title: Count Number Of Ways To Fill A N X 4 Grid Using 1 X 4 Tiles
signature: function countNumberOfWaysToFillANX4GridUsing1X4Tiles($n) {
test_code: <?php
  
  
  // SOLUTION CODE 
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  function validateSolution($actual, $expected){
      return serialize($actual) == serialize($expected);
  }
  
  function driver($n, $expected){
      try{
        if (validateSolution(PLACEHOLDER_FN_NAME($n), $expected)){
          return "PASSED";
        }
        return "FAILED";
      }
      catch (Exception $exception_obj) {
        return get_class($exception_obj);
      }
  }
  
  function main(){
      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));
      
  }
  
  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 ];
  }