#
# Question 1
#

# Part A
unic = len([x for x in set(nuts) if x[0] == 'c'])

# Part B
solo = len([x for x in set(nuts) if nuts.count(x) == 1])

# Part C
truenut = [x for x in set(nuts) if x.endswith("nut")]

# Part D
nuttiest = sorted([(nuts.count(x),x) for x in nuts])[-1][1]

# Part E:
temp = sorted([(nuts.count(x),x) for x in set(nuts)])
freqs = [x[1] for x in temp]

# ALTERNATE SOLUTION
temp = sorted(set(nuts))
freqs = [ (x,nuts.count(x) for x in temp ]  # alph ordered
freqs = sorted(freqs, key=operator.itemgetter(1))
freqs = [ x[0] for x in freqs ]


#
# Question 2
#

# 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 3
#

# #1    pass
# #2    fail    4
# #3    pass
# This code does not remove the friend's beaten card thus 
# allowing it to beaten multiple times.

# #1    pass
# #2    pass
# #3    fail    2
# This code does not sort the friend's cards thus
# it does not choose the best card to remove.

# #1    fail    3
# #2    fail    4
# #3    pass
# This code removes all the firend's cards your card can beat thus
# it does not reflect how each hand matches up overall.


#
# Question 4
#

# Part A

# A movie with 11 elements is BOTH top grossing and top rated.

# The last element is the amount of money grossed by the movie.


# Part B

# In this case, result is given as an empty dictionary and it is returned
# as a dictionary with each entry newly created such that its keys are
# (title, year) and its values are only [rank, rating].

# In this case, result is given as a dictionary already containing some data
# and it is returned as a dictionary with some of its entries updated such 
# that its value list is appended to include [ rank, rating ] 


# Part C

def mostProfitableDirectors (movies, count):
    directors = {}
    for movieInfo in movies.values():
        key = movieInfo[0]                  # director's name
        if len(movieInfo) == 10:
            value = float(movieInfo[9])  # profit
        else:
            value = float(movieInfo[7])  # rating or profit
        if value > 100:                         # not a rating
            if key not in directors:
                directors[key] = 0
            directors[key] += int(value)
    return sorted([(v, k) for (k,v) in directors.items() ], reverse=True)[:count]


# Part D

def mostMoviesPerDecades (movies): 
    movieDecades = {}
    for (title,year) in movies:
        key = year[:3]
        movieDecades[key] = movieDecades.get(key, 0) + 1
    return sorted([ (count, decade+'0s') for (decade,count) in movieDecades.items() ], reverse=True)

# ALTERNATE SOLUTION
def mostMoviesPerDecades (movies): 
    decades = [ k[1][:3]+'0s' for k in movies ]
    return sorted([ (decades.count(d), d) for d in set(decades) ], reverse=True)