finsihed mvp

names/bst.py

@@ -0,0 +1,126 @@
+import random
+"""
+Binary search trees are a data structure that enforce an ordering over 
+the data they store. That ordering in turn makes it a lot more efficient 
+at searching for a particular piece of data in the tree. 
+
+This part of the project comprises two days:
+1. Implement the methods `insert`, `contains`, `get_max`, and `for_each`
+   on the BSTNode class.
+2. Implement the `in_order_print`, `bft_print`, and `dft_print` methods
+   on the 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)
+        else:
+            if self.right is None:
+                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:
+            if self.left:
+                return self.left.contains(target)
+            else:
+                return False
+        elif target > self.value:
+            if self.right:
+                return self.right.contains(target)
+            else:
+                return False
+
+    # Return the maximum value found in the tree
+    def get_max(self):
+        if self:
+            while self.right is not None:
+                self = self.right
+            return self.value
+
+        # recursive method
+        # if not self.right:
+        #     return self.value
+        # return self.right.get_max()
+
+    # Call the function `fn` on the value of each node
+    def for_each(self, fn):
+        fn(self.value)
+
+        if self.left:
+            self.left.for_each(fn)
+        if self.right:
+            self.right.for_each(fn)
+
+    # Part 2 -----------------------
+
+    # Print all the values in order from low to high
+    # Hint:  Use a recursive, depth first traversal
+    def in_order_print(self, node):
+        if node:
+            self.in_order_print(node.left)
+            print(node.value)
+            self.in_order_print(node.right)
+
+    # Print the value of every node, starting with the given node,
+    # in an iterative breadth first traversal
+    def bft_print(self, node):
+        from collections import deque
+
+        q = deque()
+        q.append(node)
+
+        while len(q) > 0:
+            current = q.popleft()
+
+            if current.left:
+                q.append(current.left)
+            if current.right:
+                q.append(current.right)
+            print(current.value)
+
+    # Print the value of every node, starting with the given node,
+    # in an iterative depth first traversal
+    def dft_print(self, node):
+        stack = []
+        stack.append(node)
+
+        while len(stack) > 0:
+            current = stack.pop()
+
+            if current.right:
+                stack.append(current.right)
+            if current.left:
+                stack.append(current.left)
+            print(current.value)
+
+    # Stretch Goals -------------------------
+    # Note: Research may be required
+
+    # Print Pre-order recursive DFT
+    def pre_order_dft(self, node):
+        if node:
+            print(node.value)
+            self.pre_order_dft(node.left)
+            self.pre_order_dft(node.right)
+
+    # Print Post-order recursive DFT
+    def post_order_dft(self, node):
+        if node:
+            self.post_order_dft(node.left)
+            self.post_order_dft(node.right)
+            print(node.value)

names/names.py

@@ -1,4 +1,5 @@
 import time
+from bst import BSTNode
 
 start_time = time.time()
 
@@ -12,17 +13,28 @@
 
 duplicates = []  # Return the list of duplicates in this data structure
 
+# O(n) * O(n) -> O(n^2)
 # 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_1 in names_1:
+#     for name_2 in names_2:
+#         if name_1 == name_2:
+#             duplicates.append(name_1)
+
+# O(n) + O(n) -> O(2n)
+bst = BSTNode("money")
+
+for n in names_1:
+    bst.insert(n)
+
+for n in names_2:
+    if bst.contains(n):
+        duplicates.append(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
+# What's the best time you can accomplish? There are no restrictions on techniques or data
 # structures, but you may not import any additional libraries that you did not write yourself.

reverse/reverse.py

@@ -39,4 +39,13 @@ def contains(self, value):
         return False
 
     def reverse_list(self, node, prev):
-        pass
+        if node is None:
+            return None
+
+        if node.next_node == None:
+            node.next_node = prev
+            self.head = node
+            return
+
+        self.reverse_list(node.next_node, node)
+        node.next_node = prev   

ring_buffer/ring_buffer.py

@@ -1,9 +1,16 @@
 class RingBuffer:
     def __init__(self, capacity):
-        pass
+        self.index = 0
+        self.capacity = capacity
+        self.storage = []
 
     def append(self, item):
-        pass
+        if len(self.storage) == self.capacity:
+            self.storage[self.index] = item
+        else:
+            self.storage.append(item)
+
+        self.index = (self.index + 1) % self.capacity
 
     def get(self):
-        pass
+        return self.storage