LeetCode #232 — EASY

Implement Queue using Stacks

Build confidence with an intuition-first walkthrough focused on stack fundamentals.

The Problem

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 operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false

Constraints:

1 <= x <= 9
At most 100 calls will be made to push, pop, peek, and empty.
All the calls to pop and peek are valid.

Follow-up: Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.

Step 01

Brute Force Baseline

Problem summary: 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 operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid. Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Stack · Design

Example 1

["MyQueue","push","push","peek","pop","empty"]
[[],[1],[2],[],[],[]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #232: Implement Queue using Stacks
class MyQueue {
    private Deque<Integer> stk1 = new ArrayDeque<>();
    private Deque<Integer> stk2 = new ArrayDeque<>();

    public MyQueue() {
    }

    public void push(int x) {
        stk1.push(x);
    }

    public int pop() {
        move();
        return stk2.pop();
    }

    public int peek() {
        move();
        return stk2.peek();
    }

    public boolean empty() {
        return stk1.isEmpty() && stk2.isEmpty();
    }

    private void move() {
        while (stk2.isEmpty()) {
            while (!stk1.isEmpty()) {
                stk2.push(stk1.pop());
            }
        }
    }
}

/**
 * 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();
 */

// Accepted solution for LeetCode #232: Implement Queue using Stacks
type MyQueue struct {
	stk1 []int
	stk2 []int
}

func Constructor() MyQueue {
	return MyQueue{[]int{}, []int{}}
}

func (this *MyQueue) Push(x int) {
	this.stk1 = append(this.stk1, x)
}

func (this *MyQueue) Pop() int {
	this.move()
	ans := this.stk2[len(this.stk2)-1]
	this.stk2 = this.stk2[:len(this.stk2)-1]
	return ans
}

func (this *MyQueue) Peek() int {
	this.move()
	return this.stk2[len(this.stk2)-1]
}

func (this *MyQueue) Empty() bool {
	return len(this.stk1) == 0 && len(this.stk2) == 0
}

func (this *MyQueue) move() {
	if len(this.stk2) == 0 {
		for len(this.stk1) > 0 {
			this.stk2 = append(this.stk2, this.stk1[len(this.stk1)-1])
			this.stk1 = this.stk1[:len(this.stk1)-1]
		}
	}
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * obj := Constructor();
 * obj.Push(x);
 * param_2 := obj.Pop();
 * param_3 := obj.Peek();
 * param_4 := obj.Empty();
 */

# Accepted solution for LeetCode #232: Implement Queue using Stacks
class MyQueue:
    def __init__(self):
        self.stk1 = []
        self.stk2 = []

    def push(self, x: int) -> None:
        self.stk1.append(x)

    def pop(self) -> int:
        self.move()
        return self.stk2.pop()

    def peek(self) -> int:
        self.move()
        return self.stk2[-1]

    def empty(self) -> bool:
        return not self.stk1 and not self.stk2

    def move(self):
        if not self.stk2:
            while self.stk1:
                self.stk2.append(self.stk1.pop())


# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()

// Accepted solution for LeetCode #232: Implement Queue using Stacks
use std::collections::VecDeque;

struct MyQueue {
    stk1: Vec<i32>,
    stk2: Vec<i32>,
}

impl MyQueue {
    fn new() -> Self {
        MyQueue {
            stk1: Vec::new(),
            stk2: Vec::new(),
        }
    }

    fn push(&mut self, x: i32) {
        self.stk1.push(x);
    }

    fn pop(&mut self) -> i32 {
        self.move_elements();
        self.stk2.pop().unwrap()
    }

    fn peek(&mut self) -> i32 {
        self.move_elements();
        *self.stk2.last().unwrap()
    }

    fn empty(&self) -> bool {
        self.stk1.is_empty() && self.stk2.is_empty()
    }

    fn move_elements(&mut self) {
        if self.stk2.is_empty() {
            while let Some(element) = self.stk1.pop() {
                self.stk2.push(element);
            }
        }
    }
}

// Accepted solution for LeetCode #232: Implement Queue using Stacks
class MyQueue {
    stk1: number[];
    stk2: number[];

    constructor() {
        this.stk1 = [];
        this.stk2 = [];
    }

    push(x: number): void {
        this.stk1.push(x);
    }

    pop(): number {
        this.move();
        return this.stk2.pop();
    }

    peek(): number {
        this.move();
        return this.stk2.at(-1);
    }

    empty(): boolean {
        return !this.stk1.length && !this.stk2.length;
    }

    move(): void {
        if (!this.stk2.length) {
            while (this.stk1.length) {
                this.stk2.push(this.stk1.pop()!);
            }
        }
    }
}

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

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.