KevinSia

# Determine whether a string contains a Social Security number.
def has_ssn?(string)
  !!string.match(/(\d{3})-(\d{2})-(\d{4})/)
end

puts "has_ssn? returns true if it has what looks like a SSN"
puts has_ssn?("please don't share this: 234-60-1422") == true

puts "has_ssn? returns false if it doesn't have a SSN"

KevinSia

# Fill in the methods below with your solution.
# Do not change the method names

# Iteration 1: Converting one word to Pig Latin

def convert_word_to_pig_latin(word)
  if word[0].downcase == 'a' || word[0].downcase == 'e' || word[0].downcase == 'i' || word[0].downcase == 'o' || word[0].downcase == 'u'
    word
  else
    str_front = ''

KevinSia

# TODO: Refactor for elegance
def shout_backwards(string)
  # before:
  # all_caps  = string.upcase
  # backwards = all_caps.reverse
  # result    = backwards + "!!!!"

  # return result
  string.upcase.reverse + "!!!!"
end

KevinSia

def prime_factors(num)
  return [num] if num <= 1
  
  result = []
  until num == 1
    for i in (2..num)
      if num % i == 0
        result << i
        num /= i
        break

KevinSia

def linear_search(target, arr)
  for i in 0...arr.length
    return i if target == arr[i]
  end
  return nil
end

def global_linear_search(target, arr)
  indices = []
  for i in 0...arr.length

KevinSia

# Place your solutions here
# Implement a method named binary_search
def binary_search(target, arr)
  s = 0
  e = arr.length - 1
  until s < e
    m = (s + e) / 2
    # found target
    if target == arr[m]
      return m

KevinSia

# Implement an iterative version of the factorial function
def factorial_iterative(n)
  return n if n <= 1
  total = 1
  n.downto(1) { |i| total *= i }
  return total
end

# def factorial_iterative(n)
#   return n if n <= 1

KevinSia

# OLD SOLUTION
# def fibonacci_iterative(n)
#   if n <= 1
#     return n
#   else
#     arr = [0, 1]
#     index = 0
#     (n - 1).times do
#       # arr might get very big if n is a big number
#       arr << arr[index] + arr[index + 1]

KevinSia

# generates a tic tac toe array with the right ratio of 'x' and 'o'
def generate_realistic_tic_tac_toe
  # hardcode 4 'X's and 4 'O's, then pick one on random from ['X', 'O']
  one_d = (%w(X X X X O O O O) << %w(X O).sample).shuffle
  return one_d.each_slice(3).to_a
end

# turns the array of players into array of hashes
def convert_roster_format(roster)
  header = roster.shift

KevinSia

• Displaying triangles:
  • https://gist.github.com/KevinSia/1c82b8a32f79722fa968c16dac4113ef
• Deaf aunty:
  • https://gist.github.com/KevinSia/8e2b8e5cf2d9d283bde3bc4b4ded7c21
• Roman Numerals:
  • https://gist.github.com/KevinSia/fa494967a804a83636450c5bf3c314ec
• Anagrams 1 & 2:
  • https://gist.github.com/KevinSia/e9f22ebae5984e7489ac14296b86d24e
• Dictionary sort:
• https://gist.github.com/KevinSia/033f24d950cfbd7d3b5f3a80a348e61f