title: Write An Efficient Method To Check If A Number Is Multiple Of 3
signature: def writeAnEfficientMethodToCheckIfANumberIsMultipleOf3(n: Int) = {
test_code: import scala.collection.mutable.HashMap
  import scala.collection.immutable.HashSet
  
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  
  // TESTING CODE 
  // ============================================
  def validateSolution(actual: Int , expected: Int): Boolean = {
      actual.toString == expected.toString
  }
  
  def driver(n: Int, expected: Int): String = {
      try {
          if (validateSolution(PLACEHOLDER_FN_NAME(n),expected)) {
              "PASSED"
          } else{
              "FAILED"
          }
      }catch{
          case e: Exception => {e.getClass.getName}
      }
  
  }
  object QuestionEvaluator extends App {
  
      println(s"TEST-0...${driver(
        94, 
        0    
      )}")   
  
  
      println(s"TEST-1...${driver(
        94, 
        0    
      )}")   
  
  
      println(s"TEST-2...${driver(
        79, 
        0    
      )}")   
  
  
      println(s"TEST-3...${driver(
        39, 
        1    
      )}")   
  
  
      println(s"TEST-4...${driver(
        16, 
        0    
      )}")   
  
  
      println(s"TEST-5...${driver(
        90, 
        1    
      )}")   
  
  
      println(s"TEST-6...${driver(
        64, 
        0    
      )}")   
  
  
      println(s"TEST-7...${driver(
        76, 
        0    
      )}")   
  
  
      println(s"TEST-8...${driver(
        83, 
        0    
      )}")   
  
  
      println(s"TEST-9...${driver(
        47, 
        0    
      )}")   
  
  }
  
source_py: def write_an_efficient_method_to_check_if_a_number_is_multiple_of_3(n):
      odd_count = 0
      even_count = 0
      if (n < 0):
          n = (- n)
      if (n == 0):
          return 1
      if (n == 1):
          return 0
      while n:
          if (n & 1):
              odd_count += 1
          if (n & 2):
              even_count += 1
          n = (n >> 2)
      return write_an_efficient_method_to_check_if_a_number_is_multiple_of_3(abs((odd_count - even_count)))
source_cpp: int write_an_efficient_method_to_check_if_a_number_is_multiple_of_3(int n) {
    int odd_count = 0;
    int even_count = 0;
    if ( n < 0 ) n = - n;
    if ( n == 0 ) return 1;
    if ( n == 1 ) return 0;
    while ( n ) {
      if ( n & 1 ) odd_count ++;
      if ( n & 2 ) even_count ++;
      n = n >> 2;
    }
    return write_an_efficient_method_to_check_if_a_number_is_multiple_of_3 ( abs ( odd_count - even_count ) );
  }