title: Number Of Binary Trees For Given Preorder Sequence Length
signature: int numberOfBinaryTreesForGivenPreorderSequenceLength(int n) {
test_code: #include <algorithm>
  #include <assert.h>
  #include <cstdlib>
  #include <iostream>
  #include <map>
  #include <set>
  #include <string>
  #include <tuple> 
  #include <unordered_map>
  #include <vector>
  #include <math.h>
  #include <cmath>
  #include <bits/stdc++.h>
  #include <numeric>
  #include <functional>
  
  using namespace std;
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  bool validateSolution(int actual, int expected){
      return actual == expected;
  }
  
  string driver(int n, int expected){
      try
      {
          if (validateSolution(PLACEHOLDER_FN_NAME(n), expected)){
              return "PASSED";
          }
          return "FAILED";
      }
      catch (const std::overflow_error& exception_obj)
      {
          return typeid(exception_obj).name();
      }
      catch (const std::runtime_error& exception_obj)
      {
          return typeid(exception_obj).name();
      }
      catch (const std::exception& exception_obj)
      {
          return typeid(exception_obj).name();
      }
      catch (...)
      {
          return "UNK_ERROR";
      }
  }
  
  int main() {
      string result = "";
  
      result = driver(15, 9694845); 
      cout << "TEST-" << 0 << "..."
           <<  result
           << "\n";
  
      result = driver(11, 58786); 
      cout << "TEST-" << 1 << "..."
           <<  result
           << "\n";
  
      result = driver(11, 58786); 
      cout << "TEST-" << 2 << "..."
           <<  result
           << "\n";
  
      result = driver(15, 9694845); 
      cout << "TEST-" << 3 << "..."
           <<  result
           << "\n";
  
      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 ];
  }