Code (Python)

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):
        if self.q:
            return self.q.popleft()
        return None

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

    def empty(self):
        return not self.q

# Example Usage
stack = Stack()
stack.push(1)
stack.push(2)
print(stack.top())   # Output: 2
print(stack.pop())   # Output: 2
print(stack.empty()) # Output: False

Code (Java)

import java.util.LinkedList;
import java.util.Queue;

class Stack {
    private Queue<Integer> q = new LinkedList<>();

    public void push(int x) {
        q.offer(x);
        for (int i = 0; i < q.size() - 1; i++) {
            q.offer(q.poll());
        }
    }

    public int pop() {
        if (!q.isEmpty()) {
            return q.poll();
        }
        return -1; // Or throw an exception
    }

    public int top() {
        if (!q.isEmpty()) {
            return q.peek();
        }
        return -1; // Or throw an exception
    }

    public boolean empty() {
        return q.isEmpty();
    }

    public static void main(String[] args) {
        Stack stack = new Stack();
        stack.push(1);
        stack.push(2);
        System.out.println(stack.top());   // Output: 2
        System.out.println(stack.pop());   // Output: 2
        System.out.println(stack.empty()); // Output: false
    }
}

Time Complexity Analysis

push(x): O(n), where n is the number of elements already in the stack. This is because we need to move all existing elements to the end of the queue.
pop(): O(1), as we simply dequeue from the front of the queue.
top(): O(1), as we simply peek at the front of the queue.
empty(): O(1), as we check if the queue is empty.

Space Complexity Analysis

The space complexity is O(n), where n is the maximum number of elements stored in the stack. We only use one queue to store the elements.

Edge Cases

Empty Stack: When pop() or top() is called on an empty stack, it should return None or throw an exception, as handled in the provided code examples. This is crucial to avoid errors when the stack is empty.
Large Number of Operations: The efficiency of push() operation is O(n), which could become a bottleneck with frequent push operations on a large stack. The approach using one queue is generally preferred due to its simplicity.

Optimizations

While the fundamental approach of using queues to simulate a stack remains the same, optimizations are limited by the inherent nature of the problem. The O(n) complexity of the push operation when using a single queue is unavoidable since you need to reorder the queue to maintain the LIFO behavior. Using two queues doesn't fundamentally change this time complexity.

In practice, consider if a stack is truly the best data structure for your use case. If frequent pushing is needed, the overhead of reordering elements within queues may make stack implementation based on a python list or java ArrayList a better choice, because they can offer O(1) appends, as long as the array has allocated enough space.