# Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number.
# For example, if n is 10, the output should be “2, 3, 5, 7”. If n is 20, the output should be “2, 3, 5, 7, 11, 13, 17, 19”.


# Python program to print all primes smaller than or equal to
# n using Sieve of Eratosthenes

def SieveOfEratosthenes(n):
	
	# Create a boolean array "prime[0..n]" and initialize
	# all entries it as true. A value in prime[i] will
	# finally be false if i is Not a prime, else true.
	prime = [True for i in range(n + 1)]
	p = 2
	while (p * p <= n):
		
		# If prime[p] is not changed, then it is a prime
		if (prime[p] == True):
			
			# Update all multiples of p
			for i in range(p * 2, n + 1, p):
				prime[i] = False
		p += 1
	prime[0]= False
	prime[1]= False
	# Print all prime numbers
	for p in range(n + 1):
		if prime[p]:
			print (p)

# driver program
if __name__=='__main__':
    n = 30
    print("Following are the prime numbers smaller")
    print("than or equal to ", n)
    print("than or equal to ", n)
    SieveOfEratosthenes(n)

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

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)

#Sets
U = {0,1,2,3,4,5,6,7,8,9}
P = {1,2,3,4}
Q = {4,5,6}
R = {3,4,6,8,9}

def set2bits(xs,us) :
    bs=[]
    for x in us :
        if x in xs :
            bs.append(1)
        else:
            bs.append(0)
    assert len(us) == len(bs)
    return bs

def union(set1,set2) :
    finalSet = set()
    bitList1 = set2bits(set1, U)
    bitList2 = set2bits(set2, U)

    for i in range(len(U)) :
        if(bitList1[i] or bitList2[i]) :
            finalSet.add(i)

    return finalSet

def intersection(set1,set2) :
    finalSet = set()
    bitList1 = set2bits(set1, U)
    bitList2 = set2bits(set2, U)

    for i in range(len(U)) :
        if(bitList1[i] and bitList2[i]) :
            finalSet.add(i)

    return finalSet

def compliment(set1) :
    finalSet = set()
    bitList = set2bits(set1, U)

    for i in range(len(U)) :
        if(not bitList[i]) :
            finalSet.add(i)

    return finalSet

def implication(a,b):
    return union(compliment(a), b)

###########################################################################################
######################         Problems 1-6         #######################################
###########################################################################################

#p \/ (q /\ r) = (p \/ q) /\ (p \/ r)
def prob1():
    return union(P, intersection(Q,R)) == intersection(union(P,Q), union(P,R))

#p /\ (q \/ r) = (p /\ q) \/ (p /\ r)
def prob2():
    return intersection(P, union(Q,R)) == union(intersection(P,Q), intersection(P,R))

#~(p /\ q) = ~p \/ ~q
def prob3():
    return compliment(intersection(P,R)) == union(compliment(P), compliment(R))

#~(p \/ q) = ~p /\ ~q
def prob4():
    return compliment(union(P,Q)) == intersection(compliment(P), compliment(Q))

#(p=>q) = (~q => ~p)
def prob5():
    return implication(P,Q) == implication(compliment(Q), compliment(P))

#(p => q) /\ (q => r)  =>  (p => r)
def prob6():
    return implication(intersection(implication(P,Q), implication(Q,R)), implication(P,R))


print("Problem 1: ", prob1())
print("Problem 2: ", prob2())
print("Problem 3: ", prob3())
print("Problem 4: ", prob4())
print("Problem 5: ", prob5())
print("Problem 6: ", prob6())

'''
Problem 1:  True
Problem 2:  True
Problem 3:  True
Problem 4:  True
Problem 5:  True
Problem 6:  {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
'''

# 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))

#84 48 13 20 61 20 33 97 34 45 6 63 71 66 24 57 92 74 6 25 51 86 48 15 64 55 77 30 56 53 37 99 9 59 57 61 30 97 50 63 59 62 39 32 34 4 96 51 8 86 10 62 16 55 81 88 71 25 27 78 79 88 92 50 16 8 67 82 67 37 84 3 33 4 78 98 39 64 98 94 24 82 45 3 53 74 96 9 10 94 13 79 15 27 56 66 32 81 77
# xor a list of integers to find the lonely integer
res = a[0]
    for i in range(1,len(a)):
        res = res ^ a[i]