Implement Stack Using Queues

Published at23 Jul 2026

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Implement a stack using only two queues. You may only use standard queue operations: enqueue (push to back), dequeue (pop from front), peek (front), and empty check. The stack must support push(x), pop(), top(), and empty(). The key insight: a stack is LIFO while a queue is FIFO. By using an auxiliary queue and moving elements around, the newest element can be kept at the front of the main queue. This teaches how data structures can implement each other and reveals the fundamental relationship between LIFO and FIFO.

The key insight is that a stack reverses order (LIFO) while a queue preserves order (FIFO). To make a queue behave like a stack, you need to reverse the order of elements each time a new element is pushed. This can be done by pushing the new element into an empty auxiliary queue, then moving all elements from the main queue to the auxiliary queue, and finally swapping the two queues. This puts the newest element at the front of the main queue, ready to be popped first.

An alternative approach is to make the push operation O(1) and the pop operation O(n), by pushing directly and rotating elements when popping. Both are valid.

Example Input & Output

Example 1

Input

"push(1), push(2), top(), pop(), empty()"

Output

"2, 2, false"

Explanation

Push 1,2. Top is 2. Pop 2. Not empty.

Example 2

Input

"push(1), pop(), empty()"

Output

"1, true"

Explanation

Push 1, pop 1. Empty.

Algorithm Flow

Recommendation Algorithm Flow for Implement Stack Using Queues

Solution Approach

Use two queues. For push, enqueue to q2, move everything from q1 to q2, swap q1 and q2. The newest element is always at q1 front. Pop, top, and empty just work on q1.

function MyStack() {
  this.q1 = [];
  this.q2 = [];
}
MyStack.prototype.push = function(x) {
  this.q2.push(x);
  while (this.q1.length) this.q2.push(this.q1.shift());
  var t = this.q1; this.q1 = this.q2; this.q2 = t;
};
MyStack.prototype.pop = function() { return this.q1.shift(); };
MyStack.prototype.top = function() { return this.q1[0]; };
MyStack.prototype.empty = function() { return this.q1.length === 0; };

Push O(n), others O(1). Space O(n).

Best Answers
These are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.

java

import java.util.*;
class MyStack {
    Queue<Integer> q1 = new LinkedList<>();
    Queue<Integer> q2 = new LinkedList<>();
    public void push(int x) {
        q2.add(x);
        while (!q1.isEmpty()) q2.add(q1.poll());
        Queue<Integer> t = q1; q1 = q2; q2 = t;
    }
    public int pop() { return q1.poll(); }
    public int top() { return q1.peek(); }
    public boolean empty() { return q1.isEmpty(); }
}

javascript

var MyStack = function() { this.q1 = []; this.q2 = []; };
MyStack.prototype.push = function(x) {
  this.q2.push(x);
  while (this.q1.length) this.q2.push(this.q1.shift());
  var t = this.q1; this.q1 = this.q2; this.q2 = t;
};
MyStack.prototype.pop = function() { return this.q1.shift(); };
MyStack.prototype.top = function() { return this.q1[0]; };
MyStack.prototype.empty = function() { return this.q1.length === 0; };

php

<?php
class MyStack {
    private $q1 = []; private $q2 = [];
    function push($x) {
        array_push($this->q2, $x);
        while (!empty($this->q1)) array_push($this->q2, array_shift($this->q1));
        [$this->q1, $this->q2] = [$this->q2, $this->q1];
    }
    function pop() { return array_shift($this->q1); }
    function top() { return $this->q1[0]; }
    function empty() { return empty($this->q1); }
}

python

class MyStack:
    def __init__(self):
        self.q1 = []
        self.q2 = []
    def push(self, x):
        self.q2.append(x)
        while self.q1:
            self.q2.append(self.q1.pop(0))
        self.q1, self.q2 = self.q2, self.q1
    def pop(self):
        return self.q1.pop(0)
    def top(self):
        return self.q1[0]
    def empty(self):
        return len(self.q1) == 0

rust

struct MyStack {
    q1: Vec<i32>,
    q2: Vec<i32>,
}
impl MyStack {
    fn new() -> Self { MyStack { q1: Vec::new(), q2: Vec::new() } }
    fn push(&mut self, x: i32) {
        self.q2.push(x);
        while !self.q1.is_empty() { self.q2.push(self.q1.remove(0)); }
        std::mem::swap(&mut self.q1, &mut self.q2);
    }
    fn pop(&mut self) -> i32 { self.q1.remove(0) }
    fn top(&self) -> i32 { self.q1[0] }
    fn empty(&self) -> bool { self.q1.is_empty() }
}

typescript

class MyStack {
    private q1: number[] = [];
    private q2: number[] = [];
    push(x: number): void {
        this.q2.push(x);
        while (this.q1.length) this.q2.push(this.q1.shift() as number);
        var t = this.q1; this.q1 = this.q2; this.q2 = t;
    }
    pop(): number { return this.q1.shift() as number; }
    top(): number { return this.q1[0]; }
    empty(): boolean { return this.q1.length === 0; }
}

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Implement Stack Using Queues

Example Input & Output

Algorithm Flow

Solution Approach

Best AnswersThese are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.

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Best Answers
These are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.