Chris sprint2 - Data Structure Sprint challenge with Stretch Completed.

names/binary_search_tree.py

@@ -0,0 +1,53 @@
+
+
+class BinarySearchTree:
+    def __init__(self):
+        self.root_node = None
+
+    def isEmpty(self):
+        return self.root_node is None
+
+    def insert(self, value):
+        if self.isEmpty():
+            self.root_node = BSTNode(value)
+        else:
+            self.root_node.insert(value)
+
+    def contains(self, target):
+        if self.isEmpty():
+            return False
+        else:
+            return self.root_node.contains(target)
+
+
+# BSTNode class
+
+class BSTNode:
+    def __init__(self, value):
+        self.value = value
+        self.left = None
+        self.right = None
+
+    #Insert the given value into the tree
+    def insert(self, value):
+        if value < self.value:
+            if self.left is None:
+                self.left = BSTNode(value)
+            else:
+                self.left.insert(value)
+        elif value >= self.value:
+            if not self.right:
+                self.right = BSTNode(value)
+            else:
+                self.right.insert(value)
+
+
+    # Return True if the tree contains the value
+    # False if it does not
+    def contains(self, target):
+        if target == self.value:
+            return True
+        elif target < self.value:
+            return self.left.contains(target) if self.left else False
+        else:
+            return self.right.contains(target) if self.right else False

names/names.py

@@ -1,4 +1,5 @@
 import time
+from binary_search_tree import BinarySearchTree
 
 start_time = time.time()
 
@@ -13,10 +14,14 @@
 duplicates = []  # Return the list of duplicates in this data structure
 
 # Replace the nested for loops below with your improvements
+bst = BinarySearchTree()
+
 for name_1 in names_1:
-    for name_2 in names_2:
-        if name_1 == name_2:
-            duplicates.append(name_1)
+    bst.insert(name_1)
+
+for name_2 in names_2:
+    if bst.contains(name_2):
+        duplicates.append(name_2)
 
 end_time = time.time()
 print (f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")
@@ -26,3 +31,22 @@
 # 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()
+
+duplicates = [] # Returns the duplicates in this data structure
+
+# Replace the nested for loops below with your improvements
+duplicates = list(set(names_1) & set(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")

reverse/reverse.py

@@ -39,4 +39,9 @@ def contains(self, value):
         return False
 
     def reverse_list(self, node, prev):
-        pass
+        if node is None:
+            self.head = prev
+            return
+
+        self.reverse_list(node.get_next(), node)
+        node.set_next(prev)

ring_buffer/ring_buffer.py

@@ -1,9 +1,21 @@
 class RingBuffer:
-    def __init__(self, capacity):
-        pass
+    def __init__(self, capacity = 10):
+        self.capacity = capacity
+        self.size = 0
+        self.write_index = 0
+        self.storage = []
 
     def append(self, item):
-        pass
+        if self.size >= self.capacity:
+            self.storage[self.write_index] = item
+        else:
+            self.storage.append(item)
+            self.size += 1
+
+        if self.write_index < (self.capacity - 1):
+            self.write_index += 1
+        else:
+            self.write_index = 0
 
     def get(self):
-        pass
+        return self.storage