title: Number Of Binary Trees For Given Preorder Sequence Length
signature: numberOfBinaryTreesForGivenPreorderSequenceLength(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(
          15, 
          9694845)
      }");
  
      print("TEST-1...${driver(
          11, 
          58786)
      }");
  
      print("TEST-2...${driver(
          11, 
          58786)
      }");
  
      print("TEST-3...${driver(
          15, 
          9694845)
      }");
  
      return 0;
  }
source_py: def number_of_binary_trees_for_given_preorder_sequence_length(n):
      BT = ([0] * (n + 1))
      BT[0] = BT[1] = 1
      for i in range(2, (n + 1)):
          for j in range(i):
              BT[i] += (BT[j] * BT[((i - j) - 1)])
      return BT[n]
source_cpp: int number_of_binary_trees_for_given_preorder_sequence_length(int n) {
    int BT [ n + 1 ];
    memset ( BT, 0, sizeof ( BT ) );
    BT [ 0 ] = BT [ 1 ] = 1;
    for ( int i = 2;
    i <= n;
    ++ i ) for ( int j = 0;
    j < i;
    j ++ ) BT [ i ] += BT [ j ] * BT [ i - j - 1 ];
    return BT [ n ];
  }