Infix to Postfix Conversion Using Stack.

Converting an infix expression to a postfix expression involves rearranging the operators and operands to a postfix format. We can perform this operation using the Stack data structure by traversing the infix expression from left to right. But before moving to the algorithm part, let's quickly understand infix and postfix expressions.

Infix Expression:

An Infix Expression is a mathematical expression where the operators are placed between operands. For example, A + (B * C) is an Infix expression, where the operators (+ and *) are placed between the operands (A, B, and C). In infix notation, parentheses are used to determine the order of operations.

Postfix Expression:

A Postfix expression, also known as Reverse Polish Notation (RPN), is a mathematical expression where the operators follow their operands. For example, ABC*+ is a Postfix expression equivalent to the infix expression A + (B * C). In postfix notation, parentheses are not required since the order of operations is determined explicitly by the arrangement of operands and operators.

Example of Infix to Postfix Conversion.

Let’s convert the infix expression ((A + B) * C - D) / E to postfix.

1. Expression: ((A + B) * C - D) / E
2. Stack: (Empty)
3. Postfix String: (Empty)

Iterating through the expression:

• Character: ( → Push onto the stack.
• Character: ( → Push onto the stack.
• Character: A → Append to postfix string.
• Character: + → Push onto the stack.
• Character: B → Append to postfix string.
• Character: ) → Pop and append operators (+) from the stack to the postfix string.
• Character: * → Push onto the stack.
• Character: C → Append to postfix string.
• Character: - → Push onto the stack.
• Character: D → Append to postfix string.
• Character: ) → Pop and append operators (-, *) from the stack to the postfix string.
• Character: / → Push onto the stack.
• Character: E → Append to postfix string.
• Empty the Stack: Pop and append the remaining operators (/) to the postfix string.

Resulting Postfix Expression:

The infix expression ((A + B) * C - D) / E converts to the postfix expression AB+C*DE-/.

Algorithm for Infix to Postfix Conversion.

1. Initialize an empty stack to hold operators.
2. Initialize an empty string to store the postfix expression.
3. Iterate through the infix expression:
   • For each character:
   • If it’s an operand (alphabets or digits), add it to the postfix string.
   • If it’s an operator:
     • While the stack is not empty and the precedence of the current operator is less than or equal to the precedence of the operator at the top of the stack:
     • Pop and append the top of the stack to the postfix string.
     • Push the current operator onto the stack.
   • If it’s an opening parenthesis ‘(’, push it onto the stack.
   • If it’s a closing parenthesis ‘)’:
     • Pop and append operators from the stack to the postfix string until an opening parenthesis is encountered.
     • Discard the opening parenthesis.
4. Pop and append any remaining operators from the stack to the postfix string.

Below is the code implementation for the above algorithm:

C++ Java Python C#

// C++ program for infix to postfix conversion using stack.
#include <iostream>
#include <stack>
#include <string>
#include <unordered_map>
using namespace std;

bool isOperator(char c) {
    return (c == '+' || c == '-' || c == '*' || c == '/');
}

int getPrecedence(char c) {
    if (c == '+' || c == '-')
        return 1;
    if (c == '*' || c == '/')
        return 2;
    return 0;
}

string infixToPostfix(string infix) {
    stack<char> s;
    string postfix = "";
    unordered_map<char, int> precedence;

    precedence['+'] = 1;
    precedence['-'] = 1;
    precedence['*'] = 2;
    precedence['/'] = 2;

    for (char c : infix) {
        if (isalnum(c))
            postfix += c;
        else if (c == '(')
            s.push(c);
        else if (c == ')') {
            while (!s.empty() && s.top() != '(') {
                postfix += s.top();
                s.pop();
            }
            s.pop();
        } else {
            while (!s.empty() && precedence[c] <= precedence[s.top()]) {
                postfix += s.top();
                s.pop();
            }
            s.push(c);
        }
    }

    while (!s.empty()) {
        postfix += s.top();
        s.pop();
    }

    return postfix;
}

int main() {
    string infixExpression = "a+b*c-(d/e+f*g*h)";
    string postfixExpression = infixToPostfix(infixExpression);
    cout << "Infix Expression: " << infixExpression << endl;
    cout << "Postfix Expression: " << postfixExpression << endl;

    return 0;
}

// Java program for infix to postfix conversion using stack
import java.util.HashMap;
import java.util.Map;
import java.util.Stack;

public class InfixToPostfix {
    public static String infixToPostfix(String infix) {
        Stack<Character> stack = new Stack<>();
        StringBuilder postfix = new StringBuilder();
        Map<Character, Integer> precedence = new HashMap<>();

        precedence.put('+', 1);
        precedence.put('-', 1);
        precedence.put('*', 2);
        precedence.put('/', 2);

        for (char c : infix.toCharArray()) {
            if (Character.isLetterOrDigit(c))
                postfix.append(c);
            else if (c == '(')
                stack.push(c);
            else if (c == ')') {
                while (!stack.isEmpty() && stack.peek() != '(')
                    postfix.append(stack.pop());
                stack.pop();
            } else {
                while (!stack.isEmpty() && precedence.getOrDefault(c, 0) <= precedence.getOrDefault(stack.peek(), 0))
                    postfix.append(stack.pop());
                stack.push(c);
            }
        }

        while (!stack.isEmpty())
            postfix.append(stack.pop());

        return postfix.toString();
    }

    public static void main(String[] args) {
        String infixExpression = "a+b*c-(d/e+f*g*h)";
        String postfixExpression = infixToPostfix(infixExpression);
        System.out.println("Infix Expression: " + infixExpression);
        System.out.println("Postfix Expression: " + postfixExpression);
    }
}

# Python code for infix to postfix conversion using stack
def infix_to_postfix(infix):
stack = []
postfix = ""
precedence = {'+': 1, '-': 1, '*': 2, '/': 2}

for c in infix:
    if c.isalnum():
        postfix += c
    elif c == '(':
        stack.append(c)
    elif c == ')':
        while stack and stack[-1] != '(':
            postfix += stack.pop()
        stack.pop()
    else:
        while stack and precedence.get(c, 0) <= precedence.get(stack[-1], 0):
            postfix += stack.pop()
        stack.append(c)

while stack:
    postfix += stack.pop()

return postfix


# Example usage
infix_expression = "a+b*c-(d/e+f*g*h)"
postfix_expression = infix_to_postfix(infix_expression)
print("Infix Expression:", infix_expression)
print("Postfix Expression:", postfix_expression)

// C-Sharp Code for infix to postfix conversion using stack
using System;
using System.Collections.Generic;

public class InfixToPostfix {
    public static string InfixToPostfix(string infix) {
        Stack<char> stack = new Stack<char>();
        System.Text.StringBuilder postfix = new System.Text.StringBuilder();
        Dictionary<char, int> precedence = new Dictionary<char, int> {
            { '+', 1 },
            { '-', 1 },
            { '*', 2 },
            { '/', 2 }
        };

        foreach (char c in infix) {
            if (char.IsLetterOrDigit(c))
                postfix.Append(c);
            else if (c == '(')
                stack.Push(c);
            else if (c == ')') {
                while (stack.Count > 0 && stack.Peek() != '(')
                    postfix.Append(stack.Pop());
                stack.Pop();
            } else {
                while (stack.Count > 0 && precedence.GetValueOrDefault(c, 0) <= precedence.GetValueOrDefault(stack.Peek(), 0))
                    postfix.Append(stack.Pop());
                stack.Push(c);
            }
        }

        while (stack.Count > 0)
            postfix.Append(stack.Pop());

        return postfix.ToString();
    }

    public static void Main() {
        string infixExpression = "a+b*c-(d/e+f*g*h)";
        string postfixExpression = InfixToPostfix(infixExpression);
        Console.WriteLine("Infix Expression: " + infixExpression);
        Console.WriteLine("Postfix Expression: " + postfixExpression);
    }
}

Output:

Infix Expression: a+b*c-(d/e+f*g*h)
Postfix Expression: abc*+de/fg*h*+-

• Time Complexity: The time complexity of the code is O(N), where N is the length of the input infix expression.
• Space Complexity: The space complexity of the code is O(N).

How to check the technology a website is build with?

Checking the technology a website is built with involves inspecting its underlying components, such as the content management system (CMS), web server, programming languages, and frameworks. Here are several methods to uncover the technology stack of a website:

1. Using Online Tools.

Discovering the technology stack of a website becomes effortless with the assistance of online tools. Platforms like BuiltWith and Wappalyzer offer comprehensive reports by analyzing various components, unveiling the intricate web technologies shaping a website.

BuiltWith:

1. Visit BuiltWith.
2. Enter the URL of the website in question.
3. The tool will generate a detailed report outlining the technologies used.

Wappalyzer:

1. Install the Wappalyzer browser extension for Chrome or Firefox.
2. Open the website you want to inspect.
3. Click on the Wappalyzer icon in your browser to view the detected technologies.

2. Check Page Source.

Dive into the digital anatomy of a website by scrutinizing its page source. By right-clicking and exploring the HTML code, you can unveil meta tags, scripts, and comments that serve as clues, unraveling the technological tapestry behind the web interface.

• Right-click on the website page and select "View Page Source" or "Inspect" (depending on your browser).

• Look for meta tags, comments, or scripts that may indicate the use of specific technologies.

3. Loop for Generator Meta Tag.

Embark on a metadata expedition within a website's HTML source code. The presence of a generator meta tag, such as the one indicating WordPress, serves as a breadcrumb leading to the revelation of the content management system (CMS) in use.

<meta name="generator" content="WordPress">

4. WHOIS Loopup.

Unveil the digital footprint of a website through a WHOIS lookup. By exploring information about domain registration, hosting provider details, and registration history, one can glean insights into the foundational elements supporting the website's structure.

• Perform a WHOIS lookup using websites like WHOIS.net.
• Look for information about the web hosting provider, which might give clues about the technology stack.

5. Online Platform.

Peer into the collaborative world of online platforms like GitHub, GitLab, or Bitbucket. These repositories often unveil the source code and development choices behind a website, presenting a community-driven perspective on its technological foundation.

Some Tips:

• Keep in mind that not all information may be accurately disclosed, and some websites may actively hide their technology stack for security reasons.

• Use multiple methods to cross-verify results.

• Understand that the information gathered might not be exhaustive or up-to-date, especially if the website employs security measures to conceal its stack.

By employing these methods, you can gain valuable insights into the technology stack of a website, enabling a better understanding of its underlying infrastructure and development choices.

Stack Implementation Using Queues.

Stack is a linear data structure that can be implemented using many other data structures like array and linked list. In this article, we will use a Queue data structure to implement the stack operations that follow the LIFO (Last In, First Out) principle. 

Implementation of Stack Using Queue.

If you are a little bit familiar with stack and queue data structure you might know that in stack insertion and deletion of elements happen only from one top end and in queue insertion of an element happens from the rear end and deletion happens from the front. 

There are two common approaches to implementing a stack using queues:

• Using Two Queues.
• Using Single Queue.

Let's explore both the approaches,

Stack Implementation Using Two Queues.

Implementing a stack using two queues involves using one queue (Queue1) for storage and the other (Queue2) for assisting in simulating the Last In, First Out (LIFO) behavior of a stack. 

Push Operation: When pushing an element onto the stack, the new element is enqueued into Queue1, the primary storage queue. To maintain the LIFO order, elements from Queue2, if any, are dequeued and enqueued into Queue1. This step effectively reorganizes the elements to ensure the new addition becomes the topmost element in the simulated stack structure.

Pop Operation: Popping an element from the stack involves dequeuing elements from Queue1 until only one element remains. The objective is to remove the top element, mimicking the behavior of a typical stack operation. By continually dequeuing elements and enqueuing them back into Queue1, except for the last remaining element, the desired element to be removed is identified and dequeued.

Illustration:

Code for Stack Implementation Using Queues.

Below is the code implementation of the above approach:

C++ Java Python C#

// C++ program to implement stack using two Queue
#include <iostream>
#include <queue>
using namespace std;

class Stack {
private:
    queue<int> q1, q2;

public:
    // Push operation
    void push(int value) {
        q1.push(value);
    }

    // Pop operation
    void pop() {
        if (isEmpty()) {
            cout << "Stack is empty. Cannot pop." << endl;
            return;
        }

        // Move elements from q1 to q2, except the last element
        while (q1.size() > 1) {
            q2.push(q1.front());
            q1.pop();
        }

        // Pop the last element from q1
        q1.pop();

        // Swap the names of q1 and q2
        swap(q1, q2);
    }

    // Check if the stack is empty
    bool isEmpty() {
        return q1.empty();
    }

    // Display the top element of the stack
    void peek() {
        if (isEmpty()) {
            cout << "Stack is empty." << endl;
            return;
        }

        // Move elements from q1 to q2, except the last element
        while (q1.size() > 1) {
            q2.push(q1.front());
            q1.pop();
        }

        // Peek at the last element in q1
        cout << "Top element: " << q1.front() << endl;

        // Move the last element back to q1
        q2.push(q1.front());
        q1.pop();

        // Swap the names of q1 and q2
        swap(q1, q2);
    }
};

int main() {
    // Example usage of the stack
    Stack stack;

    // Push elements onto the stack
    stack.push(1);
    stack.push(2);
    stack.push(3);

    // Display the top element
    stack.peek();

    // Pop an element
    stack.pop();

    // Display the top element after pop
    stack.peek();

    return 0;
}

// Java code implementation for stack using Two Queue
import java.util.LinkedList;
import java.util.Queue;

public class Main {
    static class Stack {
        private Queue<Integer> q1 = new LinkedList<>();
        private Queue<Integer> q2 = new LinkedList<>();

        // Push operation
        void push(int value) {
            q1.add(value);
        }

        // Pop operation
        void pop() {
            if (isEmpty()) {
                System.out.println("Stack is empty. Cannot pop.");
                return;
            }

            // Move elements from q1 to q2, except the last element
            while (q1.size() > 1) {
                q2.add(q1.poll());
            }

            // Pop the last element from q1
            q1.poll();

            // Swap the names of q1 and q2
            Queue<Integer> temp = q1;
            q1 = q2;
            q2 = temp;
        }

        // Check if the stack is empty
        boolean isEmpty() {
            return q1.isEmpty();
        }

        // Display the top element of the stack
        void peek() {
            if (isEmpty()) {
                System.out.println("Stack is empty.");
                return;
            }

            // Move elements from q1 to q2, except the last element
            while (q1.size() > 1) {
                q2.add(q1.poll());
            }

            // Peek at the last element in q1
            System.out.println("Top element: " + q1.peek());

            // Move the last element back to q1
            q2.add(q1.poll());

            // Swap the names of q1 and q2
            Queue<Integer> temp = q1;
            q1 = q2;
            q2 = temp;
        }
    }

    public static void main(String[] args) {
        // Example usage of the stack
        Stack stack = new Stack();

        // Push elements onto the stack
        stack.push(1);
        stack.push(2);
        stack.push(3);

        // Display the top element
        stack.peek();

        // Pop an element
        stack.pop();

        // Display the top element after pop
        stack.peek();
    }
}

# Python program to implement stack using Two Queue
from queue import Queue

class Stack:
    def __init__(self):
        self.q1 = Queue()
        self.q2 = Queue()

    # Push operation
    def push(self, value):
        self.q1.put(value)

    # Pop operation
    def pop(self):
        if self.is_empty():
            print("Stack is empty. Cannot pop.")
            return

        # Move elements from q1 to q2, except the last element
        while self.q1.qsize() > 1:
            self.q2.put(self.q1.get())

        # Pop the last element from q1
        self.q1.get()

        # Swap the names of q1 and q2
        self.q1, self.q2 = self.q2, self.q1

    # Check if the stack is empty
    def is_empty(self):
        return self.q1.qsize() == 0

    # Display the top element of the stack
    def peek(self):
        if self.is_empty():
            print("Stack is empty.")
            return

        # Move elements from q1 to q2, except the last element
        while self.q1.qsize() > 1:
            self.q2.put(self.q1.get())

        # Peek at the last element in q1
        print("Top element:", self.q1.queue[0])

        # Move the last element back to q1
        self.q2.put(self.q1.get())

        # Swap the names of q1 and q2
        self.q1, self.q2 = self.q2, self.q1

# Example usage of the stack
stack = Stack()

# Push elements onto the stack
stack.push(1)
stack.push(2)
stack.push(3)

# Display the top element
stack.peek()

# Pop an element
stack.pop()

# Display the top element after pop
stack.peek()

// C-Sharp program to implement stack using Queues
using System;
using System.Collections.Generic;

public class StackUsingQueue {
    class Stack {
        private Queue<int> q1 = new Queue<int>();
        private Queue<int> q2 = new Queue<int>();

        // Push operation
        public void Push(int value) {
            q1.Enqueue(value);
        }

        // Pop operation
        public void Pop() {
            if (IsEmpty()) {
                Console.WriteLine("Stack is empty. Cannot pop.");
                return;
            }

            // Move elements from q1 to q2, except the last element
            while (q1.Count > 1) {
                q2.Enqueue(q1.Dequeue());
            }

            // Pop the last element from q1
            q1.Dequeue();

            // Swap the names of q1 and q2
            Queue<int> temp = q1;
            q1 = q2;
            q2 = temp;
        }

        // Check if the stack is empty
        public bool IsEmpty() {
            return q1.Count == 0;
        }

        // Display the top element of the stack
        public void Peek() {
            if (IsEmpty()) {
                Console.WriteLine("Stack is empty.");
                return;
            }

            // Move elements from q1 to q2, except the last element
            while (q1.Count > 1) {
                q2.Enqueue(q1.Dequeue());
            }

            // Peek at the last element in q1
            Console.WriteLine("Top element: " + q1.Peek());

            // Move the last element back to q1
            q2.Enqueue(q1.Dequeue());

            // Swap the names of q1 and q2
            Queue<int> temp = q1;
            q1 = q2;
            q2 = temp;
        }
    }

    static void Main() {
        // Example usage of the stack
        Stack stack = new Stack();

        // Push elements onto the stack
        stack.Push(1);
        stack.Push(2);
        stack.Push(3);

        // Display the top element
        stack.Peek();

        // Pop an element
        stack.Pop();

        // Display the top element after pop
        stack.Peek();
    }
}

Output:

Top element: 3
Top element: 2

Time and Space Complexity.

In the above code, our pop operation is costly as we have to move all the remaining elements from Queue1 to Queue2. Similarly, we can implement the same approach by making push operation costly instead of pop operation.

Stack Operation	Time Complexity	Space Complexity
Push Operation	O(1)	O(1)
Pop Operation	O(N)	O(1)
IsEmpty	O(1)	O(1)

This approach cleverly utilizes two queues to represent stack behavior by employing element movements between the queues. While it successfully emulates a stack, it incurs some space overhead and may involve extra operations, such as queue swapping which could impact performance.

Stack Implementation using Single Queue.

In the above approach, we have used two queues to implement the stack operation. In this approach, we are going to use a single queue to implement the stack by manipulating the enqueue and dequeue operations to achieve the Last In, First Out (LIFO) behavior of the stack. 

Stack Operations:

Let's understand how we will implement two important stack operations push and pop using a single Queue:

Push Operation: To push an element onto the stack, enqueue the new element into the queue. To move the new element to the top, we have to rearrange elements in the queue by dequeuing elements before the newly added element and enqueuing them back to the queue.

Pop Operation: Popping is quite simple in this approach, we just have to dequeue and return the front element, which represents the topmost element of the stack.

Code for Stack Implementation Using Single Queue.

Below is the code implementation of the above approach:

C++ Java Python C#

// C++ program to implement stack using single Queue
#include<iostream>
#include<queue>
using namespace std;

class StackWithQueue {
private:
    queue<int> myQueue;
public:
    // function to push an element onto the stack
    void push(int value) {
        //get current size of queue
        int queueSize = myQueue.size();

        //enqueue a new element into queue
        myQueue.push(value);

        // rearrange the elements in the queue to 
        // make the new element the front (top)
        for(int i = 0; i < queueSize; i++) {
            myQueue.push(myQueue.front());
            myQueue.pop();
        }
    }
    // function to pop an element from the stack
    void pop() {
        if(myQueue.empty()) {
            cout << "Stack is Empty!"<<endl;
        }
        else {
            myQueue.pop();
        }      
    }
    // function to peek top element
    int peek() {
        if(myQueue.empty()){
            cout<< "Stack is Empty!"<<endl;
            return -1;
        }
        else {
            return myQueue.front();    
        }
    }
    // function to check if stack is empty
    bool isEmpty(){
        return myQueue.empty();
    }
    // function to get the size of stack
    int size() {
        return myQueue.size();
    }
};

//main function
int main() {
    StackWithQueue stack;

    stack.push(1);
    stack.push(2);
    stack.push(5);
    stack.push(8);
    
    cout<< stack.peek() <<endl;
    stack.pop();
    cout<< stack.peek() <<endl;

    return 0;
}

// Java Program to implement stack using Single Queue
import java.util.LinkedList;
import java.util.Queue;

public class StackWithQueue {
    private Queue<Integer> myQueue = new LinkedList<>();

    // Function to push an element onto the stack
    public void push(int value) {
        int queueSize = myQueue.size();
        myQueue.add(value);

        for (int i = 0; i < queueSize; i++) {
            myQueue.add(myQueue.poll());
        }
    }

    // Function to pop an element from the stack
    public void pop() {
        if (myQueue.isEmpty()) {
            System.out.println("Stack is Empty!");
        } else {
            myQueue.poll();
        }
    }

    // Function to peek top element
    public int peek() {
        if (myQueue.isEmpty()) {
            System.out.println("Stack is Empty!");
            return -1;
        } else {
            return myQueue.peek();
        }
    }

    // Function to check if the stack is empty
    public boolean isEmpty() {
        return myQueue.isEmpty();
    }

    // Function to get the size of the stack
    public int size() {
        return myQueue.size();
    }

    // Main function
    public static void main(String[] args) {
        StackWithQueue stack = new StackWithQueue();

        stack.push(1);
        stack.push(2);
        stack.push(5);
        stack.push(8);

        System.out.println(stack.peek());
        stack.pop();
        System.out.println(stack.peek());
    }
}

# Python code Implementation of stack using Single Queue
from queue import Queue

class StackWithQueue:
    def __init__(self):
        self.my_queue = Queue()

    # Function to push an element onto the stack
    def push(self, value):
        queue_size = self.my_queue.qsize()
        self.my_queue.put(value)

        for _ in range(queue_size):
            self.my_queue.put(self.my_queue.get())

    # Function to pop an element from the stack
    def pop(self):
        if self.my_queue.empty():
            print("Stack is Empty!")
        else:
            self.my_queue.get()

    # Function to peek top element
    def peek(self):
        if self.my_queue.empty():
            print("Stack is Empty!")
            return -1
        else:
            return self.my_queue.queue[0]

    # Function to check if the stack is empty
    def is_empty(self):
        return self.my_queue.empty()

    # Function to get the size of the stack
    def size(self):
        return self.my_queue.qsize()

# Main function
if __name__ == "__main__":
    stack = StackWithQueue()

    stack.push(1)
    stack.push(2)
    stack.push(5)
    stack.push(8)

    print(stack.peek())
    stack.pop()
    print(stack.peek())

// C# Program to implement stack using single Queue
using System;
using System.Collections.Generic;

public class StackWithQueue
{
    private Queue<int> myQueue = new Queue<int>();

    // Function to push an element onto the stack
    public void Push(int value)
    {
        int queueSize = myQueue.Count;
        myQueue.Enqueue(value);

        for (int i = 0; i < queueSize; i++)
        {
            myQueue.Enqueue(myQueue.Dequeue());
        }
    }

    // Function to pop an element from the stack
    public void Pop()
    {
        if (myQueue.Count == 0)
        {
            Console.WriteLine("Stack is Empty!");
        }
        else
        {
            myQueue.Dequeue();
        }
    }

    // Function to peek top element
    public int Peek()
    {
        if (myQueue.Count == 0)
        {
            Console.WriteLine("Stack is Empty!");
            return -1;
        }
        else
        {
            return myQueue.Peek();
        }
    }

    // Function to check if the stack is empty
    public bool IsEmpty()
    {
        return myQueue.Count == 0;
    }

    // Function to get the size of the stack
    public int Size()
    {
        return myQueue.Count;
    }

    // Main function
    public static void Main()
    {
        StackWithQueue stack = new StackWithQueue();

        stack.Push(1);
        stack.Push(2);
        stack.Push(5);
        stack.Push(8);

        Console.WriteLine(stack.Peek());
        stack.Pop();
        Console.WriteLine(stack.Peek());
    }
}

Output:

8
5

Time and Space Complexity.

Stack Operation	Time Complexity	Space Complexity
Push Operation	O(N)	O(1)
Pop Operation	O(1)	O(1)
IsEmpty	O(1)	O(1)