Fixes contains() in bst to work with strings, and adds option of usin…

…g bst for names.py

names/binary_search_tree.py

@@ -48,7 +48,9 @@ def contains(self, target):
         # Return True if the tree contains the value
         # False if it does not
         while self:
-            if target is self.value:
+            # print(self.value)
+            # print(target)
+            if target == self.value:
                 return True
             elif target < self.value:
                 if not self.left:

names/names.py

@@ -5,37 +5,21 @@
 
 start_time = time.time()
 
-bst = BinarySearchTree('Heather')
-# f = open('names_1.txt', 'r')
 with open(os.path.join(sys.path[0], 'names_1.txt'), 'r') as f:
     names_1 = f.read().split("\n")  # List containing 10000 names
-    # names_1.insert(f.read().split("\n"))
-# f.close()
 
-# print(names_1[0])
-# print('Jean Velazquez' < 'Heather')
-# bst.insert(names_1[0])
-# print(bst.get_max())
-# print(bst.contains('Jean Velazquez'))
+# Uncomment below to use the bst:
 '''
+bst = BinarySearchTree('Heather')
 for name in names_1:
     bst.insert(name)
 '''
 
-# print(bst.get_max())
-# print(bst.contains('Jean Velazquez'))
-# print(bst.contains('Heather'))
-# print(bst.contains('Zoie Lloyd'))
-
-# f = open('names_2.txt', 'r')
 with open(os.path.join(sys.path[0], 'names_2.txt'), 'r') as f:
     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
-
 
 def binary_search(arr, target):
     low = 0
@@ -51,16 +35,25 @@ def binary_search(arr, target):
     return -1
 
 
+# This implementation doesn't use the bst
+#  0.08800506591796875 seconds
 names_1.sort()
 for name in names_2:
     if binary_search(names_1, name) != -1:
         duplicates.append(name)
+
+
+# This is the original implementation:
+# About 10 seconds
 '''
 for name_1 in names_1:
     for name_2 in names_2:
         if name_1 == name_2:
             duplicates.append(name_1)
 '''
+
+# This implementation uses the bst:
+# 0.13100862503051758 seconds
 '''
 for name in names_2:
     if bst.contains(name):