title: Count Number Of Ways To Fill A N X 4 Grid Using 1 X 4 Tiles
signature: countNumberOfWaysToFillANX4GridUsing1X4Tiles(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(
          61, 
          188941273)
      }");
  
      print("TEST-1...${driver(
          22, 
          657)
      }");
  
      print("TEST-2...${driver(
          65, 
          685792227)
      }");
  
      print("TEST-3...${driver(
          57, 
          52054840)
      }");
  
      print("TEST-4...${driver(
          36, 
          59864)
      }");
  
      print("TEST-5...${driver(
          25, 
          1728)
      }");
  
      print("TEST-6...${driver(
          16, 
          95)
      }");
  
      print("TEST-7...${driver(
          26, 
          2385)
      }");
  
      print("TEST-8...${driver(
          92, 
          4123691589365)
      }");
  
      print("TEST-9...${driver(
          5, 
          3)
      }");
  
      return 0;
  }
source_py: def count_number_of_ways_to_fill_a_n_x_4_grid_using_1_x_4_tiles(n):
      dp = [0 for _ in range((n + 1))]
      for i in range(1, (n + 1)):
          if (i <= 3):
              dp[i] = 1
          elif (i == 4):
              dp[i] = 2
          else:
              dp[i] = (dp[(i - 1)] + dp[(i - 4)])
      return dp[n]
source_cpp: int count_number_of_ways_to_fill_a_n_x_4_grid_using_1_x_4_tiles(int n) {
    int dp [ n + 1 ];
    dp [ 0 ] = 0;
    for ( int i = 1;
    i <= n;
    i ++ ) {
      if ( i >= 1 && i <= 3 ) dp [ i ] = 1;
      else if ( i == 4 ) dp [ i ] = 2;
      else dp [ i ] = dp [ i - 1 ] + dp [ i - 4 ];
    }
    return dp [ n ];
  }