title: Recursively Break Number 3 Parts Get Maximum Sum 1
signature: recursivelyBreakNumber3PartsGetMaximumSum1(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(
          50, 
          57)
      }");
  
      print("TEST-1...${driver(
          83, 
          93)
      }");
  
      print("TEST-2...${driver(
          18, 
          19)
      }");
  
      print("TEST-3...${driver(
          24, 
          27)
      }");
  
      print("TEST-4...${driver(
          31, 
          32)
      }");
  
      print("TEST-5...${driver(
          38, 
          41)
      }");
  
      print("TEST-6...${driver(
          94, 
          104)
      }");
  
      print("TEST-7...${driver(
          24, 
          27)
      }");
  
      print("TEST-8...${driver(
          13, 
          13)
      }");
  
      print("TEST-9...${driver(
          53, 
          57)
      }");
  
      return 0;
  }
source_py: def recursively_break_number_3_parts_get_maximum_sum_1(n):
      dp = ([0] * (n + 1))
      dp[0] = 0
      dp[1] = 1
      for i in range(2, (n + 1)):
          dp[i] = max(((dp[int((i / 2))] + dp[int((i / 3))]) + dp[int((i / 4))]), i)
      return dp[n]
source_cpp: int recursively_break_number_3_parts_get_maximum_sum_1(int n) {
    int dp [ n + 1 ];
    dp [ 0 ] = 0, dp [ 1 ] = 1;
    for ( int i = 2;
    i <= n;
    i ++ ) dp [ i ] = max ( dp [ i / 2 ] + dp [ i / 3 ] + dp [ i / 4 ], i );
    return dp [ n ];
  }