# sahoelsc's solution

## to Book Store in the Python Track

Published at Apr 18 2019 · 0 comments
Instructions
Test suite
Solution

#### Note:

This exercise has changed since this solution was written.

To try and encourage more sales of different books from a popular 5 book series, a bookshop has decided to offer discounts on multiple book purchases.

One copy of any of the five books costs \$8.

If, however, you buy two different books, you get a 5% discount on those two books.

If you buy 3 different books, you get a 10% discount.

If you buy 4 different books, you get a 20% discount.

If you buy all 5, you get a 25% discount.

Note: that if you buy four books, of which 3 are different titles, you get a 10% discount on the 3 that form part of a set, but the fourth book still costs \$8.

Your mission is to write a piece of code to calculate the price of any conceivable shopping basket (containing only books of the same series), giving as big a discount as possible.

For example, how much does this basket of books cost?

• 2 copies of the first book
• 2 copies of the second book
• 2 copies of the third book
• 1 copy of the fourth book
• 1 copy of the fifth book

One way of grouping these 8 books is:

• 1 group of 5 --> 25% discount (1st,2nd,3rd,4th,5th)
• +1 group of 3 --> 10% discount (1st,2nd,3rd)

This would give a total of:

• 5 books at a 25% discount
• +3 books at a 10% discount

Resulting in:

• 5 x (8 - 2.00) == 5 x 6.00 == \$30.00
• +3 x (8 - 0.80) == 3 x 7.20 == \$21.60

For a total of \$51.60

However, a different way to group these 8 books is:

• 1 group of 4 books --> 20% discount (1st,2nd,3rd,4th)
• +1 group of 4 books --> 20% discount (1st,2nd,3rd,5th)

This would give a total of:

• 4 books at a 20% discount
• +4 books at a 20% discount

Resulting in:

• 4 x (8 - 1.60) == 4 x 6.40 == \$25.60
• +4 x (8 - 1.60) == 4 x 6.40 == \$25.60

For a total of \$51.20

And \$51.20 is the price with the biggest discount.

## Exception messages

Sometimes it is necessary to raise an exception. When you do this, you should include a meaningful error message to indicate what the source of the error is. This makes your code more readable and helps significantly with debugging. Not every exercise will require you to raise an exception, but for those that do, the tests will only pass if you include a message.

To raise a message with an exception, just write it as an argument to the exception type. For example, instead of `raise Exception`, you should write:

``````raise Exception("Meaningful message indicating the source of the error")
``````

## Running the tests

To run the tests, run the appropriate command below (why they are different):

• Python 2.7: `py.test book_store_test.py`
• Python 3.4+: `pytest book_store_test.py`

Alternatively, you can tell Python to run the pytest module (allowing the same command to be used regardless of Python version): `python -m pytest book_store_test.py`

### Common `pytest` options

• `-v` : enable verbose output
• `-x` : stop running tests on first failure
• `--ff` : run failures from previous test before running other test cases

For other options, see `python -m pytest -h`

## Submitting Exercises

Note that, when trying to submit an exercise, make sure the solution is in the `\$EXERCISM_WORKSPACE/python/book-store` directory.

You can find your Exercism workspace by running `exercism debug` and looking for the line that starts with `Workspace`.

For more detailed information about running tests, code style and linting, please see Running the Tests.

## Source

Inspired by the harry potter kata from Cyber-Dojo. http://cyber-dojo.org

## Submitting Incomplete Solutions

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

### book_store_test.py

``````import unittest

from book_store import calculate_total

# Tests adapted from `problem-specifications//canonical-data.json` @ v1.4.0

class BookStoreTest(unittest.TestCase):
def test_only_a_single_book(self):
self.assertEqual(calculate_total([1]), 800)

def test_two_of_the_same_book(self):
self.assertEqual(calculate_total([2, 2]), 1600)

self.assertEqual(calculate_total([]), 0)

def test_two_different_books(self):
self.assertEqual(calculate_total([1, 2]), 1520)

def test_three_different_books(self):
self.assertEqual(calculate_total([1, 2, 3]), 2160)

def test_four_different_books(self):
self.assertEqual(calculate_total([1, 2, 3, 4]), 2560)

def test_five_different_books(self):
self.assertEqual(calculate_total([1, 2, 3, 4, 5]), 3000)

def test_two_groups_of_4_is_cheaper_than_group_of_5_plus_group_of_3(self):
self.assertEqual(calculate_total([1, 1, 2, 2, 3, 3, 4, 5]), 5120)

def test_two_groups_of_4_is_cheaper_than_groups_of_5_and_3(self):
self.assertEqual(calculate_total([1, 1, 2, 3, 4, 4, 5, 5]), 5120)

def test_group_of_4_plus_group_of_2_is_cheaper_than_2_groups_of_3(self):
self.assertEqual(calculate_total([1, 1, 2, 2, 3, 4]), 4080)

def test_two_each_of_first_4_books_and_1_copy_each_of_rest(self):
self.assertEqual(calculate_total([1, 1, 2, 2, 3, 3, 4, 4, 5]), 5560)

def test_two_copies_of_each_book(self):
self.assertEqual(calculate_total([1, 1, 2, 2, 3, 3, 4, 4, 5, 5]), 6000)

def test_three_copies_of_first_book_and_2_each_of_remaining(self):
self.assertEqual(
calculate_total([1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 1]), 6800)

def test_three_each_of_first_2_books_and_2_each_of_remaining_books(self):
self.assertEqual(
calculate_total([1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 1, 2]), 7520)

def test_four_groups_of_4_are_cheaper_than_2_groups_each_of_5_and_3(self):
self.assertEqual(
calculate_total([1, 1, 2, 2, 3, 3, 4, 5, 1, 1, 2, 2, 3, 3, 4, 5]),
10240)

if __name__ == '__main__':
unittest.main()``````
``````def calculate_total(books):
# One copy of any of the five books costs \$8.
# If, however, you buy two different books, you get a 5% discount on those two books.
# If you buy 3 different books, you get a 10% discount.
# If you buy 4 different books, you get a 20% discount.
# If you buy all 5, you get a 25% discount.

# so, we need to construct a complete set of the books, where the price ==
# (number of books * 8) * (1-discount[number of books-1])
# where discount = [0, .5, .10, .2, .25]

discount = [0, .05, .1, .2, .25]

combinations = [set()]
# question: does this get easier if it's sorted.
# yes? probably.
count_books = dict()
# actually want to order books by count? so most first:
# we'll do this using a dict
for book in books:
if count_books.get(book) is None:
count_books.update({book:1})
else:
count_books.update({book:count_books.get(book)+1})

count_tuples = list(count_books.items())

count_tuples.sort(key=lambda x: x[1], reverse=True)
ordered_books = []
for item in count_tuples:
ordered_books.extend([item[0] for i in range( item[1])])
# now we have a dict that contains the count of each book.
# need to make an array of each book starting with the highest count.
# i can use

for book in ordered_books:

for item in combinations:
print(item)

total_price = 0

for stack in combinations:
total_price += (len(stack)*800)*(1-discount[len(stack)-1])

# now, add to the best stack. to do this, we need to track the discount
# each possible place would get us, and then do the best one.
discount = [0, 0, .05, .1, .2, .25]
best_index = -1
index = -1
best_savings = -1
for stack in combinations:
index += 1
if book in stack:
continue
else:
savings = (len(stack)+1)*800*(discount[len(stack)+1]) - ((len(stack))*800*(discount[len(stack)]))
if savings > best_savings:
best_index = index
best_savings = savings

if best_index != -1 :
return
else:
# if we can't find a discount spot, move on.
combinations.append({book})
return

if __name__ == "__main__":
calculate_total([1, 1, 2, 2, 3, 3, 4, 5])``````