title: Recursively Break Number 3 Parts Get Maximum Sum 1
signature: function recursivelyBreakNumber3PartsGetMaximumSum1(n) {
test_code: 
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  function convertToString(value) {
      return JSON.stringify(value, function (key, val) {
          if (val instanceof Set) {
              return Array.from(val);
          } return val;});
  }
  
  function validateSolution(actual, expected){
      return convertToString(actual)===convertToString(expected);
  }
  
  function driver(n, expected){
      try{
          if (validateSolution(PLACEHOLDER_FN_NAME(n),expected)){
              return "PASSED";
          }
          return "FAILED";
      } catch (error) {
        return error.name;
      }
      
  }
      
  function main(){
    let result = '';
    result = driver(50, 57);
    console.log(`TEST-0...${result}`);
  
    result = driver(83, 93);
    console.log(`TEST-1...${result}`);
  
    result = driver(18, 19);
    console.log(`TEST-2...${result}`);
  
    result = driver(24, 27);
    console.log(`TEST-3...${result}`);
  
    result = driver(31, 32);
    console.log(`TEST-4...${result}`);
  
    result = driver(38, 41);
    console.log(`TEST-5...${result}`);
  
    result = driver(94, 104);
    console.log(`TEST-6...${result}`);
  
    result = driver(24, 27);
    console.log(`TEST-7...${result}`);
  
    result = driver(13, 13);
    console.log(`TEST-8...${result}`);
  
    result = driver(53, 57);
    console.log(`TEST-9...${result}`);
  
    process.exit(0);
  };
  
  if (require.main === module) {
    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 ];
  }