first push

README.md

@@ -2,7 +2,7 @@
 
 In this week's Sprint you implemented some classic and fundamental data structures and learned about how to go about evaluating their respective runtimes and performance. This Sprint Challenge aims to assess your comfort with these topics through exercises that build on the data structures you implemented and the algorithmic intuition you've started to build up.
 
-## Instructions
+## Instructions##
 
 **Read these instructions carefully. Understand exactly what is expected _before_ starting this Sprint Challenge.**
 
@@ -58,7 +58,7 @@ buffer.get()   # should return ['d', 'e', 'f']
 
 #### Task 2. Runtime Optimization
 
-***!Important!*** If you are running this using PowerShell by clicking on the green play button, you will get an error that `names1.txt` is not found.  To resolve this, run it, get the error, then `cd` into the `names` directory in the `python` terminal that opens in VSCode.
+**_!Important!_** If you are running this using PowerShell by clicking on the green play button, you will get an error that `names1.txt` is not found. To resolve this, run it, get the error, then `cd` into the `names` directory in the `python` terminal that opens in VSCode.
 
 Navigate into the `names` directory. Here you will find two text files containing 10,000 names each, along with a program `names.py` that compares the two files and prints out duplicate name entries. Try running the code with `python3 names.py`. Be patient because it might take a while: approximately six seconds on my laptop. What is the runtime complexity of this code?
 
@@ -71,28 +71,31 @@ A follow-up question to think about: _*once you've used one of the data structur
 Inside of the `reverse` directory, you'll find a basic implementation of a Singly Linked List. _Without_ making it a Doubly Linked List (adding a tail attribute), complete the `reverse_list()` function within `reverse/reverse.py`.
 
 For example,
+
 ```
 1->2->3->None
 ```
+
 would become...
+
 ```
 3->2->1->None
 ```
 
-While credit will be given for a functional solution, only optimal solutions will earn a ***3*** on this task.
+While credit will be given for a functional solution, only optimal solutions will earn a **_3_** on this task.
 
-#### Stretch 
-
-* Say your code from `names.py` is to run on an embedded computer with very limited RAM. Because of this, memory is extremely constrained and you are only allowed to store names in arrays (i.e. Python lists). How would you go about optimizing the code under these conditions? Try it out and compare your solution to the original runtime. (If this solution is less efficient than your original solution, include both and label the strech solution with a comment)
+#### Stretch
 
+- Say your code from `names.py` is to run on an embedded computer with very limited RAM. Because of this, memory is extremely constrained and you are only allowed to store names in arrays (i.e. Python lists). How would you go about optimizing the code under these conditions? Try it out and compare your solution to the original runtime. (If this solution is less efficient than your original solution, include both and label the strech solution with a comment)
 
 ### Rubric
-| TASK | 1 - DOES NOT MEET Expectations | 2 - MEETS Expectations | 3 - EXCEEDS Expectations | SCORE |
-| ----- | ------- | ------- | ------- | -- |
-| Task 1. Implement a Ring Buffer Data Structure | Solution in `ring_buffer.py` DOES NOT run OR it runs but has multiple logical errors, failing 2 or more tests. | Solution in `ring_buffer.py` runs, but may have one or two logical errors; passes at least 5/6 tests (Note that each function in the test file that begins with `test` is a test). | Solution in `ring_buffer.py` has no syntax or logical errors and passes all tests (Note that each function in the test file that begins with `test` is a test). | |
-| Task 2. Runtime Optimization | Student does NOT correctly identify the runtime of the starter code in `name.py` and is not able to optimize it to run in under 6 seconds using a data structure that was implemented during the week. | Student does not identify the runtime of the starter code in `name.py`, but optimizes it to run in under 6 seconds, with a solution that exhibits the appropriate runtime, using a data structure that was implemented during the week | Student does BOTH correctly identify the runtime of the starter code in `name.py` and optimizes it to run in under 6 seconds, with an appropriate runtime using a data structure that was implemented during the week. |  |
-| Task 3. Reverse the contents of a Singly Linked List | Student's solution in `reverse.py` is failing one or more tests. | Student's solution in `reverse.py` is able to correctly print out the contents of the Linked List in reverse order, passing all tests, BUT, the runtime of their solution is not optimal (requires looping through the list more than once). | Student's solution in `reverse.py` is able to correctly print out the contents of the Linked List in reverse order, passing all tests AND exhibits an appropriate runtime. |  |
 
+| TASK                                                 | 1 - DOES NOT MEET Expectations                                                                                                                                                                         | 2 - MEETS Expectations                                                                                                                                                                                                                       | 3 - EXCEEDS Expectations                                                                                                                                                                                               | SCORE |
+| ---------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----- |
+| Task 1. Implement a Ring Buffer Data Structure       | Solution in `ring_buffer.py` DOES NOT run OR it runs but has multiple logical errors, failing 2 or more tests.                                                                                         | Solution in `ring_buffer.py` runs, but may have one or two logical errors; passes at least 5/6 tests (Note that each function in the test file that begins with `test` is a test).                                                           | Solution in `ring_buffer.py` has no syntax or logical errors and passes all tests (Note that each function in the test file that begins with `test` is a test).                                                        |       |
+| Task 2. Runtime Optimization                         | Student does NOT correctly identify the runtime of the starter code in `name.py` and is not able to optimize it to run in under 6 seconds using a data structure that was implemented during the week. | Student does not identify the runtime of the starter code in `name.py`, but optimizes it to run in under 6 seconds, with a solution that exhibits the appropriate runtime, using a data structure that was implemented during the week       | Student does BOTH correctly identify the runtime of the starter code in `name.py` and optimizes it to run in under 6 seconds, with an appropriate runtime using a data structure that was implemented during the week. |       |
+| Task 3. Reverse the contents of a Singly Linked List | Student's solution in `reverse.py` is failing one or more tests.                                                                                                                                       | Student's solution in `reverse.py` is able to correctly print out the contents of the Linked List in reverse order, passing all tests, BUT, the runtime of their solution is not optimal (requires looping through the list more than once). | Student's solution in `reverse.py` is able to correctly print out the contents of the Linked List in reverse order, passing all tests AND exhibits an appropriate runtime.                                             |       |
 
 #### 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.

names/binary_search_tree.py

@@ -0,0 +1,77 @@
+''' things to remember. if you want to:
+delete - you must traverse
+on delete - smallest child becomes the parent
+>= flows right '''
+
+
+class BinarySearchTree:
+    def __init__(self, value):
+        self.value = value
+        self.left = None
+        self.right = None
+
+    ''' on insert, it adds the input value to the BST
+        it follows the rules of the order of elements
+        in order to find your insertion spot, you need to traverse'''
+
+    # Insert the given value into the tree
+    def insert(self, value):
+        if value < self.value:
+            if self.left == None:
+                self.left = BinarySearchTree(value)
+            else:
+                self.left.insert(value)
+        else:
+            if self.right == None:
+                self.right = BinarySearchTree(value)
+            else:
+                self.right.insert(value)
+
+    ''' 'contains' will search the BST for the input value
+        a boolean will be returned to confirm the value exists or not
+        Traverse the tree by starting at the root
+        On finding the first instance of the value - stop (return true)
+        if the search reaches a node that doesn't have children - then the value is not found (reutrn false)'''
+
+    # Return True if the tree contains the value
+    # False if it does not
+    def contains(self, target):
+        if self.value == target:
+            return True
+        else:
+
+            if target < self.value:
+                if self.left == None:
+                    return False
+                else:
+                    return self.left.contains(target)
+            else:
+                if self.right == None:
+                    return False
+                else:
+                    return self.right.contains(target)
+
+    ''' 'get max' will return max value in the BST
+         search moves right until it can go no further'''
+
+    # Return the maximum value found in the tree
+    def get_max(self):
+        if self.right == None:
+            return self.value
+        else:
+            return self.right.get_max()
+
+    ''' 'for each' traverses every node in the tree
+         executes the passed-in callback func on each node '''
+
+    # Call the function `cb` on the value of each node
+    # You may use a recursive or iterative approach
+    def for_each(self, cb):
+
+        cb(self.value)
+
+        if self.right != None:
+            self.right.for_each(cb)
+
+        if self.left != None:
+            self.left.for_each(cb)

names/names.py

@@ -1,6 +1,5 @@
 import time
-
-start_time = time.time()
+from binary_search_tree import BinarySearchTree
 
 f = open('names_1.txt', 'r')
 names_1 = f.read().split("\n")  # List containing 10000 names
@@ -10,17 +9,41 @@
 names_2 = f.read().split("\n")  # List containing 10000 names
 f.close()
 
-duplicates = []  # Return the list of duplicates in this data structure
+start_time = time.time()
+duplicates = []
+
+
+# Lambda initial starter Code
+
+print('\n Lambda slow poke code')
 
-# 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_time2 = time.time()
+print(f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")
+print(f"runtime: {end_time2 - start_time} seconds")
+
+
+# My code
+print("Under 1 second vroom vroom code")
+start_time2 = time.time()
+
+bst = BinarySearchTree(names_1[0])
+bstduplicates = []
+for name in names_1:
+    bst.insert(name)
+
+for name in names_2:
+    if bst.contains(name):
+        bstduplicates.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")
+print(f"{len(bstduplicates)} duplicates:\n\n{', '.join(bstduplicates)}\n\n")
+print(f"runtime: {end_time - start_time2} seconds")
 
 # ---------- Stretch Goal -----------
 # Python has built-in tools that allow for a very efficient approach to this problem

reverse/reverse.py

@@ -1,6 +1,8 @@
 class Node:
     def __init__(self, value=None, next_node=None):
+        # the value at this linked list node
         self.value = value
+        # reference to the next node in the list
         self.next_node = next_node
 
     def get_value(self):
@@ -10,15 +12,17 @@ def get_next(self):
         return self.next_node
 
     def set_next(self, new_next):
+        # set this node's next_node reference to the passed in node
         self.next_node = new_next
 
+
 class LinkedList:
     def __init__(self):
+        # reference to the head of the list
         self.head = None
 
     def add_to_head(self, value):
         node = Node(value)
-
         if self.head is not None:
             node.set_next(self.head)
 
@@ -27,16 +31,26 @@ def add_to_head(self, value):
     def contains(self, value):
         if not self.head:
             return False
-
+        # get a reference to the node we're currently at; update this as we traverse the list
         current = self.head
-
+        # check to see if we're at a valid node
         while current:
+            # return True if the current value we're looking at matches our target value
             if current.get_value() == value:
                 return True
-
+            # update our current node to the current node's next node
             current = current.get_next()
-
+        # if we've gotten here, then the target node isn't in our list
         return False
 
-    def reverse_list(self, node, prev):
-        pass
+    def reverse_list(self):
+        prev = None
+        current = self.head
+
+        while current is not None:
+            next = current.next_node
+            current.next_node = prev
+            prev = current
+            current = next
+
+        self.head = prev

reverse/test_reverse.py

@@ -1,44 +1,46 @@
 import unittest
 from reverse import LinkedList
 
+
 class LinkedListTests(unittest.TestCase):
-  def setUp(self):
-    self.list = LinkedList()
-
-  def test_add_to_head(self):
-    self.list.add_to_head(1)
-    self.assertEqual(self.list.head.value, 1)
-    self.list.add_to_head(2)
-    self.assertEqual(self.list.head.value, 2)
-
-  def test_contains(self):
-    self.list.add_to_head(1)
-    self.list.add_to_head(2)
-    self.list.add_to_head(10)
-    self.assertTrue(self.list.contains(2))
-    self.assertTrue(self.list.contains(10))
-    self.assertFalse(self.list.contains(1000))
-
-  def test_empty_reverse(self):
-    self.list.reverse_list(self.list.head, None)
-    self.assertEqual(self.list.head, None)
-
-  def test_single_reverse(self):
-    self.list.add_to_head(1)
-    self.list.reverse_list(self.list.head, None)
-    self.assertEqual(self.list.head.value, 1)
-
-  def test_longer_reverse(self):
-    self.list.add_to_head(1)
-    self.list.add_to_head(2)
-    self.list.add_to_head(3)
-    self.list.add_to_head(4)
-    self.list.add_to_head(5)
-    self.assertEqual(self.list.head.value, 5)
-    self.list.reverse_list(self.list.head, None)
-    self.assertEqual(self.list.head.value, 1)
-    self.assertEqual(self.list.head.get_next().value, 2)
-    self.assertEqual(self.list.head.get_next().get_next().value, 3)
-
+    def setUp(self):
+        self.list = LinkedList()
+
+    def test_add_to_head(self):
+        self.list.add_to_head(1)
+        self.assertEqual(self.list.head.value, 1)
+        self.list.add_to_head(2)
+        self.assertEqual(self.list.head.value, 2)
+
+    def test_contains(self):
+        self.list.add_to_head(1)
+        self.list.add_to_head(2)
+        self.list.add_to_head(10)
+        self.assertTrue(self.list.contains(2))
+        self.assertTrue(self.list.contains(10))
+        self.assertFalse(self.list.contains(1000))
+
+    def test_empty_reverse(self):
+        self.list.reverse_list()
+        self.assertEqual(self.list.head, None)
+
+    def test_single_reverse(self):
+        self.list.add_to_head(1)
+        self.list.reverse_list()
+        self.assertEqual(self.list.head.value, 1)
+
+    def test_longer_reverse(self):
+        self.list.add_to_head(1)
+        self.list.add_to_head(2)
+        self.list.add_to_head(3)
+        self.list.add_to_head(4)
+        self.list.add_to_head(5)
+        self.assertEqual(self.list.head.value, 5)
+        self.list.reverse_list()
+        self.assertEqual(self.list.head.value, 1)
+        self.assertEqual(self.list.head.get_next().value, 2)
+        self.assertEqual(self.list.head.get_next().get_next().value, 3)
+
+
 if __name__ == '__main__':
-  unittest.main()
+    unittest.main()