#
# Question 1
#
# 
# int    0
# bool   True
# int    10
# str    'bc'
# str    'z'
# list   [ 'aa', 6, [0, 0] ]
# str    'COMPSCI 101'
# float  256.0


#
# Question 2
#

# Part A
#
# Testcases 2 and 3 fail because of an extra space before &
#   Duvall, LLC
#   Terrible  & Great, LLC
#   Pearson Hardman  & Specter, LLC

# Part B
#
# Testcase 1 fails because an & is always added, even for just one name
#   & Duvall, LLC
#   Terrible & Great, LLC
#   Pearson Hardman & Specter, LLC

# Part C
#
# Testcase 1 fails because the name is not included when there is only one
#   , LLC
#   Terrible & Great, LLC
#   Pearson Hardman & Specter, LLC


#
# Question 3
#
# Part A
#
def sumPairs (values):
    if len(values) == 1:
        return values
    result = []
    for i in range(len(values)-1):
        result.append(values[i] + values[i+1])
    return result

# ALTERNATE SOLUTION
def sumPairs (values):
    if len(values) == 1:
        return values
    else:
        return [ sum(values[i-1:i+1]) for i in range(1, len(values)) ]

# Part B
#
def pascalTriangle (level):
    result = []
    row = [ 1 ]
    for k in range(level):
        result.append(row)
        row = sumPairs([0] + row + [0])
    return result

# ALTERNATE SOLUTION
def pascalTriangle (level):
    result = [ [1] ]
    for k in range(level - 1):
        result.append(sumPairs([0] + result[k] + [0]))
    return result

# Part C
#
def numCombinations (n, k):
    return pascalTriangle(n + 1)[n][k]


#
# Question 4
#

# Part A
#
# given 'hello' and 'ohell', checks for 'hello' in  'ohellohell'
#   by concatenating rotated string, the start should line up with the end, 
#   meaning the original string must be there if no letters are out of order
#   because all possible rotations can be found in there

# Part B
#
# given 'hello' and 'ohell', find rotation point, i.e., 'ello' and 'h'
#   checks every index, i.e., every possible rotation, to see where the point 
#   of rotation occurs

# Part C
#
# given 'hello' and 'ohell', compares the two lists
#     ['elloh', 'hello', 'llohe', 'lohel', 'ohell']
#     ['elloh', 'hello', 'llohe', 'lohel', 'ohell']
#  checks if both strings produce exactly the same rotations for all possible 
#  indices, sorts them to guarantee rotations they appear in the same order no
#  matter the order in which they were generated


#
# Question 5
#
# Part A
#
def isValid (ccard):
    total = 0
    for k in range(len(ccard)):
        digit = int(ccard[k])
        if k % 2 != 0:
            digit *= 2;
            if digit >= 10:
                digit -= 9;
        total += digit
    return total % 10 == 0

# Part B
#
def isExpired (useDate, expireDate):
    expired = expireDate.split("/")
    used = useDate.split("/")
    if expired[1] < used[1]:
        return True
    elif expired[1] > used[1]:
        return False
    elif expired[0] < used[0]:
        return True
    return False

# ALTERNATE SOLUTION
def isExpired (useDate, expireDate):
    (eMonth, eYear) = expireDate.split("/")
    (uMonth, uYear) = useDate.split("/")
    return eYear < uYear or (eYear == uYear and eMonth < uMonth)

# ALTERNATE SOLUTION
def isExpired3 (useDate, expireDate):
    uDate = useDate[3:] + useDate[:2] 
    eDate = expireDate[3:] + expireDate[:2]
    return eDate < uDate


# Part C
#
def checkTransaction (cardInfo, useDate):
    cardValues = cardInfo.split(":")
    if isValid(cardValues[0]):
        if isExpired(useDate, cardValues[1]):
            return "EXPIRED"
        else:
            return "APPROVED"
    else:
        return "INVALID"

# ALTERNATE SOLUTION
def checkTransaction (cardInfo, useDate):
    cardValues = cardInfo.split(":")
    if not isValid(cardValues[0]):
        return "INVALID"
    elif isExpired(useDate, cardValues[1]):
        return "EXPIRED"
    else:
        return "APPROVED"