title: Count Even Length Binary Sequences With Same Sum Of First And Second Half Bits 1
signature: countEvenLengthBinarySequencesWithSameSumOfFirstAndSecondHalfBits1(n) {
test_code: import 'dart:convert';
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  bool validateSolution(double actual, double expected){
      return (actual - expected).abs() < 1e-09;
  }
  
  String driver(n, double 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(
          25, 
          126410606437752.0)
      }");
  
      print("TEST-1...${driver(
          18, 
          9075135300.0)
      }");
  
      print("TEST-2...${driver(
          17, 
          2333606220.0)
      }");
  
      print("TEST-3...${driver(
          8, 
          12870.0)
      }");
  
      print("TEST-4...${driver(
          8, 
          12870.0)
      }");
  
      return 0;
  }
source_py: def count_even_length_binary_sequences_with_same_sum_of_first_and_second_half_bits_1(n):
      nCr = 1
      res = 1
      for r in range(1, (n + 1)):
          nCr = ((nCr * ((n + 1) - r)) / r)
          res += (nCr * nCr)
      return res
source_cpp: int count_even_length_binary_sequences_with_same_sum_of_first_and_second_half_bits_1(int n) {
    int nCr = 1, res = 1;
    for ( int r = 1;
    r <= n;
    r ++ ) {
      nCr = ( nCr * ( n + 1 - r ) ) / r;
      res += nCr * nCr;
    }
    return res;
  }