title: Write An Efficient Method To Check If A Number Is Multiple Of 3
signature: function writeAnEfficientMethodToCheckIfANumberIsMultipleOf3($n) {
test_code: <?php
  
  
  // SOLUTION CODE 
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  function validateSolution($actual, $expected){
      return serialize($actual) == serialize($expected);
  }
  
  function driver($n, $expected){
      try{
        if (validateSolution(PLACEHOLDER_FN_NAME($n), $expected)){
          return "PASSED";
        }
        return "FAILED";
      }
      catch (Exception $exception_obj) {
        return get_class($exception_obj);
      }
  }
  
  function main(){
      printf("TEST-0...%s\n",driver(
        94, 
        0));
      
      printf("TEST-1...%s\n",driver(
        94, 
        0));
      
      printf("TEST-2...%s\n",driver(
        79, 
        0));
      
      printf("TEST-3...%s\n",driver(
        39, 
        1));
      
      printf("TEST-4...%s\n",driver(
        16, 
        0));
      
      printf("TEST-5...%s\n",driver(
        90, 
        1));
      
      printf("TEST-6...%s\n",driver(
        64, 
        0));
      
      printf("TEST-7...%s\n",driver(
        76, 
        0));
      
      printf("TEST-8...%s\n",driver(
        83, 
        0));
      
      printf("TEST-9...%s\n",driver(
        47, 
        0));
      
  }
  
  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 ) );
  }