title: Count Number Of Ways To Fill A N X 4 Grid Using 1 X 4 Tiles
signature: countNumberOfWaysToFillANX4GridUsing1X4Tiles :: Integer -> Integer
  countNumberOfWaysToFillANX4GridUsing1X4Tiles n = 
test_code: import Text.Printf
  import Control.Exception (catch, SomeException)
  import Data.Typeable
  import qualified Data.List as List
  import qualified Data.Map as Map
  import qualified Data.Set as Set
  
  -- SOLUTION CODE
  -- ============================================
  
  PLACEHOLDER_CODE_BODY
  
  -- TESTING CODE 
  -- ============================================
  validateSolution :: Integer -> Integer -> Bool
  validateSolution actual expected = do
      actual == expected
  
  
  driver :: Integer -> Integer -> String
  driver n expected = do
      if (validateSolution (PLACEHOLDER_FN_NAME n) expected)
          then "PASSED"
      else "FAILED"
  
  main = do
      printf "TEST-0...%s\n" (driver (61) (188941273) )
  
      printf "TEST-1...%s\n" (driver (22) (657) )
  
      printf "TEST-2...%s\n" (driver (65) (685792227) )
  
      printf "TEST-3...%s\n" (driver (57) (52054840) )
  
      printf "TEST-4...%s\n" (driver (36) (59864) )
  
      printf "TEST-5...%s\n" (driver (25) (1728) )
  
      printf "TEST-6...%s\n" (driver (16) (95) )
  
      printf "TEST-7...%s\n" (driver (26) (2385) )
  
      printf "TEST-8...%s\n" (driver (92) (4123691589365) )
  
      printf "TEST-9...%s\n" (driver (5) (3) )
  
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 ];
  }