mydict = {'carl':40, 'alan':2, 'bob':1, 'danny':0}

# How to sort a dict by value Python 3>
sort = {key:value for key, value in sorted(mydict.items(), key=lambda kv: (kv[1], kv[0]))}
print(sort)

# How to sort a dict by key Python 3>
sort = {key:mydict[key] for key in sorted(mydict.keys())}
print(sort)

def format_timestamp(timestamp_epoch):
    """
    Convert epoch timestamp to formatted datetime string without using datetime package.
    
    Args:
        timestamp_epoch (int/float): Unix epoch timestamp (seconds since 1970-01-01 00:00:00 UTC)
        
    Returns:
        str: Formatted datetime string in 'YYYY-MM-DD HH:MM:SS' format
    """
    # Constants for time calculations
    SECONDS_PER_DAY = 86400
    SECONDS_PER_HOUR = 3600
    SECONDS_PER_MINUTE = 60
    
    # Handle negative timestamps and convert to integer
    timestamp = int(timestamp_epoch)
    
    # Calculate days since epoch and remaining seconds
    days_since_epoch = timestamp // SECONDS_PER_DAY
    remaining_seconds = timestamp % SECONDS_PER_DAY
    
    # Calculate hours, minutes, seconds
    hours = remaining_seconds // SECONDS_PER_HOUR
    remaining_seconds %= SECONDS_PER_HOUR
    minutes = remaining_seconds // SECONDS_PER_MINUTE
    seconds = remaining_seconds % SECONDS_PER_MINUTE
    
    # Calculate date (simplified, ignoring leap seconds)
    year = 1970
    days = days_since_epoch
    while days >= 365:
        is_leap = (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
        days_in_year = 366 if is_leap else 365
        if days >= days_in_year:
            days -= days_in_year
            year += 1
    
    # Month lengths (non-leap year for simplicity, adjusted later for leap years)
    month_lengths = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0):
        month_lengths[1] = 29
    
    month = 0
    while days >= month_lengths[month]:
        days -= month_lengths[month]
        month += 1
    
    # Convert to 1-based indexing for month and day
    month += 1
    day = days + 1
    
    # Format the output string
    return f"{year:04d}-{month:02d}-{day:02d} {hours:02d}:{minutes:02d}:{seconds:02d}"

# Example timestamp (Unix epoch seconds)
timestamp = 1697054700
formatted_date = format_timestamp(timestamp)
print(formatted_date + " UTC")  # Output: 2023-10-11 18:45:00

import itertools
import string
import time

def guess_password(real):
    chars = string.ascii_lowercase + string.ascii_uppercase + string.digits + string.punctuation
    attempts = 0
    for password_length in range(1, 9):
        for guess in itertools.product(chars, repeat=password_length):
            startTime = time.time()
            attempts += 1
            guess = ''.join(guess)
            if guess == real:
                return 'password is {}. found in {} guesses.'.format(guess, attempts)
            loopTime = (time.time() - startTime);
            print(guess, attempts, loopTime)

print("\nIt will take A REALLY LONG TIME to crack a long password. Try this out with a 3 or 4 letter password and see how this program works.\n")
val = input("Enter a password you want to crack that is 9 characters or below: ")
print(guess_password(val.lower()))

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"

list_1 = [1,2,3,4,5,6,7,8,9]
cubed = map(lambda x: pow(x,3), list_1)
print(list(cubed))

#Results
#[1, 8, 27, 64, 125, 216, 343, 512, 729]

# 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))