Check if an URL is valid or not using Regular Expression

Given a URL as a character string str of size N.The task is to check if the given URL is valid or not.
Examples :

Input : str = “https://www.geeksforgeeks.org/” 
Output : Yes 
Explanation :
The above URL is a valid URL.
Input : str = “https:// www.geeksforgeeks.org/” 
Output : No 
Explanation :
Note that there is a space after https://, hence the URL is invalid. 

Approach :
An approach using java.net.url class to validate a URL is discussed in the previous post. 
Here the idea is to use Regular Expression to validate a URL. 

• Get the URL.
• Create a regular expression to check the valid URL as mentioned below:

regex = “((http|https)://)(www.)?” 
+ “[a-zA-Z0-9@:%._\\+~#?&//=]{2,256}\\.[a-z]” 
+ “{2,6}\\b([-a-zA-Z0-9@:%._\\+~#?&//=]*)”

• The URL must start with either http or https and
• then followed by :// and
• then it must contain www. and
• then followed by subdomain of length (2, 256) and
• last part contains top level domain like .com, .org etc.

• Match the given URL with the regular expression. In Java, this can be done by using Pattern.matcher().
• Return true if the URL matches with the given regular expression, else return false.

Below is the implementation of the above approach:

C++

// C++ program to validate URL
// using Regular Expression
#include <iostream>
#include <regex>
using namespace std;

// Function to validate URL
// using regular expression
bool isValidURL(string url)
{

  // Regex to check valid URL
  const regex pattern("((http|https)://)(www.)?[a-zA-Z0-9@:%._\\+~#?&//=]{2,256}\\.[a-z]{2,6}\\b([-a-zA-Z0-9@:%._\\+~#?&//=]*)");

  // If the URL
  // is empty return false
  if (url.empty())
  {
     return false;
  }

  // Return true if the URL
  // matched the ReGex
  if(regex_match(url, pattern))
  {
    return true;
  }
  else
  {
    return false;
  }
}

// Driver Code
int main()
{
  string url = "https://www.geeksforgeeks.org";

  if (isValidURL(url))
  {
    cout << "YES";
  }
  else
  {
    cout << "NO";
  }
  return 0;
}

// This code is contributed by yuvraj_chandra

Java

// Java program to check URL is valid or not
// using Regular Expression

import java.util.regex.*;

class GFG {

    // Function to validate URL
    // using regular expression
    public static boolean
    isValidURL(String url)
    {
        // Regex to check valid URL
        String regex = "((http|https)://)(www.)?"
              + "[a-zA-Z0-9@:%._\\+~#?&//=]"
              + "{2,256}\\.[a-z]"
              + "{2,6}\\b([-a-zA-Z0-9@:%"
              + "._\\+~#?&//=]*)";

        // Compile the ReGex
        Pattern p = Pattern.compile(regex);

        // If the string is empty
        // return false
        if (url == null) {
            return false;
        }

        // Find match between given string
        // and regular expression
        // using Pattern.matcher()
        Matcher m = p.matcher(url);

        // Return if the string
        // matched the ReGex
        return m.matches();
    }

    // Driver code
    public static void main(String args[])
    {
        String url
            = "https://www.geeksforgeeks.org";
        if (isValidURL(url) == true) {
            System.out.println("Yes");
        }
        else
            System.out.println("NO");
    }
}

Python3

# Python3 program to check
# URL is valid or not
# using regular expression
import re

# Function to validate URL
# using regular expression
def isValidURL(str):

    # Regex to check valid URL
    regex = ("((http|https)://)(www.)?" +
             "[a-zA-Z0-9@:%._\\+~#?&//=]" +
             "{2,256}\\.[a-z]" +
             "{2,6}\\b([-a-zA-Z0-9@:%" +
             "._\\+~#?&//=]*)")

    # Compile the ReGex
    p = re.compile(regex)

    # If the string is empty
    # return false
    if (str == None):
        return False

    # Return if the string
    # matched the ReGex
    if(re.search(p, str)):
        return True
    else:
        return False

# Driver code

# Test Case 1:
url = "https://www.geeksforgeeks.org"

if(isValidURL(url) == True):
    print("Yes")
else:
    print("No")

# This code is contributed by avanitrachhadiya2155

Output:

Yes

Time Complexity: O (N)
Auxiliary Space: O (1)