old_contents: """Codewars: Are they the "same"?
  6 kyu
  
  URL: https://www.codewars.com/kata/550498447451fbbd7600041c
  
  Given two arrays a and b write a function comp(a, b) (compSame(a, b) in Clojure)
  that checks whether the two arrays have the "same" elements, with the same
  multiplicities. "Same" means, here, that the elements in b are the elements i
  a squared, regardless of the order.
  
  Examples
  Valid arrays
  a = [121, 144, 19, 161, 19, 144, 19, 11]  
  b = [121, 14641, 20736, 361, 25921, 361, 20736, 361]
  comp(a, b) returns true because in b 121 is the square of 11, 14641 is the
  square of 121, 20736 the square of 144, 361 the square of 19, 25921 the
  square of 161, and so on. It gets obvious if we write b's elements in terms of
  squares:
  a = [121, 144, 19, 161, 19, 144, 19, 11] 
  b = [11*11, 121*121, 144*144, 19*19, 161*161, 19*19, 144*144, 19*19]
  Invalid arrays
  If we change the first number to something else, comp may not return true
  anymore:
  a = [121, 144, 19, 161, 19, 144, 19, 11]  
  b = [132, 14641, 20736, 361, 25921, 361, 20736, 361]
  comp(a,b) returns false because in b 132 is not the square of any number of a.
  
  a = [121, 144, 19, 161, 19, 144, 19, 11]  
  b = [121, 14641, 20736, 36100, 25921, 361, 20736, 361]
  comp(a,b) returns false because in b 36100 is not the square of any number of a.
  
  Remarks
  - a or b might be [] (all languages except R, Shell). a or b might be nil or null
  or None or nothing (except in Haskell, Elixir, C++, Rust, R, Shell, PureScript).
  - If a or b are nil (or null or None), the problem doesn't make sense so return false.
  - If a or b are empty then the result is self-evident.
  - a or b are empty or not empty lists.
  """
  
  
  def comp(array1, array2):
      # your code
  
  
  def main():
      pass
  
  
  if __name__ == '__main__':
      main()
  
new_contents: """Codewars: Are they the "same"?
  6 kyu
  
  URL: https://www.codewars.com/kata/550498447451fbbd7600041c
  
  Given two arrays a and b write a function comp(a, b) (compSame(a, b) in Clojure)
  that checks whether the two arrays have the "same" elements, with the same
  multiplicities. "Same" means, here, that the elements in b are the elements i
  a squared, regardless of the order.
  
  Examples
  Valid arrays
  a = [121, 144, 19, 161, 19, 144, 19, 11]  
  b = [121, 14641, 20736, 361, 25921, 361, 20736, 361]
  comp(a, b) returns true because in b 121 is the square of 11, 14641 is the
  square of 121, 20736 the square of 144, 361 the square of 19, 25921 the
  square of 161, and so on. It gets obvious if we write b's elements in terms of
  squares:
  a = [121, 144, 19, 161, 19, 144, 19, 11] 
  b = [11*11, 121*121, 144*144, 19*19, 161*161, 19*19, 144*144, 19*19]
  
  Invalid arrays
  If we change the first number to something else, comp may not return true
  anymore:
  a = [121, 144, 19, 161, 19, 144, 19, 11]  
  b = [132, 14641, 20736, 361, 25921, 361, 20736, 361]
  comp(a,b) returns false because in b 132 is not the square of any number of a.
  
  a = [121, 144, 19, 161, 19, 144, 19, 11]  
  b = [121, 14641, 20736, 36100, 25921, 361, 20736, 361]
  comp(a,b) returns false because in b 36100 is not the square of any number of a.
  
  Remarks
  - a or b might be [] (all languages except R, Shell). a or b might be nil or null
  or None or nothing (except in Haskell, Elixir, C++, Rust, R, Shell, PureScript).
  - If a or b are nil (or null or None), the problem doesn't make sense so return false.
  - If a or b are empty then the result is self-evident.
  - a or b are empty or not empty lists.
  """
  
  
  def comp(a1, a2):
      from collections import defaultdict
  
      # Edge cases.
      if a1 is None or a2 is None:
          return False
  
      # Create dict:a1->count & dict:a2->count.
      a1_squared_count_d = defaultdict(int)
      for i in a1:
          a1_squared_count_d[i ** 2] += 1
  
      a2_count_d = defaultdict(int)
      for i in a2:
          a2_count_d[i] += 1
  
      # Iterate through a2 numbers to check its existence and count match.
      for i, count in a2_count_d.items():
          if i not in a1_squared_count_d or a1_squared_count_d[i] != count:
              return False
              
      return True
  
  
  def main():
      a1 = [121, 144, 19, 161, 19, 144, 19, 11]  
      a2 = [121, 14641, 20736, 361, 25921, 361, 20736, 361]
      assert comp(a1, a2) == True
  
      a1 = [121, 144, 19, 161, 19, 144, 19, 11]  
      a2 = [132, 14641, 20736, 361, 25921, 361, 20736, 361]
      assert comp(a1, a2) == False
  
      a1 = [121, 144, 19, 161, 19, 144, 19, 11]  
      a2 = [121, 14641, 20736, 36100, 25921, 361, 20736, 361]
      assert comp(a1, a2) == False
  
  
  if __name__ == '__main__':
      main()
  
subject: Complete dict sol