Implement Stack using Queues
Last Updated :
05 Aug, 2024
Given a Queue data structure that supports standard operations like enqueue() and dequeue(). The task is to implement a Stack data structure using Queue.
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;
}
/* 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
# 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
/*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
Outputcurrent 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;
}
/* 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
# 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
/*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
Outputcurrent 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 single 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)
{
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();
}
}
void pop()
{
if (q.empty())
cout << "No elements\n";
else
q.pop();
}
int top() { return (q.empty()) ? -1 : q.front(); }
int size() { return q.size(); }
bool empty() { return (q.empty()); }
};
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;
}
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
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# 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 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
Outputcurrent 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.
