Implement Stack Using Queues Leetcode Problem 225 [Python Solution]

Let’s delve into the problem of implementing a stack using queues, Implement Stack Using Queues Leetcode Problem 225.

It’s a great problem for beginners to familiarize themselves with stacks and queues, as well as the relationship between them.

We’ll explore the concept of a Last-In-First-Out (LIFO) stack and how to achieve this using queues.

Problem Overview

Problem Statement:
You are tasked with implementing a Last-In-First-Out (LIFO) stack using only two queues.

The implemented stack should support all the functions of a standard stack, which includes push, top, pop, and empty operations.

To accomplish this task, you need to implement the MyStack class with the following methods:

1. void push(int x): This method pushes element x to the top of the stack.
2. int pop(): Removes the element on the top of the stack and returns it.
3. int top(): Returns the element on the top of the stack.
4. boolean empty(): Returns true if the stack is empty, false otherwise.

Notes:

• You must use only standard operations of a queue.

This means that only push to back, peek/pop from front, size, and is empty operations are valid.

• Depending on your language, the queue may not be supported natively.

You may simulate a queue using a list or deque (double-ended queue) as long as you use only a queue’s standard operations.

Let’s dive into the problem with a concrete example:

Example 1:

Input:

["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]

Output:

[null, null, null, 2, 2, false]

Explanation:
We start by initializing the MyStack and pushing two values, 1 and 2, onto the stack.

When we call top(), it returns 2, which is the top element of the stack.

We then call pop(), which removes and returns the top element (2).

Finally, we check if the stack is empty using the empty() method, which returns false because there is still one element (1) in the stack.

Now that we understand the problem, let’s proceed with the solution.

We will cover the following aspects:

1. Understanding Constraints
2. Edge Cases a Valid Solution Must Consider
3. Bruteforce Approach with Python Code Solution
4. Efficient Approach with Explanation
5. Reasoning Behind Our Efficient Approach
6. Time and Space Complexity
7. Conclusion

Understanding Constraints

Before we jump into solving the problem, it’s essential to understand the constraints mentioned in the problem statement:

• The input value x for the push operation will be within the range 1 to 9.
• At most 100 calls will be made to push, pop, top, and empty.
• All calls to pop and top are valid.

Edge Cases a Valid Solution Must Consider

When solving this problem, we need to consider the following edge cases to ensure our solution works correctly:

1. Pushing elements onto the stack.
2. Popping elements from the stack.
3. Calling top when the stack is empty.
4. Checking if the stack is empty.
5. Handling multiple operations in sequence.

Bruteforce Approach

Before we delve into the efficient approach, let’s consider a brute-force method for solving this problem.

In this approach, we will use two queues, but the pop operation will be time-inefficient.

Pseudo Code:

class MyStack:
    def __init__(self):
        # Initialize two queues q1 and q2
        self.q1 = []
        self.q2 = []

    def push(self, x: int) -> None:
        # Push the new element into q1
        self.q1.append(x)

    def pop(self) -> int:
        # Move elements from q1 to q2, leaving one element in q1
        while len(self.q1) > 1:
            self.q2.append(self.q1.pop(0))

        # Store the last element from q1
        top_element = self.q1[0]

        # Swap q1 and q2 to make q2 empty
        self.q1, self.q2 = self.q2, self.q1

        return top_element

    def top(self) -> int:
        return self.q1[-1]

    def empty(self) -> bool:
        return len(self.q1) == 0

In this approach, we maintain two queues, q1 and q2. When we push an element, we add it to q1. To perform the pop operation, we move all elements from q1 to q2, leaving only one element in q1, which will be the element to pop.

After that, we swap q1 and q2 to make q2 empty.

While this approach works, it’s not efficient for the pop operation since it requires moving elements between queues, resulting in a time complexity of O(n), where n is the number of elements in the stack.

We’ll explore a more efficient approach next.

Efficient Approach with Explanation

The efficient approach to solving this problem is to use a single queue while making the push operation efficient.

We will keep the pop operation as time-inefficient as before.

Here’s a breakdown of how this approach works:

1. We initialize a single queue q in the constructor.
2. For the push operation, we directly append the new element to the queue, making it an efficient operation.
3. To perform the pop operation, we need to simulate popping the top element from the stack.

To do this, we iterate through the elements in the queue and remove them one by one, except for the last element.

4. While iterating through the queue, we remove each element from the front and append it back to the queue.

This process continues until there is only one element left, which is the one we want to pop.

5. After identifying the top element, we return it.

The time complexity of this operation is O(n), where n is the number of elements in the queue.

6. The top operation is straightforward.

We return the last element in the queue, which represents the top of the stack.

7. The empty operation checks if the queue is empty and returns true if it is and false otherwise.

Let’s implement this efficient approach in Python code.

Python Code Solution

class MyStack:
    def __init__(self):
        self.q = deque()

    def push(self, x: int) -> None:
        self.q.append(x)

    def pop(self) -> int:
        for _ in range(len(self.q) - 1):
            self.q.append(self.q.popleft())
        return self.q.popleft()

    def top(self) -> int:
        for _ in range(len(self.q) - 1):
            self.q.append(self.q.popleft())
        res = self.q[0]
        self.q.append(self.q.popleft())
        return res

    def empty(self) -> bool:
        return len(self.q) == 0

This Python code defines the MyStack class that implements the efficient approach we discussed earlier.

The pop operation, although time-inefficient, is necessary to simulate the stack’s behavior using a single queue.

The other operations, push, top, and empty, are efficient and perform in O(1) time.

Reasoning Behind Our Efficient Approach

The reasoning behind this approach is to make the push operation efficient by directly appending elements to the queue while keeping the pop operation time-inefficient, as we need to simulate the stack’s behavior.

To achieve this, we iterate through the elements in the queue, removing each element from the front and appending it back until only the top element remains.

This top element is the one we want to pop.

While the pop operation is time-inefficient (O(n)), it’s an acceptable trade-off for the efficiency gained in the push operation (O(1)).

Time and Space Complexity

Let’s analyze the time and space complexity of our solution:

• Time Complexity:
• push(x): O(1) – Appending elements to the queue is an O(1) operation.
• pop(): O(n) – Iterating through the queue to find and remove the top element takes O(n) time.
• top(): O(n) – Similar to the pop() operation, it iterates through the queue, so it’s O(n).
• empty(): O(1) – Checking if the queue is empty is an O(1) operation.
• Space Complexity:
• The space complexity is O(n) as we use a single queue to store the elements.

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Conclusion

In this article, we tackled the problem of implementing a stack using queues.

We explored an efficient approach that makes the push operation efficient by directly appending elements to a single queue while keeping the pop operation time-inefficient.

This approach allows us to maintain the stack’s behavior while adhering to the constraints of the problem.

We covered edge cases, provided a Python code solution, and discussed the time and space complexity of our approach.

This problem is a great way for beginners to gain a better understanding of stack and queue data structures and their relationship.

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Happy coding!

LeetCode Problem Link

Now, it’s your turn to practice and explore this problem further.

Happy coding!