title: Recursively Break Number 3 Parts Get Maximum Sum
signature: fun recursivelyBreakNumber3PartsGetMaximumSum(n: Int): Int {
test_code: package main
  
  import kotlin.text.*
  import kotlin.math.*
  
  
  // SOLUTION CODE
  // ===================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE
  // ===================================
  fun validateSolution(actual:Int, expected:Int):Boolean{
    return actual.equals(expected)
  }
  fun driver(n: Int, expected:Int):String {
      try{
        if (validateSolution(PLACEHOLDER_FN_NAME(n), expected)) {
                return "PASSED"
        }
        return "FAILED"
      } catch (e: Exception){
        return e::class.simpleName as String
      }
  
  }
  
  fun main() {
      var result: String = ""
      result = driver(39, 41)
      println("TEST-0...${result}")
      
      result = driver(79, 87)
      println("TEST-1...${result}")
      
      result = driver(7, 7)
      println("TEST-2...${result}")
      
      result = driver(76, 87)
      println("TEST-3...${result}")
      
      result = driver(48, 57)
      println("TEST-4...${result}")
      
      result = driver(18, 19)
      println("TEST-5...${result}")
      
      result = driver(58, 63)
      println("TEST-6...${result}")
      
      result = driver(17, 17)
      println("TEST-7...${result}")
      
      result = driver(36, 41)
      println("TEST-8...${result}")
      
      result = driver(5, 5)
      println("TEST-9...${result}")
      
          
  }
  
  main()
source_py: def recursively_break_number_3_parts_get_maximum_sum(n):
      if ((n == 0) or (n == 1)):
          return n
      return max(((recursively_break_number_3_parts_get_maximum_sum((n // 2)) + recursively_break_number_3_parts_get_maximum_sum((n // 3))) + recursively_break_number_3_parts_get_maximum_sum((n // 4))), n)
source_cpp: int recursively_break_number_3_parts_get_maximum_sum(int n) {
    if ( n == 0 || n == 1 ) return n;
    return max ( ( recursively_break_number_3_parts_get_maximum_sum ( n / 2 ) + recursively_break_number_3_parts_get_maximum_sum ( n / 3 ) + recursively_break_number_3_parts_get_maximum_sum ( n / 4 ) ), n );
  }