Question

Data Structures

Java Language

include also (Test class ) to test the methods

Q2) Using a single queue (linkedQueue), re-implement the concept of Stack ADT, what is the complexity of the method push, pop, top, isEmpty, and size. You should not use any extra data structure.

Answer 1

// If u have any doubt feel free to ask in the comments

// I will be very happy to solve them

// Standard libraries
import java.util.*; 
import java.util.Scanner; 
public class Main 
{ 
        Queue<Integer> q = new LinkedList<Integer>(); 
        // Push operation 
        void push(int val) 
        { 
                // get previous size of queue 
                int size = q.size(); 
                
                // Add current element 
                q.add(val); 
                // Removing all previous elements and adding them after this element 
                for (int i = 0; i < size; i++) 
                { 
                        int x = q.remove(); 
                        q.add(x); 
                } 
        }
        // Removes the top element i.e pop() 
        int pop() 
        { 
                if (q.isEmpty()) 
                { 
                        System.out.println("No elements"); 
                        return -1; 
                } 
                int x = q.remove(); 
                return x; 
        } 
    void peekMethod()
    {
      if(q.isEmpty())
          System.out.println("Underflow");
      else
          System.out.println("Topmost element is without removing it : "+q.peek());}

        // Returns top of stack 
        int top() 
        { 
                if (q.isEmpty()) 
                        return -1; 
                return q.peek(); 
        } 
        // Returns true if Stack is empty else false 
        boolean isEmpty() 
        { 
                return q.isEmpty(); 
        } 
        // Test class to test all above methods
        public static void main(String[] args) 
        { 
                Main s = new Main(); 
                s.push(11); // calling push function for inserting a new element
                s.push(26); 
        s.peekMethod();
                System.out.println("Top element :" + s.top()); 
                s.pop(); 
                s.push(98); 
                s.pop(); 
                System.out.println("Top element :" + s.top()); 
        } 
} 

SCREENSHOTS:

8 14 Main.java 1 // Standard Libraries 2 import java.util.*; 3 import java.util.Scanner; 4 public class Main 5-{| 6 Queue Integer> q = new LinkedList<Integer>(); 7 // Push operation void push(int val) 9 { 10 // get previous size of queue 11 int size q.size(); 12 13 // Add current element q.add(val); 15 // Removing all previous elements and adding them after this element 16 for (int i = 0; i < size; i++) 17 { 18 int x = q.remove(); 19 q.add(x); 20 } 21 } 22 // Removes the top element i.e pop() 23 int pop() 24 { 25 if (q.isEmpty()) 26 { 27 System.out.println("No elements"); 28 return -1; 29 } 30 int x = 4.rem remove();

31 32 33 34 35 36 37 38 39 40 41 42 43 return x; } // Returns top of stack int top() { if (q.isEmpty()) return -1; return q. peek(); } // Returns true if Stack is empty else false boolean isEmpty() { return q.isEmpty(); } // Test class to test all above methods public static void main(String[] args) { Main s = new Main(); s.push(11); // calling push function for inserting a new element s.push(26); System.out.println("Top element :" + s. s.top()); s.pop(); 5.push(98); s.pop(); System.out.println("Top element :" + s. s.top()); 44 45 46 47 48 49 50 51 52 53 54 55 56 57 } 58
OUTPUT:

Top element :26 Top element :11 .Program finished with exit code 0 Press ENTER to exit console.

Complexity Analysis::::::::

Time Complexity-----

push() function is taking O(n) time while pop() and top() functions require constant time only i.e. O(1). Because for pushing an element we remove and add the number of elements which were already present in it. This makes the operation to run in polynomial time.

Space Complexity-----

O(n) because we are using space to store n number of elements.