Implement Stack using Queues
Last Updated :
06 Feb, 2024
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.
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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:Â
- 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:Â
- Dequeue an item from q1 and return it.
Below is the implementation of the above approach.Â
C++
#include <bits/stdc++.h>
using namespace std;
class Stack {
queue< int > q1, q2;
public :
void push( int x)
{
q2.push(x);
while (!q1.empty()) {
q2.push(q1.front());
q1.pop();
}
queue< int > q = q1;
q1 = q2;
q2 = q;
}
void pop()
{
if (q1.empty())
return ;
q1.pop();
}
int top()
{
if (q1.empty())
return -1;
return q1.front();
}
int size() { return q1.size(); }
};
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
import java.util.*;
class GfG {
static class Stack {
static Queue<Integer> q1
= new LinkedList<Integer>();
static Queue<Integer> q2
= new LinkedList<Integer>();
static int curr_size;
static void push( int x)
{
q2.add(x);
while (!q1.isEmpty()) {
q2.add(q1.peek());
q1.remove();
}
Queue<Integer> q = q1;
q1 = q2;
q2 = q;
}
static void pop()
{
if (q1.isEmpty())
return ;
q1.remove();
}
static int top()
{
if (q1.isEmpty())
return - 1 ;
return q1.peek();
}
static int size() { return q1.size(); }
}
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());
}
}
|
Python3
from _collections import deque
class Stack:
def __init__( self ):
self .q1 = deque()
self .q2 = deque()
def push( self , x):
self .q2.append(x)
while ( self .q1):
self .q2.append( self .q1.popleft())
self .q1, self .q2 = self .q2, self .q1
def pop( self ):
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)
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())
|
C#
using System;
using System.Collections;
class GfG {
public class Stack {
public Queue q1 = new Queue();
public Queue q2 = new Queue();
public void push( int x)
{
q2.Enqueue(x);
while (q1.Count > 0) {
q2.Enqueue(q1.Peek());
q1.Dequeue();
}
Queue q = q1;
q1 = q2;
q2 = q;
}
public void pop()
{
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; }
};
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());
}
}
|
Javascript
class Stack {
constructor() {
this .q1 = [];
this .q2 = [];
}
push(x) {
this .q2.push(x);
while ( this .q1.length != 0) {
this .q2.push( this .q1[0]);
this .q1.shift();
}
this .q = this .q1;
this .q1 = this .q2;
this .q2 = this .q;
}
pop() {
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 this .q1.length == 0;
}
front() {
return this .q1[0];
}
}
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();
|
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:Â
- Enqueue x to q1 (assuming the size of q1 is unlimited).
- Follow the below steps to implement the pop(s) operation:Â
- 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++
#include <bits/stdc++.h>
using namespace std;
class Stack {
queue< int > q1, q2;
public :
void pop()
{
if (q1.empty())
return ;
while (q1.size() != 1) {
q2.push(q1.front());
q1.pop();
}
q1.pop();
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();
}
int temp = q1.front();
q1.pop();
q2.push(temp);
queue< int > q = q1;
q1 = q2;
q2 = q;
return temp;
}
int size() { return q1.size(); }
};
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
import java.util.*;
class Stack {
Queue<Integer> q1 = new LinkedList<>(),
q2 = new LinkedList<>();
void remove()
{
if (q1.isEmpty())
return ;
while (q1.size() != 1 ) {
q2.add(q1.peek());
q1.remove();
}
q1.remove();
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();
}
int temp = q1.peek();
q1.remove();
q2.add(temp);
Queue<Integer> q = q1;
q1 = q2;
q2 = q;
return temp;
}
int size() { return q1.size(); }
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());
}
}
|
Python3
from _collections import deque
class Stack:
def __init__( self ):
self .q1 = deque()
self .q2 = deque()
def push( self , x):
self .q1.append(x)
def pop( self ):
if ( not self .q1):
return
while ( len ( self .q1) ! = 1 ):
self .q2.append( self .q1.popleft())
self .q1, self .q2 = self .q2, self .q1
def top( self ):
if ( not self .q1):
return
while ( len ( self .q1) ! = 1 ):
self .q2.append( self .q1.popleft())
top = self .q1[ 0 ]
self .q2.append( self .q1.popleft())
self .q1, self .q2 = self .q2, self .q1
return top
def size( self ):
return len ( self .q1)
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())
|
C#
using System;
using System.Collections;
class GfG {
public class Stack {
public Queue q1 = new Queue();
public Queue q2 = new Queue();
public void Push( int x) = > q1.Enqueue(x);
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());
}
}
|
Javascript
class Stack {
constructor() {
this .q1 = [];
this .q2 = [];
}
pop()
{
if ( this .q1.length == 0)
return ;
while ( this .q1.length != 1){
this .q2.push( this .q1[0]);
this .q1.shift();
}
this .q1.shift();
this .q = this .q1;
this .q1 = this .q2;
this .q2 = this .q;
}
push(x) {
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();
}
let temp = this .q1[0];
this .q1.shift();
this .q2.push(temp);
this .q = this .q1;
this .q1 = this .q2;
this .q2 = this .q;
return temp;
}
size() {
console.log( this .q1.length);
}
isEmpty() {
return this .q1.length == 0;
}
front() {
return this .q1[0];
}
}
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();
|
Output
current size: 3
3
2
1
current size: 1
Time Complexity:Â
- 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:Â
- The idea behind this approach is to make one queue and push the first element in it.Â
- After the first element, we push the next element and then push the first element again and finally pop the first element.Â
- 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.Â
- 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;
class Stack {
queue< int > q;
public :
void push( int data);
void pop();
int top();
int size();
bool empty();
};
void Stack::push( int data)
{
int s = q.size();
q.push(data);
for ( int i = 0; i < s; i++) {
q.push(q.front());
q.pop();
}
}
void Stack::pop()
{
if (q.empty())
cout << "No elements\n" ;
else
q.pop();
}
int Stack::top() { return (q.empty()) ? -1 : q.front(); }
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.*;
class Stack {
Queue<Integer> q1 = new LinkedList<Integer>();
void push( int x)
{
int s = q1.size();
q1.add(x);
for ( int i = 0 ; i < s; i++) {
q1.add(q1.remove());
}
}
void pop()
{
if (q1.isEmpty())
return ;
q1.remove();
}
int top()
{
if (q1.isEmpty())
return - 1 ;
return q1.peek();
}
int size() { return q1.size(); }
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());
}
}
|
Python3
from _collections import deque
class Stack:
def __init__( self ):
self .q = deque()
def push( self , data):
s = len ( self .q)
self .q.append(data)
for i in range (s):
self .q.append( self .q.popleft())
def pop( self ):
if ( not self .q):
print ( "No elements" )
else :
self .q.popleft()
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#
using System;
using System.Collections;
class GfG {
public class Stack
{
public Queue q = new Queue();
public void push( int x)
{
int s = q.Count;
q.Enqueue(x);
for ( int i = 0; i < s; i++) {
q.Enqueue(q.Peek());
q.Dequeue();
}
}
public void pop()
{
if (q.Count == 0)
Console.WriteLine( "No elements" );
else
q.Dequeue();
}
public int top()
{
if (q.Count == 0)
return -1;
return ( int )q.Peek();
}
public int size() { return q.Count; }
};
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());
}
}
|
Javascript
class Stack {
constructor() {
this .q = [];
}
push(data) {
let s = this .q.length;
this .q.push(data);
for (let i = 0; i < s; i++) {
this .q.push( this .q[0]);
this .q.shift();
}
}
pop() {
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 this .q.length == 0;
}
front() {
return this .q[0];
}
}
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();
|
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++
#include <bits/stdc++.h>
using namespace std;
class Stack {
queue< int > q;
public :
void push( int data, int c);
void pop();
int top();
int size();
bool empty();
};
void Stack::push( int data, int c)
{
q.push(data);
if (c <= 0)
return ;
int x = q.front();
q.pop();
c--;
Stack::push(x, c);
}
void Stack::pop()
{
if (q.empty())
cout << "No elements\n" ;
else
q.pop();
}
int Stack::top() { return (q.empty()) ? -1 : q.front(); }
bool Stack::empty() { return (q.empty()); }
int Stack::size() { return q.size(); }
int main()
{
Stack st;
st.push(1, st.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;
}
|
Java
import java.util.*;
class Stack {
Queue<Integer> q1 = new LinkedList<Integer>();
void push( int data, int c)
{
q1.add(data);
if (c <= 0 )
return ;
c--;
push(q1.remove(), c);
}
void pop()
{
if (q1.isEmpty())
return ;
q1.remove();
}
int top()
{
if (q1.isEmpty())
return - 1 ;
return q1.peek();
}
int size() { return q1.size(); }
public static void main(String[] args)
{
Stack s = new Stack();
s.push( 1 , s.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());
}
}
|
Python3
from _collections import deque
class Stack:
def __init__( self ):
self .q = deque()
def push( self , data, c):
self .q.append(data)
if c < = 0 :
return
x = self .q.popleft()
c = c - 1
self .push(x, c)
def pop( self ):
if ( not self .q):
print ( "No elements" )
else :
self .q.popleft()
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())
|
C#
using System;
using System.Collections;
class GfG {
public class Stack {
public Queue q = new Queue();
public void push( int x, int c)
{
q.Enqueue(x);
if (c <= 0)
return ;
int p = ( int )q.Peek();
q.Dequeue();
c--;
push(p, c);
}
public void pop()
{
if (q.Count == 0)
Console.WriteLine( "No elements" );
else
q.Dequeue();
}
public int top()
{
if (q.Count == 0)
return -1;
return ( int )q.Peek();
}
public int size() { return q.Count; }
};
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());
}
}
|
Javascript
class Stack {
constructor() {
this .q = [];
}
push(data, c) {
this .q.push(data);
if (c <= 0) {
return ;
}
let x = this .q[0];
this .q.shift();
c--;
this .push(x, c);
}
pop() {
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];
}
isEmpty() {
return this .q.length == 0;
}
size() {
return this .q.length;
}
}
let st = new Stack();
st.push(1, st.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());
|
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.
Additional Methods:
- Using a Deque (Double Ended Queue):
A Deque is a data structure that supports adding and removing elements from both ends in constant time. To implement a Stack using a Deque, we can make use of the addFirst and removeFirst methods to implement push and pop operations respectively.
C++
#include <bits/stdc++.h>
using namespace std;
class Stack {
private :
deque< int > my_deque;
public :
void push( int item)
{
my_deque.push_back(item);
}
int pop()
{
int item = my_deque.back();
my_deque.pop_back();
return item;
}
int size()
{
return my_deque.size();
}
bool is_empty()
{
return my_deque.empty();
}
int top()
{
if (is_empty()) {
return -1;
}
else {
return my_deque.back();
}
}
};
int main()
{
Stack st;
st.push(1);
st.push(2);
st.push(3);
cout << "current size: " << st.size() << endl;
cout << st.top() << endl;
st.pop();
cout << st.top() << endl;
st.pop();
cout << st.top() << endl;
cout << "current size: " << st.size() << endl;
return 0;
}
|
Java
import java.util.*;
class Stack {
Deque<Integer> myDeque = new LinkedList<>();
void push( int item) {
myDeque.addLast(item);
}
int pop() {
int item = myDeque.getLast();
myDeque.removeLast();
return item;
}
int size() {
return myDeque.size();
}
boolean isEmpty() {
return myDeque.isEmpty();
}
int top() {
if (isEmpty()) {
return - 1 ;
}
else {
return myDeque.getLast();
}
}
}
class GFG {
public static void main(String[] args) {
Stack st = new Stack();
st.push( 1 );
st.push( 2 );
st.push( 3 );
System.out.println( "current size: " + st.size());
System.out.println(st.top());
st.pop();
System.out.println(st.top());
st.pop();
System.out.println(st.top());
System.out.println( "current size: " + st.size());
}
}
|
Python3
from collections import deque
class Stack:
def __init__( self ):
self .dequeue = deque()
def push( self , item):
self .dequeue.append(item)
def pop( self ):
return self .dequeue.pop()
def size( self ):
return len ( self .dequeue)
def is_empty( self ):
return not self .dequeue
def top( self ):
if self .is_empty():
return None
else :
return self .dequeue[ - 1 ]
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#
using System;
using System.Collections.Generic;
class Stack {
private LinkedList< int > list = new LinkedList< int >();
public void Push( int item)
{
list.AddLast(item);
}
public int Pop()
{
int item = list.Last.Value;
list.RemoveLast();
return item;
}
public int Size()
{
return list.Count;
}
public bool IsEmpty()
{
return list.Count == 0;
}
public int Top()
{
if (IsEmpty()) {
return -1;
}
else {
return list.Last.Value;
}
}
}
class Program {
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());
}
}
|
Javascript
<script>
class Stack {
constructor() {
this .dequeue = [];
}
push(item) {
this .dequeue.push(item);
}
pop() {
return this .dequeue.pop();
}
size() {
return this .dequeue.length;
}
is_empty() {
return this .dequeue.length == 0;
}
top() {
if ( this .is_empty()) {
return null ;
} else {
return this .dequeue[ this .dequeue.length - 1];
}
}
}
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());
</script>
|
Output
current size: 3
3
2
1
current size: 1
- Using a Circular Queue:
In this method, we use a Circular Queue to implement the Stack. We keep track of the front and rear indices, and whenever we need to push an element, we simply increase the rear index and add the element to the rear position. To pop an element, we simply decrease the rear index.
C++
#include <bits/stdc++.h>
using namespace std;
class Stack {
int front, rear, size;
unsigned capacity;
int * arr;
public :
Stack(unsigned capacity)
{
this ->capacity = capacity;
front = size = 0;
rear = capacity - 1;
arr = new int [ this ->capacity];
}
bool isFull()
{
return (size == capacity);
}
bool isEmpty()
{
return (size == 0);
}
void push( int x)
{
if (isFull())
return ;
rear = (rear + 1) % capacity;
arr[rear] = x;
size++;
}
void pop()
{
if (isEmpty())
return ;
front = (front + 1) % capacity;
size--;
}
int top()
{
if (isEmpty())
return -1;
return arr[front];
}
int getSize()
{
return size;
}
};
int main()
{
Stack s(3);
s.push(1);
s.push(2);
s.push(3);
cout << "current size: " << s.getSize() << endl;
cout << s.top() << endl;
s.pop();
cout << s.top() << endl;
s.pop();
cout << s.top() << endl;
cout << "current size: " << s.getSize() << endl;
return 0;
}
|
Java
import java.util.*;
class Stack {
private int front, rear, size;
private int capacity;
private int [] arr;
public Stack( int capacity)
{
this .capacity = capacity;
front = size = 0 ;
rear = capacity - 1 ;
arr = new int [ this .capacity];
}
public boolean isFull()
{
return (size == capacity);
}
public boolean isEmpty()
{
return (size == 0 );
}
public void push( int x)
{
if (isFull())
return ;
rear = (rear + 1 ) % capacity;
arr[rear] = x;
size++;
}
public void pop()
{
if (isEmpty())
return ;
front = (front + 1 ) % capacity;
size--;
}
public int top()
{
if (isEmpty())
return - 1 ;
return arr[front];
}
public int getSize()
{
return size;
}
}
public class GFG {
public static void main(String[] args) {
Stack s = new Stack( 3 );
s.push( 1 );
s.push( 2 );
s.push( 3 );
System.out.println( "current size: " + s.getSize());
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.getSize());
}
}
|
Python3
class Stack:
def __init__( self , capacity):
self .capacity = capacity
self .front = self .size = 0
self .rear = capacity - 1
self .arr = [ 0 ] * self .capacity
def isFull( self ):
return self .size = = self .capacity
def isEmpty( self ):
return self .size = = 0
def push( self , x):
if self .isFull():
return
self .rear = ( self .rear + 1 ) % self .capacity
self .arr[ self .rear] = x
self .size + = 1
def pop( self ):
if self .isEmpty():
return
self .front = ( self .front + 1 ) % self .capacity
self .size - = 1
def top( self ):
if self .isEmpty():
return - 1
return self .arr[ self .front]
def getSize( self ):
return self .size
if __name__ = = "__main__" :
s = Stack( 3 )
s.push( 1 )
s.push( 2 )
s.push( 3 )
print ( "current size:" , s.getSize())
print (s.top())
s.pop()
print (s.top())
s.pop()
print (s.top())
print ( "current size:" , s.getSize())
|
C#
using System;
class Stack
{
int front, rear, size;
uint capacity;
int [] arr;
public Stack( uint capacity)
{
this .capacity = capacity;
front = size = 0;
rear = ( int )capacity - 1;
arr = new int [capacity];
}
public bool IsFull()
{
return (size == capacity);
}
public bool IsEmpty()
{
return (size == 0);
}
public void Push( int x)
{
if (IsFull())
return ;
rear = (rear + 1) % ( int )capacity;
arr[rear] = x;
size++;
}
public void Pop()
{
if (IsEmpty())
return ;
front = (front + 1) % ( int )capacity;
size--;
}
public int Top()
{
if (IsEmpty())
return -1;
return arr[front];
}
public int GetSize()
{
return size;
}
}
class Program
{
static void Main( string [] args)
{
Stack s = new Stack(3);
s.Push(1);
s.Push(2);
s.Push(3);
Console.WriteLine( "current size: " + s.GetSize());
Console.WriteLine(s.Top());
s.Pop();
Console.WriteLine(s.Top());
s.Pop();
Console.WriteLine(s.Top());
Console.WriteLine( "current size: " + s.GetSize());
}
}
|
Javascript
class Stack {
constructor(capacity) {
this .capacity = capacity;
this .front = this .size = 0;
this .rear = this .capacity - 1;
this .arr = new Array( this .capacity);
}
isFull() {
return this .size === this .capacity;
}
isEmpty() {
return this .size === 0;
}
push(x) {
if ( this .isFull())
return ;
this .rear = ( this .rear + 1) % this .capacity;
this .arr[ this .rear] = x;
this .size++;
}
pop() {
if ( this .isEmpty())
return ;
this .front = ( this .front + 1) % this .capacity;
this .size--;
}
top() {
if ( this .isEmpty())
return -1;
return this .arr[ this .front];
}
getSize() {
return this .size;
}
}
let s = new Stack(3);
s.push(1);
s.push(2);
s.push(3);
console.log( "current size: " + s.getSize());
console.log(s.top());
s.pop();
console.log(s.top());
s.pop();
console.log(s.top());
console.log( "current size: " + s.getSize());
|
Output
current size: 3
1
2
3
current size: 1
References:Â
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.
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