Implement Stack using Queues

Given a Queue data structure that supports standard operations like enqueue() and dequeue(). The task is to implement a Stack data structure using only instances of Queue and Queue operations allowed on the instances.

A Stack can be implemented using two queues. Let Stack to be implemented be ‘s’ and queues used to implement are ‘q1’ and ‘q2’. Stack ‘s’ can be implemented in two ways:

Implement Stack using Queues By making push() operation costly:

Below is the idea to solve the problem:

The idea is to keep newly entered element at the front of ‘q1’ so that pop operation dequeues from ‘q1’. ‘q2’ is used to put every new element in front of ‘q1’.

• Follow the below steps to implement the push(s, x) operation: 
  • Enqueue x to q2.
  • One by one dequeue everything from q1 and enqueue to q2.
  • Swap the queues of q1 and q2.
• Follow the below steps to implement the pop(s) operation: 
  • Dequeue an item from q1 and return it.

Below is the implementation of the above approach. 

C++

/* Program to implement a stack using
two queue */
#include <bits/stdc++.h>

using namespace std;

class Stack {
    // Two inbuilt queues
    queue<int> q1, q2;

public:
    void push(int x)
    {
        // Push x first in empty q2
        q2.push(x);

        // Push all the remaining
        // elements in q1 to q2.
        while (!q1.empty()) {
            q2.push(q1.front());
            q1.pop();
        }

        // swap the names of two queues
        queue<int> q = q1;
        q1 = q2;
        q2 = q;
    }

    void pop()
    {
        // if no elements are there in q1
        if (q1.empty())
            return;
        q1.pop();
    }

    int top()
    {
        if (q1.empty())
            return -1;
        return q1.front();
    }

    int size() { return q1.size(); }
};

// Driver code
int main()
{
    Stack s;
    s.push(1);
    s.push(2);
    s.push(3);

    cout << "current size: " << s.size() << endl;
    cout << s.top() << endl;
    s.pop();
    cout << s.top() << endl;
    s.pop();
    cout << s.top() << endl;

    cout << "current size: " << s.size() << endl;
    return 0;
}
// This code is contributed by Chhavi

Java

/* Java Program to implement a stack using
two queue */
import java.util.*;

class GfG {

    static class Stack {
        // Two inbuilt queues
        static Queue<Integer> q1
            = new LinkedList<Integer>();
        static Queue<Integer> q2
            = new LinkedList<Integer>();

        // To maintain current number of
        // elements
        static int curr_size;

        static void push(int x)
        {
            // Push x first in empty q2
            q2.add(x);

            // Push all the remaining
            // elements in q1 to q2.
            while (!q1.isEmpty()) {
                q2.add(q1.peek());
                q1.remove();
            }

            // swap the names of two queues
            Queue<Integer> q = q1;
            q1 = q2;
            q2 = q;
        }

        static void pop()
        {

            // if no elements are there in q1
            if (q1.isEmpty())
                return;
            q1.remove();
        }

        static int top()
        {
            if (q1.isEmpty())
                return -1;
            return q1.peek();
        }

        static int size() { return q1.size(); }
    }

    // driver code
    public static void main(String[] args)
    {
        Stack s = new Stack();
        s.push(1);
        s.push(2);
        s.push(3);

        System.out.println("current size: " + s.size());
        System.out.println(s.top());
        s.pop();
        System.out.println(s.top());
        s.pop();
        System.out.println(s.top());

        System.out.println("current size: " + s.size());
    }
}
// This code is contributed by Prerna

Python3

# Program to implement a stack using
# two queue
from _collections import deque


class Stack:

    def __init__(self):

        # Two inbuilt queues
        self.q1 = deque()
        self.q2 = deque()

    def push(self, x):

        # Push x first in empty q2
        self.q2.append(x)

        # Push all the remaining
        # elements in q1 to q2.
        while (self.q1):
            self.q2.append(self.q1.popleft())

        # swap the names of two queues
        self.q1, self.q2 = self.q2, self.q1

    def pop(self):

        # if no elements are there in q1
        if self.q1:
            self.q1.popleft()

    def top(self):
        if (self.q1):
            return self.q1[0]
        return None

    def size(self):
        return len(self.q1)


# Driver Code
if __name__ == '__main__':
    s = Stack()
    s.push(1)
    s.push(2)
    s.push(3)

    print("current size: ", s.size())
    print(s.top())
    s.pop()
    print(s.top())
    s.pop()
    print(s.top())

    print("current size: ", s.size())

# This code is contributed by PranchalK

C#

/* C# Program to implement a stack using
two queue */
using System;
using System.Collections;

class GfG {

    public class Stack {
        // Two inbuilt queues
        public Queue q1 = new Queue();
        public Queue q2 = new Queue();

        public void push(int x)
        {
            // Push x first in empty q2
            q2.Enqueue(x);

            // Push all the remaining
            // elements in q1 to q2.
            while (q1.Count > 0) {
                q2.Enqueue(q1.Peek());
                q1.Dequeue();
            }

            // swap the names of two queues
            Queue q = q1;
            q1 = q2;
            q2 = q;
        }

        public void pop()
        {

            // if no elements are there in q1
            if (q1.Count == 0)
                return;
            q1.Dequeue();
        }

        public int top()
        {
            if (q1.Count == 0)
                return -1;
            return (int)q1.Peek();
        }

        public int size() { return q1.Count; }
    };

    // Driver code
    public static void Main(String[] args)
    {
        Stack s = new Stack();
        s.push(1);
        s.push(2);
        s.push(3);
        Console.WriteLine("current size: " + s.size());
        Console.WriteLine(s.top());
        s.pop();
        Console.WriteLine(s.top());
        s.pop();
        Console.WriteLine(s.top());
        Console.WriteLine("current size: " + s.size());
    }
}

// This code is contributed by Arnab Kundu

Javascript

/*Javascript Program to implement a stack using
two queue */

// Two inbuilt queues
class Stack {
    constructor() {
        this.q1 = [];
        this.q2 = [];
    }

    push(x) {

        // Push x first in isEmpty q2
        this.q2.push(x);
        // Push all the remaining
        // elements in q1 to q2.
        while (this.q1.length != 0) {
            this.q2.push(this.q1[0]);
            this.q1.shift();

        }

        // swap the names of two queues
        this.q = this.q1;
        this.q1 = this.q2;
        this.q2 = this.q;
    }

    pop() {
        // if no elements are there in q1
        if (this.q1.length == 0)
            return;
        this.q1.shift();
    }

    top() {
        if (this.q1.length == 0)
            return -1;
        return this.q1[0];
    }

    size() {
        console.log(this.q1.length);
    }

    isEmpty() {
        // return true if the queue is empty.
        return this.q1.length == 0;
    }

    front() {
        return this.q1[0];
    }
}

// Driver code


let s = new Stack();
s.push(1);
s.push(2);
s.push(3);

console.log("current size: ");
s.size();
console.log(s.top());
s.pop();
console.log(s.top());
s.pop();
console.log(s.top());

console.log("current size: ");
s.size();

// This code is contributed by adityamaharshi21

Output

current size: 3
3
2
1
current size: 1

Time Complexity:

• Push operation: O(N), As all the elements need to be popped out from the Queue (q1) and push them back to Queue (q2).
• Pop operation: O(1), As we need to remove the front element from the Queue.

Auxiliary Space: O(N), As we use two queues for the implementation of a Stack.

Implement Stack using Queues by making pop() operation costly:

Below is the idea to solve the problem:

The new element is always enqueued to q1. In pop() operation, if q2 is empty then all the elements except the last, are moved to q2. Finally, the last element is dequeued from q1 and returned.

• Follow the below steps to implement the push(s, x) operation: 
  • Enqueue x to q1 (assuming the size of q1 is unlimited).
• Follow the below steps to implement the pop(s) operation: 
  • One by one dequeue everything except the last element from q1 and enqueue to q2.
  • Dequeue the last item of q1, the dequeued item is the result, store it.
  • Swap the names of q1 and q2
  • Return the item stored in step 2.

Below is the implementation of the above approach:

C++

// Program to implement a stack
// using two queue
#include <bits/stdc++.h>
using namespace std;

class Stack {
    queue<int> q1, q2;

public:
    void pop()
    {
        if (q1.empty())
            return;

        // Leave one element in q1 and
        // push others in q2.
        while (q1.size() != 1) {
            q2.push(q1.front());
            q1.pop();
        }

        // Pop the only left element
        // from q1
        q1.pop();

        // swap the names of two queues
        queue<int> q = q1;
        q1 = q2;
        q2 = q;
    }

    void push(int x) { q1.push(x); }

    int top()
    {
        if (q1.empty())
            return -1;

        while (q1.size() != 1) {
            q2.push(q1.front());
            q1.pop();
        }

        // last pushed element
        int temp = q1.front();

        // to empty the auxiliary queue after
        // last operation
        q1.pop();

        // push last element to q2
        q2.push(temp);

        // swap the two queues names
        queue<int> q = q1;
        q1 = q2;
        q2 = q;
        return temp;
    }

    int size() { return q1.size(); }
};

// Driver code
int main()
{
    Stack s;
    s.push(1);
    s.push(2);
    s.push(3);

    cout << "current size: " << s.size() << endl;
    cout << s.top() << endl;
    s.pop();
    cout << s.top() << endl;
    s.pop();
    cout << s.top() << endl;
    cout << "current size: " << s.size() << endl;
    return 0;
}
// This code is contributed by Chhavi

Java

/* Java Program to implement a stack
using two queue */
import java.util.*;

class Stack {
    Queue<Integer> q1 = new LinkedList<>(),
                   q2 = new LinkedList<>();

    void remove()
    {
        if (q1.isEmpty())
            return;

        // Leave one element in q1 and
        // push others in q2.
        while (q1.size() != 1) {
            q2.add(q1.peek());
            q1.remove();
        }

        // Pop the only left element
        // from q1
        q1.remove();

        // swap the names of two queues
        Queue<Integer> q = q1;
        q1 = q2;
        q2 = q;
    }

    void add(int x) { q1.add(x); }

    int top()
    {
        if (q1.isEmpty())
            return -1;

        while (q1.size() != 1) {
            q2.add(q1.peek());
            q1.remove();
        }

        // last pushed element
        int temp = q1.peek();

        // to empty the auxiliary queue after
        // last operation
        q1.remove();

        // push last element to q2
        q2.add(temp);

        // swap the two queues names
        Queue<Integer> q = q1;
        q1 = q2;
        q2 = q;
        return temp;
    }

    int size() { return q1.size(); }

    // Driver code
    public static void main(String[] args)
    {
        Stack s = new Stack();
        s.add(1);
        s.add(2);
        s.add(3);

        System.out.println("current size: " + s.size());
        System.out.println(s.top());
        s.remove();
        System.out.println(s.top());
        s.remove();
        System.out.println(s.top());
        System.out.println("current size: " + s.size());
    }
}

// This code is contributed by Princi Singh

Python3

# Program to implement a stack using
# two queue
from _collections import deque


class Stack:

    def __init__(self):

        # Two inbuilt queues
        self.q1 = deque()
        self.q2 = deque()

    def push(self, x):
        self.q1.append(x)

    def pop(self):
        # if no elements are there in q1
        if (not self.q1):
            return
        # Leave one element in q1 and push others in q2
        while(len(self.q1) != 1):
            self.q2.append(self.q1.popleft())

        # swap the names of two queues
        self.q1, self.q2 = self.q2, self.q1

    def top(self):
        # if no elements are there in q1
        if (not self.q1):
            return
        # Leave one element in q1 and push others in q2
        while(len(self.q1) != 1):
            self.q2.append(self.q1.popleft())

        # Pop the only left element from q1 to q2
        top = self.q1[0]
        self.q2.append(self.q1.popleft())

        # swap the names of two queues
        self.q1, self.q2 = self.q2, self.q1

        return top

    def size(self):
        return len(self.q1)


# Driver Code
if __name__ == '__main__':
    s = Stack()
    s.push(1)
    s.push(2)
    s.push(3)

    print("current size: ", s.size())
    print(s.top())
    s.pop()
    print(s.top())
    s.pop()
    print(s.top())

    print("current size: ", s.size())

# This code is contributed by jainlovely450

C#

using System;
using System.Collections;
class GfG {
    public class Stack {
        public Queue q1 = new Queue();
        public Queue q2 = new Queue();
        // Just enqueue the new element to q1
        public void Push(int x) = > q1.Enqueue(x);

        // move all elements from q1 to q2 except the rear
        // of q1. Store the rear of q1 swap q1 and q2 return
        // the stored result
        public int Pop()
        {
            if (q1.Count == 0)
                return -1;
            while (q1.Count > 1) {
                q2.Enqueue(q1.Dequeue());
            }
            int res = (int)q1.Dequeue();
            Queue temp = q1;
            q1 = q2;
            q2 = temp;
            return res;
        }

        public int Size() = > q1.Count;

        public int Top()
        {
            if (q1.Count == 0)
                return -1;
            while (q1.Count > 1) {
                q2.Enqueue(q1.Dequeue());
            }
            int res = (int)q1.Dequeue();
            q2.Enqueue(res);
            Queue temp = q1;
            q1 = q2;
            q2 = temp;
            return res;
        }
    };
    public static void Main(String[] args)
    {
        Stack s = new Stack();
        s.Push(1);
        s.Push(2);
        s.Push(3);
        Console.WriteLine("current size: " + s.Size());
        Console.WriteLine(s.Top());
        s.Pop();
        Console.WriteLine(s.Top());
        s.Pop();
        Console.WriteLine(s.Top());
        Console.WriteLine("current size: " + s.Size());
    }
}

// Submitted by Sakti Prasad

Javascript

/*Javascript Program to implement a stack using
two queue */

// Two inbuilt queues
class Stack {
    constructor() {
        this.q1 = [];
        this.q2 = [];
    }

    pop()
    {
        if (this.q1.length == 0)
            return;

        // Leave one element in q1 and
        // push others in q2.
        while (this.q1.length != 1){
            this.q2.push(this.q1[0]);
            this.q1.shift();
        }

        // Pop the only left element
        // from q1f
        this.q1.shift();

        // swap the names of two queues
        this.q = this.q1;
        this.q1 = this.q2;
        this.q2 = this.q;
    }

    push(x) {
        // if no elements are there in q1
        this.q1.push(x);
    }

    top() {
        if (this.q1.length == 0)
            return -1;

        while (this.q1.length != 1) {
            this.q2.push(this.q1[0]);
            this.q1.shift();
        }

        // last pushed element
        let temp = this.q1[0];

        // to empty the auxiliary queue after
        // last operation
        this.q1.shift();

        // push last element to q2
        this.q2.push(temp);

        // swap the two queues names
        this.q = this.q1;
        this.q1 = this.q2;
        this.q2 = this.q;
        return temp;
    }

    size() {
        console.log(this.q1.length);
    }

    isEmpty() {
        // return true if the queue is empty.
        return this.q1.length == 0;
    }

    front() {
        return this.q1[0];
    }
}

// Driver code
let s = new Stack();
s.push(1);
s.push(2);
s.push(3);
console.log("current size: ");
s.size();
console.log(s.top());
s.pop();
console.log(s.top());
s.pop();
console.log(s.top());

console.log("current size: ");
s.size();

// This code is contributed by Susobhan Akhuli

Output

current size: 3
3
2
1
current size: 1

Time Complexity:

• Push operation: O(1), As, on each push operation the new element is added at the end of the Queue.
• Pop operation: O(N), As, on each pop operation, all the elements are popped out from the Queue (q1) except the last element and pushed into the Queue (q2).

Auxiliary Space: O(N) since 2 queues are used.

Implement Stack using 1 queue:

Below is the idea to solve the problem:

Using only one queue and make the queue act as a Stack in modified way of the above discussed approach.

Follow the below steps to implement the idea: 

• The idea behind this approach is to make one queue and push the first element in it. 
• After the first element, we push the next element and then push the first element again and finally pop the first element. 
• So, according to the FIFO rule of the queue, the second element that was inserted will be at the front and then the first element as it was pushed again later and its first copy was popped out. 
• So, this acts as a Stack and we do this at every step i.e. from the initial element to the second last element, and the last element will be the one that we are inserting and since we will be pushing the initial elements after pushing the last element, our last element becomes the first element.

Below is the implementation for the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

// Stack Class that acts as a queue
class Stack {

    queue<int> q;

public:
    void push(int data);
    void pop();
    int top();
    int size();
    bool empty();
};

// Push operation
void Stack::push(int data)
{
    //  Get previous size of queue
    int s = q.size();

    // Push the current element
    q.push(data);

    // Pop all the previous elements and put them after
    // current element

    for (int i = 0; i < s; i++) {
        // Add the front element again
        q.push(q.front());

        // Delete front element
        q.pop();
    }
}

// Removes the top element
void Stack::pop()
{
    if (q.empty())
        cout << "No elements\n";
    else
        q.pop();
}

// Returns top of stack
int Stack::top() { return (q.empty()) ? -1 : q.front(); }

// Returns true if Stack is empty else false
bool Stack::empty() { return (q.empty()); }

int Stack::size() { return q.size(); }

int main()
{
    Stack st;
    st.push(1);
    st.push(2);
    st.push(3);
    cout << "current size: " << st.size() << "\n";
    cout << st.top() << "\n";
    st.pop();
    cout << st.top() << "\n";
    st.pop();
    cout << st.top() << "\n";
    cout << "current size: " << st.size();
    return 0;
}

Java

import java.util.*;

/* Java Program to implement a stack
using only one queue */

class Stack {
    // One queue
    Queue<Integer> q1 = new LinkedList<Integer>();

    void push(int x)
    {
        //  Get previous size of queue
        int s = q1.size();

        // Push the current element
        q1.add(x);

        // Pop all the previous elements and put them after
        // current element
        for (int i = 0; i < s; i++) {
            q1.add(q1.remove());
        }
    }

    void pop()
    {
        // if no elements are there in q1
        if (q1.isEmpty())
            return;
        q1.remove();
    }

    int top()
    {
        if (q1.isEmpty())
            return -1;
        return q1.peek();
    }

    int size() { return q1.size(); }

    // driver code
    public static void main(String[] args)
    {
        Stack s = new Stack();
        s.push(1);
        s.push(2);
        s.push(3);

        System.out.println("current size: " + s.size());
        System.out.println(s.top());
        s.pop();
        System.out.println(s.top());
        s.pop();
        System.out.println(s.top());

        System.out.println("current size: " + s.size());
    }
}

// This code is contributed by Vishal Singh Shekhawat

Python3

from _collections import deque

# Stack Class that acts as a queue


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

    # Push operation
    def push(self, data):
        # Get previous size of queue
        s = len(self.q)

        # Push the current element
        self.q.append(data)

        # Pop all the previous elements and put them after
        # current element
        for i in range(s):
            self.q.append(self.q.popleft())

    # Removes the top element
    def pop(self):
        if (not self.q):
            print("No elements")
        else:
            self.q.popleft()

    # Returns top of stack
    def top(self):
        if (not self.q):
            return
        return self.q[0]

    def size(self):
        return len(self.q)


if __name__ == '__main__':
    st = Stack()
    st.push(1)
    st.push(2)
    st.push(3)
    print("current size: ", st.size())
    print(st.top())
    st.pop()
    print(st.top())
    st.pop()
    print(st.top())
    print("current size: ", st.size())

C#

/* C# Program to implement a stack using only one queue */
using System;
using System.Collections;

class GfG {

  public class Stack
  {

    // One inbuilt queue
    public Queue q = new Queue();

    public void push(int x)
    {
      // Get previous size of queue
      int s = q.Count;

      // Push the current element
      q.Enqueue(x);

      // Pop all the previous elements and put them
      // afte current element
      for (int i = 0; i < s; i++) {
        // Add the front element again
        q.Enqueue(q.Peek());

        // Delete front element
        q.Dequeue();
      }
    }

    // Removes the top element
    public void pop()
    {
      // if no elements are there in q
      if (q.Count == 0)
        Console.WriteLine("No elements");
      else
        q.Dequeue();
    }

    // Returns top of stack
    public int top()
    {
      if (q.Count == 0)
        return -1;
      return (int)q.Peek();
    }

    public int size() { return q.Count; }
  };

  // Driver code
  public static void Main(String[] args)
  {
    Stack st = new Stack();
    st.push(1);
    st.push(2);
    st.push(3);
    Console.WriteLine("current size: " + st.size());
    Console.WriteLine(st.top());
    st.pop();
    Console.WriteLine(st.top());
    st.pop();
    Console.WriteLine(st.top());
    Console.WriteLine("current size: " + st.size());
  }
}

// This code is contributed by Susobhan Akhuli

Javascript

/*Javascript Program to implement a stack using
only one queue */

// One inbuilt queue
class Stack {
    constructor() {
        this.q = [];
    }

    // Push operation
    push(data) {

        //  Get previous size of queue
        let s = this.q.length;

        // Push the current element
        this.q.push(data);

        // Pop all the previous elements and put them after
        // current element
        for (let i = 0; i < s; i++) {
            // Add the front element again
            this.q.push(this.q[0]);

            // Delete front element
            this.q.shift();

        }
    }

    // Removes the top element
    pop() {
        // if no elements are there in q1
        if (this.q.length == 0)
            console.log("No elements");
        else
            this.q.shift();
    }

    top() {
        if (this.q.length == 0)
            return -1;
        return this.q[0];
    }

    size() {
        console.log(this.q.length);
    }

    isEmpty() {
        // return true if the queue is empty.
        return this.q.length == 0;
    }

    front() {
        return this.q[0];
    }
}

// Driver code


let st = new Stack();
st.push(1);
st.push(2);
st.push(3);

console.log("current size: ");
st.size();
console.log(st.top());
st.pop();
console.log(st.top());
st.pop();
console.log(st.top());

console.log("current size: ");
st.size();

// This code is contributed by Susobhan Akhuli

Output

current size: 3
3
2
1
current size: 1

Time Complexity:

• Push operation: O(N)
• Pop operation: O(1)

Auxiliary Space: O(N) since 1 queue is used.

Recursive Method:

Below is the implementation for the above approach using recursion –

C++

// CPP Program to implement a stack
// using one queue and recursion
#include <bits/stdc++.h>
using namespace std;

// Stack Class that acts as a queue
class Stack {
    queue<int> q;

public:
    void push(int data, int c);
    void pop();
    int top();
    int size();
    bool empty();
};

// Push operation
void Stack::push(int data, int c)
{
    // Push the current element first and
    // After every recursion add the front element again
    q.push(data);

    // Return if size becomes 0
    if (c <= 0)
        return;

    // Store current front
    int x = q.front();

    // Delete front element
    q.pop();

    // Decrement size by 1 in every recursion
    c--;
    Stack::push(x, c);
}

// Removes the top element
void Stack::pop()
{
    if (q.empty())
        cout << "No elements\n";
    else
        q.pop();
}

// Returns top of stack
int Stack::top() { return (q.empty()) ? -1 : q.front(); }

// Returns true if Stack is empty else false
bool Stack::empty() { return (q.empty()); }

int Stack::size() { return q.size(); }

int main()
{
    Stack st;
    st.push(1, st.size()); // Value and size
    st.push(2, st.size());
    st.push(3, st.size());
    cout << "current size: " << st.size() << "\n";
    cout << st.top() << "\n";
    st.pop();
    cout << st.top() << "\n";
    st.pop();
    cout << st.top() << "\n";
    cout << "current size: " << st.size();
    return 0;
}

// This code is contributed by Susobhan Akhuli

Java

import java.util.*;

/* Java Program to implement a stack
using only one queue */

class Stack {
    // One queue
    Queue<Integer> q1 = new LinkedList<Integer>();

    void push(int data, int c)
    {

        // Push the current element first and
        // After every recursion add the front element again
        q1.add(data);

        // Return if size becomes 0
        if (c <= 0)
            return;

        // Decrement size by 1 in every recursion
        c--;

        // remove front element from queue and return it
        // using q1.remove() and call recursive function
        push(q1.remove(), c);
    }

    void pop()
    {
        // if no elements are there in q1
        if (q1.isEmpty())
            return;
        q1.remove();
    }

    int top()
    {
        if (q1.isEmpty())
            return -1;
        return q1.peek();
    }

    int size() { return q1.size(); }

    // driver code
    public static void main(String[] args)
    {
        Stack s = new Stack();
        s.push(1, s.size()); // Value and current size
        s.push(2, s.size());
        s.push(3, s.size());

        System.out.println("current size: " + s.size());
        System.out.println(s.top());
        s.pop();
        System.out.println(s.top());
        s.pop();
        System.out.println(s.top());

        System.out.println("current size: " + s.size());
    }
}

// This code is contributed by Susobhan Akhuli

Python3

from _collections import deque

# Stack Class that acts as a queue


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

    # Push operation
    def push(self, data, c):

        # Push the current element
        self.q.append(data)

        # Return if size becomes 0
        if c <= 0:
            return

        # Store and then pop the current front
        x = self.q.popleft()

        # Decrement size by 1 in every recursion
        c = c-1
        self.push(x, c)

    # Removes the top element
    def pop(self):
        if (not self.q):
            print("No elements")
        else:
            self.q.popleft()

    # Returns top of stack
    def top(self):
        if (not self.q):
            return
        return self.q[0]

    def size(self):
        return len(self.q)


if __name__ == '__main__':
    st = Stack()
    st.push(1, st.size())
    st.push(2, st.size())
    st.push(3, st.size())
    print("current size: ", st.size())
    print(st.top())
    st.pop()
    print(st.top())
    st.pop()
    print(st.top())
    print("current size: ", st.size())

# This code is contributed by Susobhan Akhuli

C#

// C# Program to implement a stack
// using one queue and recursion

using System;
using System.Collections;

class GfG {

    public class Stack {
        // One inbuilt queue
        public Queue q = new Queue();

        // Push operation
        public void push(int x, int c)
        {

            // Push the current element first and
            // After every recursion add the front element
            // again
            q.Enqueue(x);

            // Return if size becomes 0
            if (c <= 0)
                return;

            // Store current front
            int p = (int)q.Peek();

            // Delete front element
            q.Dequeue();

            // Decrement size by 1 in every recursion
            c--;
            push(p, c);
        }

        // Removes the top element
        public void pop()
        {
            // if no elements are there in q
            if (q.Count == 0)
                Console.WriteLine("No elements");
            else
                q.Dequeue();
        }

        // Returns top of stack
        public int top()
        {
            if (q.Count == 0)
                return -1;
            return (int)q.Peek();
        }

        public int size() { return q.Count; }
    };

    // Driver code
    public static void Main(String[] args)
    {
        Stack st = new Stack();
        st.push(1, st.size());
        st.push(2, st.size());
        st.push(3, st.size());
        Console.WriteLine("current size: " + st.size());
        Console.WriteLine(st.top());
        st.pop();
        Console.WriteLine(st.top());
        st.pop();
        Console.WriteLine(st.top());
        Console.WriteLine("current size: " + st.size());
    }
}

// This code is contributed by Susobhan Akhuli

Javascript

// Javascript Program to implement a stack using one queue and recursion

      // Stack Class that acts as a queue
      class Stack {
        constructor() {
          this.q = [];
        }

        // Push operation
        push(data, c) {
          // Push the current element first and
          //After every recursion add the front element again
          this.q.push(data);

          //Returns if size becomes 0
          if (c <= 0) {
            return;
          }

          //Store Current Front
          let x = this.q[0];

          //Delete front element
          this.q.shift();

          //Decrease size by 1 in every recursion
          c--;
          this.push(x, c);
        }

        // Removes the top element
        pop() {
          if (this.q.length == 0) console.log("No elements");
          else this.q.shift();
        }

        //Return top of stack
        top() {
          if (this.q.length == 0) return -1;
          return this.q[0];
        }

        // return true if the stack is empty else false.
        isEmpty() {
          return this.q.length == 0;
        }

        size() {
          return this.q.length;
        }
      }

      // Driver code
      let st = new Stack();
      st.push(1, st.size()); //value and size
      st.push(2, st.size());
      st.push(3, st.size());

      console.log("current size: " + st.size());
      console.log(st.top());
      st.pop();
      console.log(st.top());
      st.pop();
      console.log(st.top());

      console.log("current size: " + st.size());

      // This code is contributed by satwiksuman.

Output

current size: 3
3
2
1
current size: 1

Time Complexity:

• Push operation: O(N)
• Pop operation: O(1)

Auxiliary Space: O(N) since 1 queue is used and also for the stack used for recursion.
References:
Implement Stack using Two Queues
This article was compiled by Sumit Jain and reviewed by the GeeksforGeeks team. Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.