Sprint Challenge: Algorithms | Oliver Abreu

README.md

@@ -112,3 +112,4 @@ Uncomment the `test_stretch_times()` test in `test_robot.py`. Can you optimize y
 
 ### 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 **average of at least 2** for all items on the rubric.
+<!-- Initial commit -->

Short-Answer/Algorithms_Answers.md

@@ -2,14 +2,93 @@
 
 ## Exercise I
 
-a)
+a) O(n) - Linear runtime, in the worst case
 
+<div style="-webkit-column-count: 2; -moz-column-count: 2; column-count: 2; -webkit-column-rule:">
+<div style="display: inline-block;">
+<h3>Original Code</h3>
+<pre>
+<code class="language-python">a = 0
+  while (a < n * n * n):    
+    a = a + n * n </code>
+</pre>
+</div>
+<div style="display: inline-block;">
+<h3>Justification</h3>
+<pre>
+<code class="language-python">O(1) <!-- simple variable assignment --->
+O(n) <!-- while loop, determined by size of `n`
+        since `a` is known to be 0 --->
+O(n) <!-- variable assigment again --->
+</code></pre>
+    </div>
+</div>
 
-b)
+b) O(n^2) - Quadratic runtime, in the worst case
 
+<div style="-webkit-column-count: 2; -moz-column-count: 2; column-count: 2; -webkit-column-rule: 1px dotted #e0e0e0; -moz-column-rule: 1px dotted #e0e0e0; column-rule: 1px dotted #e0e0e0;">
+<div style="display: inline-block;">
+<h3>Original Code</h3>
+<pre>
+<code class="language-python">sum = 0
+  for i in range(n):
+    j = 1
+    while j < n:
+      j *= 2
+      sum += 1  </code>
+</pre>
+</div>
+<div style="display: inline-block;">
+<h3>Justification</h3>
+<pre>
+<code class="language-python">O(1) <!-- simple variable assignment --->
+O(n) <!-- `for loop`, determined by size of `n`
+        since `a` is known to be 0 --->
+O(1) <!-- simple variable assignment  --->
+O(n^2) <!-- nested while loop --->
+O(2n) <!-- value is mutliped by 2 -->
+O(1) <!-- value is added to by 1 --></code>
+</pre>
+    </div>
+</div>
 
-c)
+c) O(n) - Linear runtime, in the worst case
 
-## Exercise II
+<div style="-webkit-column-count: 2; -moz-column-count: 2; column-count: 2; -webkit-column-rule:">
+<div style="display: inline-block;">
+<h3>Original Code</h3>
+<pre>
+<code class="language-python">def bunnyEars(bunnies):
+  if bunnies == 0:
+    return 0
 
+  return 2 + bunnyEars(bunnies-1)</code>
+</pre>
+</div>
+<div style="display: inline-block;">
+<h3>Justification</h3>
+<pre>
+<code class="language-python"> <!-- funtion declaration, no runtime -->
+<!-- `if statement`, no runtime -->  
+O(1) <!-- constant runtime -->
+O(2 + n) <!-- 2 + call of recursive function, dependent on size of variable --></code>
+</pre>
+</div>
+</div>
 
+## Exercise II
+```
+Go to the middle floor of the building by dividing `n` by two:
+  drop the egg from current floor:
+    if the egg does not break
+        go halfway up the building (by adding current floor to total floors (n) and dividing sum by two)
+        repeat step 2
+    if the egg does break
+      go up one floor
+        drop the egg from current floor
+          if the egg breaks
+            return current floor minus one as `f`
+          if the egg does not break
+            go to step 3
+```
+O(log n) - Logarithmic runtime, in the worst case, because the problem is halved on each pass, making the solution relatively quick to obtain.

Short-Answer/Algorithms_Questions.md

@@ -33,4 +33,4 @@ c)  def bunnyEars(bunnies):
 
 Suppose that you have an n-story building and plenty of eggs. Suppose also that an egg gets broken if it is thrown off floor f or higher, and doesn't get broken if dropped off a floor less than floor f. Devise a strategy to determine the value of f such that the number of dropped + broken eggs is minimized.
 
-Write out your proposed algorithm in plain English or pseudocode AND give the runtime complexity of your solution.
+Write out your proposed algorithm in plain English or pseudocode AND give the runtime complexity of your solution.

recursive_count_th/count_th.py

@@ -4,7 +4,19 @@
 Your function must utilize recursion. It cannot contain any loops.
 '''
 def count_th(word):
-
-    # TBC
-
-    pass
+
+    # If the overall length of the input word is less than 2, 
+    # the word can contain no occurences of our substring
+    if len(word) < 2:
+        return 0
+
+
+    # If the input word starts with input, add 1 to count,
+    # and continue to count remaining occurences of substring
+    elif word[:2] == "th":
+        return 1 + count_th(word[1:])
+
+    # Count all occurences of the substring
+    # starting from the second index
+    else:
+        return count_th(word[1:])

robot_sort/robot_sort.py

@@ -96,8 +96,23 @@ def sort(self):
         """
         Sort the robot's list.
         """
-        # Fill this out
-        pass
+        # if the robot cannot move left or right, 
+            # there is no `list` (there are no items to compare) - make no swap,
+            # and return 0
+        # else if the robot cannot move left, but can move right, begin comparing items and moving right
+            # [compare next item] if held item's value is `None` compared to item in front of robot ( == `None`),
+                # swap currently held item with the list item in front
+                # and move right
+            # if held item's value is equal to item in front of robot (== 0),
+                # make no swap,
+                # and move right
+            # else if held item's value is less than the item in front of robot (== -1),
+                # swap currently held item with the list item in front
+                # and move right
+            # else, the held item's value must be greater than item in front of robot (== 1)
+                # make no swap,
+                # and move right
+        # or else the robot can move left, but cannot move right, `list` must be already sorted
 
 
 if __name__ == "__main__":