title: Recursively Break Number 3 Parts Get Maximum Sum 1
signature: fun recursivelyBreakNumber3PartsGetMaximumSum1(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(50, 57)
      println("TEST-0...${result}")
      
      result = driver(83, 93)
      println("TEST-1...${result}")
      
      result = driver(18, 19)
      println("TEST-2...${result}")
      
      result = driver(24, 27)
      println("TEST-3...${result}")
      
      result = driver(31, 32)
      println("TEST-4...${result}")
      
      result = driver(38, 41)
      println("TEST-5...${result}")
      
      result = driver(94, 104)
      println("TEST-6...${result}")
      
      result = driver(24, 27)
      println("TEST-7...${result}")
      
      result = driver(13, 13)
      println("TEST-8...${result}")
      
      result = driver(53, 57)
      println("TEST-9...${result}")
      
          
  }
  
  main()
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 ];
  }