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++
#include <iostream>
#include <regex>
using namespace std;
bool isValidURL(string url)
{
const regex pattern( "((http|https)://)(www.)?[a-zA-Z0-9@:%._\\+~#?&//=]{2,256}\\.[a-z]{2,6}\\b([-a-zA-Z0-9@:%._\\+~#?&//=]*)" );
if (url.empty())
{
return false ;
}
if (regex_match(url, pattern))
{
return true ;
}
else
{
return false ;
}
}
int main()
{
if (isValidURL(url))
{
cout << "YES" ;
}
else
{
cout << "NO" ;
}
return 0;
}
|
Java
import java.util.regex.*;
class GFG {
public static boolean
isValidURL(String url)
{
String regex = "((http|https)://)(www.)?"
+ "[a-zA-Z0-9@:%._\\+~#?&//=]"
+ "{2,256}\\.[a-z]"
+ "{2,6}\\b([-a-zA-Z0-9@:%"
+ "._\\+~#?&//=]*)" ;
Pattern p = Pattern.compile(regex);
if (url == null ) {
return false ;
}
Matcher m = p.matcher(url);
return m.matches();
}
public static void main(String args[])
{
String url
if (isValidURL(url) == true ) {
System.out.println( "Yes" );
}
else
System.out.println( "NO" );
}
}
|
Python3
import re
def isValidURL( str ):
regex = ( "((http|https)://)(www.)?" +
"[a-zA-Z0-9@:%._\\+~#?&//=]" +
"{2,256}\\.[a-z]" +
"{2,6}\\b([-a-zA-Z0-9@:%" +
"._\\+~#?&//=]*)" )
p = re. compile (regex)
if ( str = = None ):
return False
if (re.search(p, str )):
return True
else :
return False
if (isValidURL(url) = = True ):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
using System.Text.RegularExpressions;
class GFG
{
static void Main( string [] args)
{
foreach ( string s in str) {
Console.WriteLine( isValidURL(s) ? "true" : "false" );
}
Console.ReadKey(); }
public static bool isValidURL( string str)
{
string strRegex = @"((http|https)://)(www.)?" +
"[a-zA-Z0-9@:%._\\+~#?&//=]" +
"{2,256}\\.[a-z]" +
"{2,6}\\b([-a-zA-Z0-9@:%" +
"._\\+~#?&//=]*)" ;
Regex re = new Regex(strRegex);
if (re.IsMatch(str))
return ( true );
else
return ( false );
}
}
|
Javascript
function isValidURL(str) {
if (/^(http(s):\/\/.)[-a-zA-Z0-9@:%._\+~ #=]{2,256}\.[a-z]{2,6}\b([-a-zA-Z0-9@:%_\+.~#?&//=]*)$/g.test(str)) {
console.log( 'YES' );
} else {
console.log( 'NO' );
}
}
|
Time Complexity: O (N)
Auxiliary Space: O (1)
Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!
Last Updated :
07 Dec, 2022
Like Article
Save Article