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]

import sys

# sample Tuples
Tuple1 = ("A", 1, "B", 2, "C", 3)
Tuple2 = ("Geek1", "Raju", "Geek2", "Nikhil", "Geek3", "Deepanshu")
Tuple3 = ((1, "Lion"), ( 2, "Tiger"), (3, "Fox"), (4, "Wolf"))

# print the sizes of sample Tuples
print("Size of Tuple1: " + str(sys.getsizeof(Tuple1)) + "bytes")
print("Size of Tuple2: " + str(sys.getsizeof(Tuple2)) + "bytes")
print("Size of Tuple3: " + str(sys.getsizeof(Tuple3)) + "bytes")

#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}
'''

#You are given a two-digit integer n. Return the sum of its digits.
#Example
#For n = 29 the output should be solution (n) = 11
def solution(n):
    return (n//10 + n%10)

from collections import Counter

def find_parity_outliers(nums):
  return [
    x for x in nums
    if x % 2 != Counter([n % 2 for n in nums]).most_common()[0][0]
  ]

find_parity_outliers([1, 2, 3, 4, 6]) # [1, 3]

def sum_of_powers(end, power = 2, start = 1):
  return sum([(i) ** power for i in range(start, end + 1)])

sum_of_powers(10) # 385
sum_of_powers(10, 3) # 3025
sum_of_powers(10, 3, 5) # 2925