def max_n(lst, n = 1):
  return sorted(lst, reverse = True)[:n]

max_n([1, 2, 3]) # [3]
max_n([1, 2, 3], 2) # [3, 2]

def print_x_pattern(size):
    i,j = 0,size - 1

    while j >= 0 and i < size:

        initial_spaces = ' '*min(i,j)
        middle_spaces = ' '*(abs(i - j) - 1)
        final_spaces = ' '*(size - 1 - max(i,j))

        if j == i:
            print(initial_spaces + '*' + final_spaces)
        else:
            print(initial_spaces + '*' + middle_spaces + '*' + final_spaces)

        i += 1
        j -= 1

print_x_pattern(7)

def print_pyramid_pattern(n):
     
    # outer loop to handle number of rows
    # n in this case
    for i in range(0, n):
     
        # inner loop to handle number of columns
        # values changing acc. to outer loop
        for j in range(0, i+1):
         
            # printing stars
            print("* ",end="")
      
        # ending line after each row
        print("\r")
 
print_pyramid_pattern(10)

# 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))

my_list = [1, 2, 3, 4, 5]
removed_element = my_list.pop(2)  # Remove and return element at index 2
print(removed_element)  # 3
print(my_list)  # [1, 2, 4, 5]

last_element = my_list.pop()  # Remove and return the last element
print(last_element)  # 5
print(my_list)  # [1, 2, 4]