225. Implement Stack using Queues

Description

Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (push, top, pop, and empty).

Implement the MyStack class:

void push(int x) Pushes element x to the top of the stack.
int pop() Removes the element on the top of the stack and returns it.
int top() Returns the element on the top of the stack.
boolean empty() Returns true if the stack is empty, false otherwise.

Notes:

You must use only standard operations of a queue, which 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.

Example 1:

Input
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 2, 2, false]

Explanation
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.empty(); // return False

Constraints:

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

Follow-up: Can you implement the stack using only one queue?

Solutions

Solution 1: Two Queues

We use two queues $q_1$ and $q_2$, where $q_1$ is used to store the elements in the stack, and $q_2$ is used to assist in implementing the stack operations.

push operation: Push the element into $q_2$, then pop the elements in $q_1$ one by one and push them into $q_2$, finally swap the references of $q_1$ and $q_2$. The time complexity is $O(n)$.
pop operation: Directly pop the front element of $q_1$. The time complexity is $O(1)$.
top operation: Directly return the front element of $q_1$. The time complexity is $O(1)$.
empty operation: Check whether $q_1$ is empty. The time complexity is $O(1)$.

The space complexity is $O(n)$, where $n$ is the number of elements in the stack.

import java.util.Deque;

class MyStack {
    private Deque<Integer> q1 = new ArrayDeque<>();
    private Deque<Integer> q2 = new ArrayDeque<>();

    public MyStack() {
    }

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

    public int pop() {
        return q1.poll();
    }

    public int top() {
        return q1.peek();
    }

    public boolean empty() {
        return q1.isEmpty();
    }
}

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

class MyStack {
public:
    MyStack() {
    }

    void push(int x) {
        q2.push(x);
        while (!q1.empty()) {
            q2.push(q1.front());
            q1.pop();
        }
        swap(q1, q2);
    }

    int pop() {
        int x = q1.front();
        q1.pop();
        return x;
    }

    int top() {
        return q1.front();
    }

    bool empty() {
        return q1.empty();
    }

private:
    queue<int> q1;
    queue<int> q2;
};

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack* obj = new MyStack();
 * obj->push(x);
 * int param_2 = obj->pop();
 * int param_3 = obj->top();
 * bool param_4 = obj->empty();
 */

# one queue

'''
push(1), [1]
push(2), [1,2] => [2,1]
push(3), [2,1,3] => [1,3,2] => [3,2,1]
push(4), [3,2,1,4] => switch 3 time to get [4,3,2,1]
push(5), [4,3,2,1,5] => switch 4 time to get [5,4,3,2,1]
'''
class Stack:

    def __init__(self):
        self._queue = collections.deque()

    def push(self, x):
        q = self._queue
        q.append(x)
        for _ in range(len(q) - 1):
            q.append(q.popleft())
        
    def pop(self):
        return self._queue.popleft()

    def top(self):
        return self._queue[0]
    
    def empty(self):
        return not len(self._queue)

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

############

from collections import deque

# two queues
class MyStack:

	def __init__(self):
		self.q1 = deque()
		self.q2 = deque()

	def push(self, x: int) -> None:
		self.q1.append(x)

	def pop(self) -> int:
		while len(self.q1) != 1:
			self.q2.append(self.q1.popleft())

		val = self.q1.popleft()
		self.q1, self.q2 = self.q2, self.q1
		return val

	def top(self) -> int:
		while len(self.q1) != 1:
			self.q2.append(self.q1.popleft())

        # tried to re-use while part, but seems not achievable, since val is retrieved in-between 
		val = self.q1[0]
		self.q2.append(self.q1.popleft())  # note: add back to q2, so q1 will always be empty

		self.q1, self.q2 = self.q2, self.q1
		return val


	def empty(self) -> bool:
		return not self.q1

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

type MyStack struct {
	q1 []int
	q2 []int
}

func Constructor() MyStack {
	return MyStack{}
}

func (this *MyStack) Push(x int) {
	this.q2 = append(this.q2, x)
	for len(this.q1) > 0 {
		this.q2 = append(this.q2, this.q1[0])
		this.q1 = this.q1[1:]
	}
	this.q1, this.q2 = this.q2, this.q1
}

func (this *MyStack) Pop() int {
	x := this.q1[0]
	this.q1 = this.q1[1:]
	return x
}

func (this *MyStack) Top() int {
	return this.q1[0]
}

func (this *MyStack) Empty() bool {
	return len(this.q1) == 0
}

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

class MyStack {
    q1: number[] = [];
    q2: number[] = [];

    constructor() {}

    push(x: number): void {
        this.q2.push(x);
        while (this.q1.length) {
            this.q2.push(this.q1.shift()!);
        }
        [this.q1, this.q2] = [this.q2, this.q1];
    }

    pop(): number {
        return this.q1.shift()!;
    }

    top(): number {
        return this.q1[0];
    }

    empty(): boolean {
        return this.q1.length === 0;
    }
}

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

use std::collections::VecDeque;

struct MyStack {
    /// There could only be two status at all time
    /// 1. One contains N elements, the other is empty
    /// 2. One contains N - 1 elements, the other contains exactly 1 element
    q_1: VecDeque<i32>,
    q_2: VecDeque<i32>,
    // Either 1 or 2, originally begins from 1
    index: i32,
}

impl MyStack {
    fn new() -> Self {
        Self {
            q_1: VecDeque::new(),
            q_2: VecDeque::new(),
            index: 1,
        }
    }

    fn move_data(&mut self) {
        // Always move from q1 to q2
        assert!(self.q_2.len() == 1);
        while !self.q_1.is_empty() {
            self.q_2.push_back(self.q_1.pop_front().unwrap());
        }
        let tmp = self.q_1.clone();
        self.q_1 = self.q_2.clone();
        self.q_2 = tmp;
    }

    fn push(&mut self, x: i32) {
        self.q_2.push_back(x);
        self.move_data();
    }

    fn pop(&mut self) -> i32 {
        self.q_1.pop_front().unwrap()
    }

    fn top(&mut self) -> i32 {
        *self.q_1.front().unwrap()
    }

    fn empty(&self) -> bool {
        self.q_1.is_empty()
    }
}

225 - Implement Stack using Queues

225. Implement Stack using Queues

Description

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225. Implement Stack using Queues

Description

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All Problems

All Solutions