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
Explaination :
The above URL is a valid URL.

Input : str = “https:// www.geeksforgeeks.org/”
Output : No
Explaination :
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 discuss in previous post.
Here the idea is to use Regular Expression to validate a URL.

Get the URL.
Create a regular expression to check 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:

// 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"); 
    } 
} 

Output:

Yes

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

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