# Implement Stack using Queues

• Difficulty Level : Medium
• Last Updated : 25 Jul, 2022

We are given a Queue data structure that supports standard operations like enqueue() and dequeue(). We need 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 be ‘q1’ and ‘q2’. Stack ‘s’ can be implemented in two ways:
Method 1 (By making push operation costly)
This method ensures that the newly entered element is always at the front of ‘q1’ so that pop operation dequeues from ‘q1’. ‘q2’ is used to put every new element in front of ‘q1’.

1. push(s, x) operation’s steps are described below:
• Enqueue x to q2
• One by one dequeue everything from q1 and enqueue to q2.
• Swap the names of q1 and q2
2. pop(s) operation’s function is described below:
• 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 ` `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 q1 = ``new` `LinkedList();``        ``static` `Queue q2 = ``new` `LinkedList();` `        ``// 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 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`

Output

```current size: 3
3
2
1
current size: 1```

Complexity Analysis:

• 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.

Method 2 (By making pop operation costly):
In a push operation, 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.

1. push(s, x) operation:
• Enqueue x to q1 (assuming size of q1 is unlimited).
2. 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 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 ``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);``    ``s.push(4);` `    ``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 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 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 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``);``        ``s.add(``4``);` `        ``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``)``    ``s.push(``4``)` `    ``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);``        ``s.Push(4);``        ``Console.WriteLine(``"Size of Stack: "` `+ s.Size());``        ``Console.WriteLine(s.Top());``        ``s.Pop();``        ``Console.WriteLine(s.Top());``        ``s.Pop();``        ``Console.WriteLine(s.Top());``        ``Console.WriteLine(``"Size of Stack: "` `+ s.Size());``    ``}``}` `// Submitted by Sakti Prasad`

Output

```current size: 4
4
3
2
current size: 2```

Complexity Analysis:

• 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 it into the Queue (q2).
• Auxiliary Space: O(N) since 2 queues are used.

Method 3 (Using only 1 queue):

In this method, we will be using only one queue and make the queue act as a stack. 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 queue, 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 which we are inserting and since we will be pushing the initial elements after pushing the last element, our last element becomes the first element.

## C++

 `#include ``using` `namespace` `std;` `// Stack Class that acts as a queue``class` `Stack {` `    ``queue<``int``> q;` `public``:``    ``void` `push(``int` `data);``    ``void` `pop();``    ``int` `top();``    ``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` `main()``{``    ``Stack st;``    ``st.push(40);``    ``st.push(50);``    ``st.push(70);``    ``cout << st.top() << ``"\n"``;``    ``st.pop();``    ``cout << st.top() << ``"\n"``;``    ``st.pop();``    ``cout << st.top() << ``"\n"``;``    ``st.push(80);``    ``st.push(90);``    ``st.push(100);``    ``cout << st.top() << ``"\n"``;``    ``st.pop();``    ``cout << st.top() << ``"\n"``;``    ``st.pop();``    ``cout << st.top() << ``"\n"``;``    ``return` `0;``}`

## 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``]` `      ` `if` `__name__ ``=``=` `'__main__'``:``    ``st ``=` `Stack()``    ``st.push(``40``)``    ``st.push(``50``)``    ``st.push(``70``)``    ``print``(st.top())``    ``st.pop()``    ``print``(st.top())``    ``st.pop()``    ``print``(st.top())``    ``st.push(``80``)``    ``st.push(``90``)``    ``st.push(``100``)``    ``print``(st.top())``    ``st.pop()``    ``print``(st.top())``    ``st.pop()``    ``print``(st.top())`

Output

```70
50
40
100
90
80```

Complexity Analysis:

• Time Complexity:
• Push operation: O(N)
• Pop operation: O(1)
• Space Complexity: O(N) since 1 queue is used.

References:
Implement Stack using Two Queues