title: Write An Efficient Method To Check If A Number Is Multiple Of 3
signature: def write_an_efficient_method_to_check_if_a_number_is_multiple_of_3(n):
test_code: import array
  import bisect
  import collections
  import datetime
  import functools
  import heapq
  import itertools
  import math
  import queue
  import re
  import string
  import sys
  from typing import Any, Dict, List, Optional, Set
  
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  def validate_solution(actual, expected):
      return actual == expected
  
  def driver(n, expected):
      try:
          if validate_solution(PLACEHOLDER_FN_NAME(
              n),
              expected
          ):
              return "PASSED"
          return "FAILED"
      except Exception as exception_obj:
          return type(exception_obj).__name__
      
  if __name__ == "__main__":
      result = driver(94, 0)
      print(f"TEST-0...{result}")
  
      result = driver(94, 0)
      print(f"TEST-1...{result}")
  
      result = driver(79, 0)
      print(f"TEST-2...{result}")
  
      result = driver(39, 1)
      print(f"TEST-3...{result}")
  
      result = driver(16, 0)
      print(f"TEST-4...{result}")
  
      result = driver(90, 1)
      print(f"TEST-5...{result}")
  
      result = driver(64, 0)
      print(f"TEST-6...{result}")
  
      result = driver(76, 0)
      print(f"TEST-7...{result}")
  
      result = driver(83, 0)
      print(f"TEST-8...{result}")
  
      result = driver(47, 0)
      print(f"TEST-9...{result}")
  
      sys.exit(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 ) );
  }