What is the time complexity of Implement Queue using Stacks?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Implement Queue using Stacks?

This problem uses Stack, Design, Queue. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Implement Queue using Stacks?

Using two stacks allows us to efficiently simulate queue behavior. One stack (`stack1`) is used primarily for pushing elements, and the other (`stack2`) is used for popping elements. The key to achieving amortized O(1) time complexity is to only transfer elements between stacks when necessary (i.e., when `pop()` or `peek()` is called and `stack2` is empty).

Which companies ask LeetCode 232. Implement Queue using Stacks in interviews?

This problem is frequently asked at Infosys, Netflix, Oracle, Qualcomm.

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Implement Queue using Stacks - LeetCode 232 Solution

Implement Queue using Stacks - Complete Solution Guide

Implement Queue using Stacks is LeetCode problem 232, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue ( push , peek , pop , and empty ). Implement the MyQueue class: void push(int x) Pushes element x to the back of the queue. int pop() Removes the element from the front of the queue and returns it. int peek() Returns the element at the front of the queue. boolean empty() Returns true if the queue is empty, false otherwise. Notes: You must use only standard o

Detailed Explanation

The problem asks you to design a queue data structure using only two stacks. A queue follows the FIFO (First-In, First-Out) principle, meaning the first element added is the first element removed. The challenge lies in simulating this behavior using stacks, which follow the LIFO (Last-In, First-Out) principle. You need to implement four methods: `push(x)` to add an element `x` to the rear of the queue, `pop()` to remove and return the element at the front, `peek()` to return the element at the front without removing it, and `empty()` to check if the queue is empty.

Solution Approach

The solution utilizes two stacks, `stack1` and `stack2`. Elements are pushed onto `stack1`. When `pop()` or `peek()` is called and `stack2` is empty, all elements from `stack1` are transferred to `stack2`. This effectively reverses the order of elements, making the element that was at the bottom of `stack1` (the front of the queue) now at the top of `stack2`. Then, the top element of `stack2` can be easily popped or peeked.

Step-by-Step Algorithm

**push(x):** Push the element `x` onto `stack1`.
**pop():** If `stack2` is empty, transfer all elements from `stack1` to `stack2`. Then, pop and return the top element from `stack2`.
**peek():** If `stack2` is empty, transfer all elements from `stack1` to `stack2`. Then, return the top element of `stack2`.
**empty():** Check if both `stack1` and `stack2` are empty. Return `true` if both are empty, `false` otherwise.

Key Insights

Using two stacks allows us to efficiently simulate queue behavior. One stack (`stack1`) is used primarily for pushing elements, and the other (`stack2`) is used for popping elements.
The key to achieving amortized O(1) time complexity is to only transfer elements between stacks when necessary (i.e., when `pop()` or `peek()` is called and `stack2` is empty).
Understanding that the cost of transferring elements between stacks is distributed over multiple operations is crucial for understanding the amortized complexity.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(n)

Topics

This problem involves: Stack, Design, Queue.

Companies

Asked at: Infosys, Netflix, Oracle, Qualcomm.

Implement Queue using Stacks | LeetCode 232

Easy

Stack Design Queue

Time: O(1)Space: O(n)

Solution Code

Available in:

import java.util.Stack;

class MyQueue {

    Stack<Integer> stack1;
    Stack<Integer> stack2;

    public MyQueue() {
        stack1 = new Stack<>();
        stack2 = new Stack<>();
    }
    
    public void push(int x) {
        stack1.push(x);
    }
    
    public int pop() {
        if(stack2.isEmpty()){
            while(!stack1.isEmpty()){
                stack2.push(stack1.pop());
            }
        }
        return stack2.pop();
    }
    
    public int peek() {
        if(stack2.isEmpty()){
            while(!stack1.isEmpty()){
                stack2.push(stack1.pop());
            }
        }
        return stack2.peek();
    }
    
    public boolean empty() {
        return stack1.isEmpty() && stack2.isEmpty();
    }
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue obj = new MyQueue();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.peek();
 * boolean param_4 = obj.empty();
 */

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Problem Info

Description

Problem 232: Implement Queue using Stacks - Detailed explanation and solution approaches.

Topics

Stack Design Queue

Companies

InfosysNetflixOracleQualcomm

Stats

Difficulty:Easy

Languages:4

Companies:4

Problem #:232

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

4 steps

1
**push(x):** Push the element `x` onto `stack1`.
2
**pop():** If `stack2` is empty, transfer all elements from `stack1` to `stack2`. Then, pop and return the top element from `stack2`.
3
**peek():** If `stack2` is empty, transfer all elements from `stack1` to `stack2`. Then, return the top element of `stack2`.
4
**empty():** Check if both `stack1` and `stack2` are empty. Return `true` if both are empty, `false` otherwise.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Pushing an element is always O(1). Popping and peeking have a worst-case time complexity of O(n) when all elements from `stack1` need to be transferred to `stack2`. However, this transfer is only done once for every n elements that are pushed. Therefore, distributing the cost of transferring n elements over n operations gives us an amortized time complexity of O(1) for each push, pop, and peek operation. Space complexity is O(n) because, in the worst case, both stacks can hold up to n elements.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Single Stack Approach (Less Efficient)

It's possible to implement a queue using a single stack, but it would require more complex operations and not achieve amortized O(1) time complexity for all operations.

Time: O(n) for pop and peekSpace: O(n)

Trade-offs: This approach is less efficient than the two-stack approach in terms of time complexity, making it unsuitable for performance-critical applications.

✨

Key Insights

1Using two stacks allows us to efficiently simulate queue behavior. One stack (`stack1`) is used primarily for pushing elements, and the other (`stack2`) is used for popping elements.
2The key to achieving amortized O(1) time complexity is to only transfer elements between stacks when necessary (i.e., when `pop()` or `peek()` is called and `stack2` is empty).
3Understanding that the cost of transferring elements between stacks is distributed over multiple operations is crucial for understanding the amortized complexity.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Queue

FIFO structure - First In, First Out. Like a line at a store!

Edge Cases

Empty Queue: The solution correctly handles an empty queue. `empty()` will return `true`, and `pop()` and `peek()` will handle the empty `stack2` by transferring from `stack1`.
Queue with one element: This case is handled efficiently. When you push one element, then pop or peek, only one element transfer is performed, making it still amortized O(1).

⚠️

Common Mistakes to Avoid

✗Forgetting to handle the case where `stack2` is empty when `pop()` or `peek()` is called. This leads to incorrect results or crashes.

✗Not understanding amortized analysis. Beginners might incorrectly calculate the time complexity as O(n) because of the potential for transferring elements between stacks, overlooking the fact that the transfer is amortized over multiple operations.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Implementing basic queue functionality in systems with limited resources or where memory management is crucial. Using stacks can provide a more space-efficient method than dedicated queue implementations.

🏦Simulating a queue in embedded systems or real-time systems where you might have only stack-based memory management.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Stacks are perfect for problems involving matching (parentheses) or "next greater element" patterns. If you need to look back at recent elements, think stack.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Popping from an empty stack. Always check if the stack is non-empty before popping.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

Frequently Asked Questions

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Implement Queue using Stacks"?

For this easy-level Stack problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether design techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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