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 |
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 |
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 |
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. |
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.
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