title: Count Number Of Ways To Fill A N X 4 Grid Using 1 X 4 Tiles
signature: pub fn count_number_of_ways_to_fill_a_n_x_4_grid_using_1_x_4_tiles(n: i32) -> i64 {
test_code: use std::collections::HashMap;
  use std::collections::HashSet;
  use std::cmp::*;
  use std::collections::*;
  use std::fmt::*; 
  
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  pub fn validate_solution(actual: i64, expected: i64)->bool{
  return actual==expected
  }
  pub fn driver(n: i32, expected: i64)->String{
      if validate_solution(PLACEHOLDER_FN_NAME(n), expected) {
        return "PASSED".to_string();
      };
      return "FAILED".to_string();
  
  }
  
  fn main() {
      println!("TEST-0...{}", driver(
        61, 
        188941273
      ));
      
      println!("TEST-1...{}", driver(
        22, 
        657
      ));
      
      println!("TEST-2...{}", driver(
        65, 
        685792227
      ));
      
      println!("TEST-3...{}", driver(
        57, 
        52054840
      ));
      
      println!("TEST-4...{}", driver(
        36, 
        59864
      ));
      
      println!("TEST-5...{}", driver(
        25, 
        1728
      ));
      
      println!("TEST-6...{}", driver(
        16, 
        95
      ));
      
      println!("TEST-7...{}", driver(
        26, 
        2385
      ));
      
      println!("TEST-8...{}", driver(
        92, 
        4123691589365
      ));
      
      println!("TEST-9...{}", 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 ];
  }