class BinarySearchTreeTests(unittest.TestCase):
def setUp(self):
self.bst = BSTNode(5)
def test_insert(self):
self.bst.insert(2)
self.bst.insert(3)
self.bst.insert(7)
self.bst.insert(6)
self.assertEqual(self.bst.left.right.value, 3)
self.assertEqual(self.bst.right.left.value, 6)
def test_handle_dupe_insert(self):
self.bst2 = BSTNode(1)
self.bst2.insert(1)
self.assertEqual(self.bst2.right.value, 1)
def test_contains(self):
self.bst.insert(2)
self.bst.insert(3)
self.bst.insert(7)
self.assertTrue(self.bst.contains(7))
self.assertFalse(self.bst.contains(8))
def test_get_max(self):
self.assertEqual(self.bst.get_max(), 5)
self.bst.insert(30)
self.assertEqual(self.bst.get_max(), 30)
self.bst.insert(300)
self.bst.insert(3)
self.assertEqual(self.bst.get_max(), 300)
def test_for_each(self):
arr = []
cb = lambda x: arr.append(x)
v1 = random.randint(1, 101)
v2 = random.randint(1, 101)
v3 = random.randint(1, 101)
v4 = random.randint(1, 101)
v5 = random.randint(1, 101)
self.bst.insert(v1)
self.bst.insert(v2)
self.bst.insert(v3)
self.bst.insert(v4)
self.bst.insert(v5)
self.bst.for_each(cb)
self.assertTrue(5 in arr)
self.assertTrue(v1 in arr)
self.assertTrue(v2 in arr)
self.assertTrue(v3 in arr)
self.assertTrue(v4 in arr)
self.assertTrue(v5 in arr)
duplicates = [] # Return the list of duplicates in this data structure
# O (n^2)
# This Method is 5.24 Seconds
# 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)
# 0(n log n) ?
# BST Method Done in 0.11 Seconds
bst = BSTNode("")
# Build BST
for name_2 in names_2:
bst.insert(name_2)
for name_1 in names_1:
if bst.contains(name_1):
duplicates.append(name_1)
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.
f = open('names_1.txt', 'r')
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
new_data = BSTNode("Jean")
for i in names_1:
new_data.insert(i)
for i in names_2:
if new_data.contains(i):
duplicates.append(i)
# O(n^2) -> O(n) & look up is O(log(n))
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.
class BinarySearchTreeTests(unittest.TestCase):
def setUp(self):
self.bst = BSTNode(5)
def test_insert(self):
self.bst.insert(2)
self.bst.insert(3)
self.bst.insert(7)
self.bst.insert(6)
self.assertEqual(self.bst.left.right.value, 3)
self.assertEqual(self.bst.right.left.value, 6)
def test_handle_dupe_insert(self):
self.bst2 = BSTNode(1)
self.bst2.insert(1)
self.assertEqual(self.bst2.right.value, 1)
def test_contains(self):
self.bst.insert(2)
self.bst.insert(3)
self.bst.insert(7)
self.assertTrue(self.bst.contains(7))
self.assertFalse(self.bst.contains(8))
def test_get_max(self):
self.assertEqual(self.bst.get_max(), 5)
self.bst.insert(30)
self.assertEqual(self.bst.get_max(), 30)
self.bst.insert(300)
self.bst.insert(3)
self.assertEqual(self.bst.get_max(), 300)
def test_get_min(self):
self.assertEqual(self.bst.get_min(), 5)
self.bst.insert(30)
self.assertEqual(self.bst.get_min(), 5)
self.bst.insert(300)
self.bst.insert(3)
self.assertEqual(self.bst.get_min(), 3)
def test_for_each(self):
arr = []
cb = lambda x: arr.append(x)
v1 = random.randint(1, 101)
v2 = random.randint(1, 101)
v3 = random.randint(1, 101)
v4 = random.randint(1, 101)
v5 = random.randint(1, 101)
self.bst.insert(v1)
self.bst.insert(v2)
self.bst.insert(v3)
self.bst.insert(v4)
self.bst.insert(v5)
self.bst.for_each(cb)
self.assertTrue(5 in arr)
self.assertTrue(v1 in arr)
self.assertTrue(v2 in arr)
self.assertTrue(v3 in arr)
self.assertTrue(v4 in arr)
self.assertTrue(v5 in arr)
def test_print_traversals(self):
# WARNING: Tests are for Print()
# Debug calls to Print() in functions will cause failure
stdout_ = sys.stdout # Keep previous value
sys.stdout = io.StringIO()
self.bst = BSTNode(1)
self.bst.insert(8)
self.bst.insert(5)
self.bst.insert(7)
self.bst.insert(6)
self.bst.insert(3)
self.bst.insert(4)
self.bst.insert(2)
self.bst.in_order_print(self.bst)
output = sys.stdout.getvalue()
self.assertEqual(output, "1\n2\n3\n4\n5\n6\n7\n8\n")
sys.stdout = io.StringIO()
self.bst.bft_print(self.bst)
output = sys.stdout.getvalue()
self.assertTrue(output == "1\n8\n5\n3\n7\n2\n4\n6\n" or
output == "1\n8\n5\n7\n3\n6\n4\n2\n")
sys.stdout = io.StringIO()
self.bst.dft_print(self.bst)
output = sys.stdout.getvalue()
self.assertTrue(output == "1\n8\n5\n7\n6\n3\n4\n2\n" or
output == "1\n8\n5\n3\n2\n4\n7\n6\n")
sys.stdout = io.StringIO()
self.bst.pre_order_dft(self.bst)
output = sys.stdout.getvalue()
self.assertEqual(output, "1\n8\n5\n3\n2\n4\n7\n6\n")
sys.stdout = io.StringIO()
self.bst.post_order_dft(self.bst)
output = sys.stdout.getvalue()
self.assertEqual(output, "2\n4\n3\n6\n7\n5\n8\n1\n")
sys.stdout = stdout_ # Restore stdout
start_time = time.time()
f = open('names_1.txt', 'r')
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
BSTNode(None)
for name_1 in names_1:
BSTNode.insert(name_1)
for name_2 in names_2:
if BSTNode.contains(name_2):
duplicates.append(name_1)
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.
for name_1 in names_1:
for name_2 in names_2:
if name_1 == name_2:
duplicates.append(name_1)
end_time = time.time()
print(f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")
print(f"runtime: {end_time - start_time} seconds")
start2 = time.time()
duplicates2 = []
search = BSTNode(names_1[0])
for n1 in names_1:
search.insert(n1)
for n2 in names_2:
if search.contains(n2):
duplicates2.append(n2)
end2 = time.time()
print(f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")
print(f"runtime: {end2 - start2} 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.
names_2_bst = BSTNode("")
for name in names_2:
names_2_bst.insert(name)
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)
for n1 in names_1:
if names_2_bst.contains(n1):
duplicates.append(n1)
end_time = time.time()
print("BST Version:")
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.
start_time2 = time.time()
dupes = []
names2_set = set(names_2)
f = open('names_1.txt', 'r')
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()
# CREATE TREE WITH names_1 LIST
start_time = time.time()
tree = BSTNode(names_1[0])
for name in names_1:
tree.insert(name)
duplicates = [name for name in names_2 if tree.contains(name)]
end_time = time.time()
print(f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")
print(f"runtime: {end_time - start_time} seconds")
# CREATE TREE WITH names_2 LIST (just in case order is better)
# ... it's not. it's the same
start_time = time.time()
tree = BSTNode(names_2[0])
for name in names_2:
tree.insert(name)
duplicates = [name for name in names_1 if tree.contains(name)]
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)
names_tree = BSTNode(names_1[0])
for name in names_1[1:]:
names_tree.insert(name)
for name in names_2:
if names_tree.contains(name):
duplicates.append(name)
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.
# ----------------------------------------------------------------------------
# --------------------T-H-I-S--I-S--F-A-S-T-----------------------------------
# duplicates = numpy.intersect1d(names_1, names_2)
# ----------------------------------------------------------------------------
# ----------------------------------------------------------------------------
# Instantiate a tree/node with the first name in the `name_1` file
b_s_tree = BSTNode(names_1[0])
# Put all the names in the file in the tree
for n in names_1:
b_s_tree.insert(n)
# List comprehension
duplicates = [n for n in names_2 if b_s_tree.contains(n)]
# ----------------------------------------------------------------------------
print('\nTime complexity: O(n log n)\n')
print('How many are common:', len(duplicates), '\n')
print('What\'s_common', duplicates)
end_time = time.time()
# print (f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")
print(f"\nRuntime: {end_time - start_time:.03} 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.
start_time = time.time()
f = open('names_1.txt', 'r')
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()
tree = BSTNode(names_1[0])
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:
tree.insert(name_1)
for name_2 in names_2:
if tree.contains(name_2):
duplicates.append(name_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.
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
bst_node = BSTNode("")
# We'll need to loop through all the names in 'names_1' file
for name in names_1:
# use the 'insert' method to add each name
bst_node.insert(name)
# Loop through all the names in 'name_2' file
for name in names_2:
# Use an if statement to check if the tree has identical names
if bst_node.contains(name):
# in the case it does, append those names to the duplicated list
duplicates.append(name)
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.
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
bst = BSTNode("")
# 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)
for name in names_2:
bst.insert(name)
for other_name in names_1:
if bst.contains(other_name):
duplicates.append(other_name)
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.
# 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)
# my implementation
# maybe use bst for one of the lists.
# method on bst to maybe compare the second method.
bst = BSTNode(names_1[0])
for name in names_1:
bst.insert(name)
for name in names_2:
if bst.contains(name):
duplicates.append(name)
end_time = time.time()
print (f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")
print (f"runtime: {end_time - start_time} seconds")
# runtime complexity of original answer
# quadratic time or O(n squared) because for every value visited in names1.txt, we have to
# visit every single value of names2.txt. thats every input of names1 multiplied by
# every value of names2. hence big o of n squared time. As the input increases, the number
# of computations is exponentially growing.
# ---------- Stretch Goal -----------
# Python has built-in tools that allow for a very efficient approach to this problem
f.close()
f = open('./names/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)
# runtime: 12.790003776550293 seconds
BST = BSTNode(names_1[0])
for name in names_1:
BST.insert(name)
for name in names_2:
if BST.contains(name):
duplicates.append(name)
# runtime: 0.16200661659240723 seconds
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.
f = open('names_1.txt', 'r')
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
bst = BSTNode(names_1[0])
# Replace the nested for loops below with your improvements
for name_1 in names_1:
bst.insert(name_1)
for name_2 in names_2:
if bst.contains(name_2):
duplicates.append(name_2)
# for name_1 in names_1:
# for name_2 in names_2:
# if name_1 == name_2:
# duplicates.append(name_1)
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.
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
tree = BSTNode(names_1[0])
duplicateCheck = [name for name in names_1[1:]]
for name in names_1:
tree.insert(name)
for name in names_2:
if tree.contains(name):
duplicates.append(name)
# 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)
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
