title: Count Number Of Ways To Fill A N X 4 Grid Using 1 X 4 Tiles
signature: function count_number_of_ways_to_fill_a_n_x_4_grid_using_1_x_4_tiles(n)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(n, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(n), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver(61, 188941273);
      @printf("TEST-0...%s\n",result);
      
      result = driver(22, 657);
      @printf("TEST-1...%s\n",result);
      
      result = driver(65, 685792227);
      @printf("TEST-2...%s\n",result);
      
      result = driver(57, 52054840);
      @printf("TEST-3...%s\n",result);
      
      result = driver(36, 59864);
      @printf("TEST-4...%s\n",result);
      
      result = driver(25, 1728);
      @printf("TEST-5...%s\n",result);
      
      result = driver(16, 95);
      @printf("TEST-6...%s\n",result);
      
      result = driver(26, 2385);
      @printf("TEST-7...%s\n",result);
      
      result = driver(92, 4123691589365);
      @printf("TEST-8...%s\n",result);
      
      result = driver(5, 3);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
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 ];
  }