title: Count Ofdifferent Ways Express N Sum 1 3 4
signature: function count_ofdifferent_ways_express_n_sum_1_3_4(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(44, 821223649);
      @printf("TEST-0...%s\n",result);
      
      result = driver(9, 40);
      @printf("TEST-1...%s\n",result);
      
      result = driver(19, 4895);
      @printf("TEST-2...%s\n",result);
      
      result = driver(10, 64);
      @printf("TEST-3...%s\n",result);
      
      result = driver(78, 10472279279564025);
      @printf("TEST-4...%s\n",result);
      
      result = driver(70, 222915410843904);
      @printf("TEST-5...%s\n",result);
      
      result = driver(81, 44361286907595736);
      @printf("TEST-6...%s\n",result);
      
      result = driver(81, 44361286907595736);
      @printf("TEST-7...%s\n",result);
      
      result = driver(49, 9107509825);
      @printf("TEST-8...%s\n",result);
      
  end
  
  main()
source_py: def count_ofdifferent_ways_express_n_sum_1_3_4(n):
      DP = [0 for i in range(0, (n + 1))]
      DP[0] = DP[1] = DP[2] = 1
      DP[3] = 2
      for i in range(4, (n + 1)):
          DP[i] = ((DP[(i - 1)] + DP[(i - 3)]) + DP[(i - 4)])
      return DP[n]
source_cpp: int count_ofdifferent_ways_express_n_sum_1_3_4(int n) {
    int DP [ n + 1 ];
    DP [ 0 ] = DP [ 1 ] = DP [ 2 ] = 1;
    DP [ 3 ] = 2;
    for ( int i = 4;
    i <= n;
    i ++ ) DP [ i ] = DP [ i - 1 ] + DP [ i - 3 ] + DP [ i - 4 ];
    return DP [ n ];
  }