title: Count Digits Factorial Set 2
signature: countDigitsFactorialSet2(n) {
test_code: import 'dart:convert';
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  bool validateSolution(int actual, int expected){
      return actual.toString() == expected.toString();
  }
  
  String driver(n, int expected){
      try
      {
          if (validateSolution(PLACEHOLDER_FN_NAME(n), expected)){
              return "PASSED";
          }
          return "FAILED";
      }catch(exception_object){
          return exception_object.runtimeType.toString();
      }
  }
  
  int main() {
  
      print("TEST-0...${driver(
          56, 
          75)
      }");
  
      print("TEST-1...${driver(
          92, 
          143)
      }");
  
      print("TEST-2...${driver(
          52, 
          68)
      }");
  
      print("TEST-3...${driver(
          96, 
          150)
      }");
  
      print("TEST-4...${driver(
          96, 
          150)
      }");
  
      print("TEST-5...${driver(
          63, 
          88)
      }");
  
      print("TEST-6...${driver(
          51, 
          67)
      }");
  
      print("TEST-7...${driver(
          79, 
          117)
      }");
  
      print("TEST-8...${driver(
          70, 
          101)
      }");
  
      print("TEST-9...${driver(
          9, 
          6)
      }");
  
      return 0;
  }
source_py: def count_digits_factorial_set_2(n):
      if (n < 0):
          return 0
      if (n <= 1):
          return 1
      x = ((n * math.log10((n / math.e))) + (math.log10(((2 * math.pi) * n)) / 2.0))
      return (math.floor(x) + 1)
source_cpp: long long count_digits_factorial_set_2(int n) {
    if ( n < 0 ) return 0;
    if ( n <= 1 ) return 1;
    double x = ( ( n * log10 ( n / M_E ) + log10 ( 2 * M_PI * n ) / 2.0 ) );
    return floor ( x ) + 1;
  }