title: How To Print Maximum Number Of A Using Given Four Keys
signature: def howToPrintMaximumNumberOfAUsingGivenFourKeys(n: Int) = {
test_code: import scala.collection.mutable.HashMap
  import scala.collection.immutable.HashSet
  
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  
  // TESTING CODE 
  // ============================================
  def validateSolution(actual: Long , expected: Long): Boolean = {
      actual.toString == expected.toString
  }
  
  def driver(n: Int, expected: Long): String = {
      try {
          if (validateSolution(PLACEHOLDER_FN_NAME(n),expected)) {
              "PASSED"
          } else{
              "FAILED"
          }
      }catch{
          case e: Exception => {e.getClass.getName}
      }
  
  }
  object QuestionEvaluator extends App {
  
      println(s"TEST-0...${driver(
        41, 
        110592L    
      )}")   
  
  
      println(s"TEST-1...${driver(
        94, 
        274877906944L    
      )}")   
  
  
      println(s"TEST-2...${driver(
        80, 
        5435817984L    
      )}")   
  
  
      println(s"TEST-3...${driver(
        40, 
        82944L    
      )}")   
  
  
      println(s"TEST-4...${driver(
        76, 
        1811939328L    
      )}")   
  
  
      println(s"TEST-5...${driver(
        5, 
        5L    
      )}")   
  
  
      println(s"TEST-6...${driver(
        43, 
        196608L    
      )}")   
  
  
      println(s"TEST-7...${driver(
        67, 
        150994944L    
      )}")   
  
  
      println(s"TEST-8...${driver(
        24, 
        1024L    
      )}")   
  
  
      println(s"TEST-9...${driver(
        90, 
        86973087744L    
      )}")   
  
  }
  
source_py: def how_to_print_maximum_number_of_a_using_given_four_keys(N):
      if (N <= 6):
          return N
      screen = ([0] * N)
      for n in range(1, 7):
          screen[(n - 1)] = n
      for n in range(7, (N + 1)):
          screen[(n - 1)] = max((2 * screen[(n - 4)]), max((3 * screen[(n - 5)]), (4 * screen[(n - 6)])))
      return screen[(N - 1)]
source_cpp: int how_to_print_maximum_number_of_a_using_given_four_keys(int N) {
    if ( N <= 6 ) return N;
    int screen [ N ];
    int b;
    int n;
    for ( n = 1;
    n <= 6;
    n ++ ) screen [ n - 1 ] = n;
    for ( n = 7;
    n <= N;
    n ++ ) {
      screen [ n - 1 ] = max ( 2 * screen [ n - 4 ], max ( 3 * screen [ n - 5 ], 4 * screen [ n - 6 ] ) );
    }
    return screen [ N - 1 ];
  }