# Python code to find the URL from an input string
# Using the regular expression
import re

def Find(string):

	# findall() has been used
	# with valid conditions for urls in string
	regex = r"(?i)\b((?:https?://|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'\".,<>?«»“”‘’]))"
	url = re.findall(regex,string)	
	return [x[0] for x in url]
	
# Driver Code
string = 'My Profile: https://codecatch.net'
print("Urls: ", Find(string))

print('hello, world')

import math

def factorial(n):
    print(math.factorial(n))
    return (math.factorial(n))
    
factorial(5)
factorial(10)
factorial(15)

import re
_proposal_regex = r'(?:(?:(IKE|ESP):)?[\w/]+(?:/NO_EXT_SEQ)?(?:, ?(IKE|ESP):[\w/]+(?:/NO_EXT_SEQ)?)*)?'
_proposals_re = rf'(?P<proposals>{_proposal_regex}|)'
pattern = rf'received proposals: {_proposals_re}'
match = re.match(pattern, 'received proposals: ')
print(match.group('proposals') if match else "No match")  # Prints "No match"

# Python Program to calculate the square root

num = float(input('Enter a number: '))

num_sqrt = num ** 0.5
print('The square root of %0.3f is %0.3f'%(num ,num_sqrt))

from typing import Optional

from datetime import datetime

def convert_timestamp_string_to_epoch(timestamp: str) -> Optional[int]:
    epoch_time = None
    time_obj = datetime.strptime(timestamp, "%Y-%m-%d %H:%M:%S.%f")
    epoch_time = int((time_obj - datetime(1970, 1, 1)).total_seconds() * 1000)
    return epoch_time


print(int(convert_timestamp_string_to_epoch("2025-08-01 13:11:47.171")))
#above outputs 1754053907171.0

#how to I remove the .0 ?