title: Write An Efficient Method To Check If A Number Is Multiple Of 3
signature: function write_an_efficient_method_to_check_if_a_number_is_multiple_of_3(n)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(n, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(n), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver(94, 0);
      @printf("TEST-0...%s\n",result);
      
      result = driver(94, 0);
      @printf("TEST-1...%s\n",result);
      
      result = driver(79, 0);
      @printf("TEST-2...%s\n",result);
      
      result = driver(39, 1);
      @printf("TEST-3...%s\n",result);
      
      result = driver(16, 0);
      @printf("TEST-4...%s\n",result);
      
      result = driver(90, 1);
      @printf("TEST-5...%s\n",result);
      
      result = driver(64, 0);
      @printf("TEST-6...%s\n",result);
      
      result = driver(76, 0);
      @printf("TEST-7...%s\n",result);
      
      result = driver(83, 0);
      @printf("TEST-8...%s\n",result);
      
      result = driver(47, 0);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
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 ) );
  }