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Recursion-writing Gessica Mohr #29

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6 changes: 6 additions & 0 deletions .idea/.gitignore

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4 changes: 4 additions & 0 deletions .idea/misc.xml

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8 changes: 8 additions & 0 deletions .idea/modules.xml

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27 changes: 27 additions & 0 deletions .idea/recursion-writing.iml

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36 changes: 36 additions & 0 deletions Gemfile.lock
Original file line number Diff line number Diff line change
@@ -0,0 +1,36 @@
GEM
remote: https://rubygems.org/
specs:
ansi (1.5.0)
builder (3.2.4)
coderay (1.1.3)
method_source (1.0.0)
minitest (5.14.2)
minitest-reporters (1.4.2)
ansi
builder
minitest (>= 5.0)
ruby-progressbar
minitest-skip (0.0.3)
minitest (~> 5.0)
minitest-spec (0.0.2.1)
minitest (>= 3.0)
pry (0.13.1)
coderay (~> 1.1)
method_source (~> 1.0)
rake (13.0.1)
ruby-progressbar (1.10.1)

PLATFORMS
ruby

DEPENDENCIES
minitest
minitest-reporters
minitest-skip
minitest-spec
pry
rake

BUNDLED WITH
2.1.4
98 changes: 70 additions & 28 deletions lib/recursive-methods.rb
Original file line number Diff line number Diff line change
@@ -1,49 +1,91 @@
# Authoring recursive algorithms. Add comments including time and space complexity for each method.

# Time complexity: ?
# Space complexity: ?
# Time complexity: O(n) it depends on the size of n to run
# Space complexity: O(n) it creates a stack every time it calls the recursive method
def factorial(n)
Comment on lines +3 to 5

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👍

raise NotImplementedError, "Method not implemented"
raise ArgumentError if n < 0
return 1 if n == 0
return n * factorial(n - 1)

end

# Time complexity: ?
# Space complexity: ?
# Time complexity: O(n^2) - the slice method in the reverse method creates another array so that counts for O(n) and then the method runs n times depending on the size of array, O(n). Total of O(n^2).
# Space complexity: O(n^2) - because we are creating a new array when using the slice method and it creates a stack to run the recursion method.
def reverse(s)
Comment on lines +12 to 14

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👍

raise NotImplementedError, "Method not implemented"
if s.length <= 1
return s
end

return s[-1] + reverse(s[1..-2]) + s[0]
end

# Time complexity: ?
# Space complexity: ?
def reverse_inplace(s)
raise NotImplementedError, "Method not implemented"

# Time complexity: O(n) it depends on the size of s to run
# Space complexity: O(n) it creates a stack every time if calls the recursive method
def reverse_inplace(s, i = 0, j = s.length - 1)
Comment on lines +23 to +25

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👍

return s if s.length == 0 || s.length == 1
return s if i >= j
temp = s[i]
s[i] = s[j]
s[j] = temp

return reverse_inplace(s, i + 1, j - 1)
end

# Time complexity: ?
# Space complexity: ?

# Time complexity: O(n) it depends on the size of n to run
# Space complexity: O(n) it creates a stack every time if calls the recursive method
def bunny(n)
Comment on lines +36 to 38

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👍

raise NotImplementedError, "Method not implemented"
return 0 if n <= 0
return 2 + bunny(n - 1)
end

# Time complexity: ?
# Space complexity: ?
def nested(s)
raise NotImplementedError, "Method not implemented"
# Time complexity: O(n) it depends on the size of s to run
# Space complexity: O(n) it creates a stack every time if calls the recursive method
def nested(s, i = 0, j = s.length - 1)
Comment on lines +43 to +45

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👍 Very compact!

return true if s.empty? || i > j && s[i] != s[j]
return false if s.length == 1 || s[i] == s[j]
return nested(s, i + 1, j - 1)

end

# Time complexity: ?
# Space complexity: ?
def search(array, value)
raise NotImplementedError, "Method not implemented"
# Time complexity: O(n) it depends on the size of the array to run
# Space complexity: O(n) it creates a stack every time if calls the recursive method
def search(array, value, index = 0)
Comment on lines +52 to +54

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👍

return false if index >= array.length
return true if array[index] == value
return search(array, value, index + 1)
end

# Time complexity: ?
# Space complexity: ?
def is_palindrome(s)
raise NotImplementedError, "Method not implemented"
# Time complexity: O(n) it depends on the size of s to run
# Space complexity: O(n) it creates a stack every time if calls the recursive method
def is_palindrome(s, i = 0, j = s.length - 1)
Comment on lines +60 to +62

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👍

return true if s.empty? || i >= j
return false if s[i] != s[j]

return is_palindrome(s, i + 1, j - 1)
end

# Time complexity: ?
# Space complexity: ?
# Time complexity: O(log n)
# Space complexity: O(log n)
def digit_match(n, m)
Comment on lines +69 to 71

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👍

raise NotImplementedError, "Method not implemented"

return 1 if n == 0 && m == 0
return 0 if n <= 1 || m <= 1

if n % 10 == m % 10
return 1 + digit_match(n / 10, m / 10)
else
return digit_match(n / 10, m / 10)
end
end



# Time complexity: O(2^n) it depends on the size of n to run
# Space complexity: O(2^n) it creates a stack every time if calls the recursive method
def fibonacci(n)
Comment on lines +85 to +87

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👍 However the space complexity is O(n)

return 0 if n == 0
return 1 if n == 1
return fibonacci(n - 1) + fibonacci(n - 2)
end
35 changes: 31 additions & 4 deletions test/recursion_writing_test.rb
Original file line number Diff line number Diff line change
Expand Up @@ -167,7 +167,7 @@
end
end

xdescribe "nested" do
describe "nested" do
it "will return true for empystring" do
# Arrange
string = ""
Expand Down Expand Up @@ -213,7 +213,7 @@
end
end

xdescribe "search" do
describe "search" do
it "will return false for empty array" do
# Arrange
item = "a"
Expand Down Expand Up @@ -263,7 +263,7 @@
end
end

xdescribe "is_palindrome" do
describe "is_palindrome" do
it "will return true for emptystring" do
# Arrange
string = ""
Expand Down Expand Up @@ -298,7 +298,7 @@
end
end

xdescribe "digit_match" do
describe "digit_match" do
it "returns 4 for 1072503891 and 62530841" do
# Arrange
num1 = 1072503891
Expand Down Expand Up @@ -359,3 +359,30 @@
expect(answer).must_equal 1
end
end

describe "fib" do
it "returns 0 for fib(0)" do
n = 0
answer = fibonacci(n)
expect(answer).must_equal 0
end

it "returns 1 for fibonnaci(1)" do

n = 1
answer = fibonacci(n)
expect(answer).must_equal 1
end

it "returns 3 for fibonnaci(4)" do
n = 4
answer = fibonacci(n)
expect(answer).must_equal 3
end

it "returns 34 for (9)" do
n = 9
answer = fibonacci(n)
expect(answer).must_equal 34
end
end