Oliver Abreu | Python Data Structures - Sprint Challenge

README.md

@@ -96,3 +96,4 @@ While credit will be given for a functional solution, only optimal solutions wil
 
 #### Passing the Sprint
 Score ranges for a 1, 2, and 3 are shown in the rubric above. For a student to have _passed_ a sprint challenge, they need to earn an **at least 2** for all items on the rubric.
+<!-- Initial Commit  -->

names/binary_search_tree.py

@@ -0,0 +1,31 @@
+class BSTNode:
+    def __init__(self, value):
+        self.value = value
+        self.left = None
+        self.right = None
+
+    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)
+
+    def contains(self, value):
+        if self.value == value:
+            return True
+        if value < self.value:
+            if not self.left:
+                return False
+            else:
+                return self.left.contains(value)                                            
+        else:
+            if not self.right:
+                return False
+            else:
+                return self.right.contains(value) 

names/names.py

@@ -1,22 +1,25 @@
 import time
+from binary_search_tree import BSTNode
 
 start_time = time.time()
 
-f = open('names_1.txt', 'r')
+f = open('./names/names_1.txt', 'r')
 names_1 = f.read().split("\n")  # List containing 10000 names
 f.close()
 
-f = open('names_2.txt', 'r')
+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)
+search_tree = BSTNode("None")
+for name in names_1:
+    search_tree.insert(name)
+for name in names_2:
+    if search_tree.contains(name):
+        duplicates.append(name)
 
 end_time = time.time()
 print (f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")

reverse/reverse.py

@@ -39,4 +39,12 @@ def contains(self, value):
         return False
 
     def reverse_list(self, node, prev):
-        pass
+        self.prev = None
+        current = self.head
+
+        while current is not None:
+            next_node = current.next_node
+            current.next_node = prev
+            prev = current
+            current = next_node
+        self.head = prev

ring_buffer/ring_buffer.py

@@ -1,9 +1,19 @@
 class RingBuffer:
     def __init__(self, capacity):
-        pass
+        self.capacity = capacity
+        self.current_index = 0
+        self.storage = []
 
     def append(self, item):
-        pass
+        """  
+            When buffer is full, place the value of `item` at the current index;
+            otherwise, append the item to the end of the buffer
+        """
+        if len(self.storage) == self.capacity:
+            self.storage[self.current_index] = item
+            self.current_index = (self.current_index + 1) % self.capacity
+        else:
+            self.storage.append(item)
 
     def get(self):
-        pass
+        return self.storage