class BinarySearchTest(TestCase):
"""Get the index of the item with an expected number of loops in\
array [1, 2 . . . 20]
Returns a dictionary containing {count: value, index: value}
"""
def setUp(self):
self.one_to_twenty = BinarySearch(20, 1)
self.two_to_forty = BinarySearch(20, 2)
self.ten_to_thousand = BinarySearch(100, 10)
def test_small_list_search(self):
search = self.one_to_twenty.search(16)
self.assertGreater(5,
search['count'],
msg='should return {count: 4, index: 15} for 16')
self.assertEqual(15,
search['index'],
msg='should return {count: 4, index: 15} for 16')
def test_medium_list_search(self):
search1 = self.two_to_forty.search(16)
search2 = self.two_to_forty.search(40)
search3 = self.two_to_forty.search(33)
self.assertGreater(5,
search1['count'],
msg='should return {count: 4, index: 7} for 16')
self.assertEqual(7,
search1['index'],
msg='should return {count: 4, index: 7} for 16')
self.assertEqual(0,
search2['count'],
msg='should return {count: 0, index: 19} for 40')
self.assertEqual(19,
search2['index'],
msg='should return {count: 5, index: 19} for 40')
self.assertGreater(4,
search3['count'],
msg='should return {count: 3, index: -1} for 33')
self.assertEqual(-1,
search3['index'],
msg='should return {count: 3, index: -1} for 33')
def test_large_list_search(self):
search1 = self.ten_to_thousand.search(40)
search2 = self.ten_to_thousand.search(880)
search3 = self.ten_to_thousand.search(10000)
self.assertGreater(
7,
search1['count'],
msg='should return {count: # <= 7, index: 3} for 40')
self.assertEqual(3,
search1['index'],
msg='should return {count: # <= 7, index: 3} for 40')
self.assertGreater(
4,
search2['count'],
msg='should return {count: # <= 3, index: 87} for 880')
self.assertEqual(
87,
search2['index'],
msg='should return {count: # <= 3, index: 87} for 880')
self.assertGreater(7,
search3['count'],
msg='should return {count: 3, index: -1} for 10000')
self.assertEqual(-1,
search3['index'],
msg='should return {count: 3, index: -1} for 10000')
def setUp(self):
self.one_to_twenty = BinarySearch(20, 1)
self.two_to_forty = BinarySearch(20, 2)
self.ten_to_thousand = BinarySearch(100, 10)
names_1 = f.read().split("\n") # List containing 10000 names
f.close()
f = open('names_2.txt', 'r')
names_2 = f.read().split("\n") # List containing 10000 names
f.close()
duplicates = [] # Return the list of duplicates in this data structure
# Replace the nested for loops below with your improvements
# for name_1 in names_1:
# for name_2 in names_2:
# if name_1 == name_2:
# duplicates.append(name_1)
tree_name = BinarySearch(names_1[0])
for name in range(1, len(names_1)):
tree_name.insert(names_1[name])
for names_2 in names_2:
if (tree_name.contains(names_2)):
duplicates.append(names_2)
end_time = time.time()
print(f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")
print(f"runtime: {end_time - start_time} seconds")
# ---------- Stretch Goal -----------
# Python has built-in tools that allow for a very efficient approach to this problem
# What's the best time you can accomplish? Thare are no restrictions on techniques or data
# structures, but you may not import any additional libraries that you did not write yourself.
