title: Write An Efficient Method To Check If A Number Is Multiple Of 3
signature: write_an_efficient_method_to_check_if_a_number_is_multiple_of_3 <- function(n) {
test_code: 
  
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  validate_solution <- function(actual, expected) {
      toString(actual)==toString(expected)
  }
  
  driver <- function(n, expected){
      if (validate_solution(PLACEHOLDER_FN_NAME(n), expected)){
          "PASSED"
      } else{
          "FAILED"
      }
  }
  
  main <- function() {
      cat(sprintf("TEST-0...%s\n",driver(
          94L, 
          0L
      )))
  
      cat(sprintf("TEST-1...%s\n",driver(
          94L, 
          0L
      )))
  
      cat(sprintf("TEST-2...%s\n",driver(
          79L, 
          0L
      )))
  
      cat(sprintf("TEST-3...%s\n",driver(
          39L, 
          1L
      )))
  
      cat(sprintf("TEST-4...%s\n",driver(
          16L, 
          0L
      )))
  
      cat(sprintf("TEST-5...%s\n",driver(
          90L, 
          1L
      )))
  
      cat(sprintf("TEST-6...%s\n",driver(
          64L, 
          0L
      )))
  
      cat(sprintf("TEST-7...%s\n",driver(
          76L, 
          0L
      )))
  
      cat(sprintf("TEST-8...%s\n",driver(
          83L, 
          0L
      )))
  
      cat(sprintf("TEST-9...%s\n",driver(
          47L, 
          0L
      )))
  
  }
  
  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 ) );
  }