Leaves - Elizabeth

lib/fibonacci.rb

@@ -1,8 +1,25 @@
 # Improved Fibonacci
 
-# Time Complexity - ?
-# Space Complexity - ?  (should be O(n))
+# Time Complexity - O(n) n being the number given
+# Space Complexity - O(n)   (should be O(n))
 # Hint, you may want a recursive helper method
 def fibonacci(n)
-
+  return fibonacci_helper([0,1], 2, n)
 end
+
+def fibonacci_helper(solutions, current, n)
+  if n < 0
+    raise ArgumentError
+  end
+
+  if n == 0 || n == 1
+    return n
+  end
+
+  if current == n
+    return solutions[n-1] + solutions[n-2]
+  end
+
+  solutions << solutions[current-1] + solutions[current-2]
+  return fibonacci_helper(solutions, current + 1, n)
+end

lib/super_digit.rb

@@ -1,10 +1,29 @@
+require "pry"
 # Superdigit
 
-# Time Complexity - ?
-# Space Complexity - ?
-def super_digit(n)
-
+# Time Complexity - super_digit_helper is O(log(n)) and this gets called O(n) in super_digit where n is the number of integers in num. So it's O(n).
+# Space Complexity - O(1)
+
+
+def super_digit_helper(n)
+  if n == 0
+    return 0
+  end
+
+  return n % 10 + super_digit_helper(n/10)
 end
+
+def super_digit(num)
+  # binding.pry
+  super_num = super_digit_helper(num)
+ if super_num > 9
+  super_digit(super_num)
+ else
+  return super_num
+ end
+
+end
+
 
 
 # Time Complexity - ?

test/super_digit_test.rb

@@ -1,6 +1,6 @@
 require_relative "test_helper"
 
-xdescribe "super_digit" do
+describe "super_digit" do
   it "will return 2 for super_digit(9875)" do
     # Act
     answer = super_digit(9875)
@@ -33,7 +33,7 @@
     expect(answer).must_equal 6
   end
 
-  describe "refined superdigit" do 
+  xdescribe "refined superdigit" do 
     it "will return 1 for n = 1 and k = 1" do
       # Act
       answer = refined_super_digit(1, 1)