Daily Challenge #201 - Complete the Pattern

Implement a function pattern, which returns the following pattern for up to n number of rows. If n < 1 then it should return " " i.e. empty string. There are no whitespaces in the pattern.

Pattern:
1
22
333
....
.....
nnnnnn

Examples

pattern(5):
1
22
333
4444
55555

pattern(11):
1
22
333
4444
55555
666666
7777777
88888888
999999999
10101010101010101010
1111111111111111111111

Tests

pattern(4)
pattern(8)
pattern(0.5)

Good luck!

---

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Comments

[NecroTechno]

Rust :)

Playground link

fn pattern(input: f32) -> String {
    if input < 1.0 {
        return "".to_string();
    }
    let input_u32 = input.round() as u32;
    let mut response: String = "".to_string();
    for a in 0..input_u32 + 1 {
        for _b in 0..a {
            response.push_str(&a.to_string());
        }
        response.push_str("\n")
    }
    response
}

fn main() {
    println!("{}", pattern(4_f32));
    println!("{}", pattern(8_f32));
    println!("{}", pattern(0.5));
    println!("{}", pattern(2.5));
}

[kesprit]

My Swift solution :

func pattern(number: Double) -> String {
    guard number > 1 else { return " " }
    let numbers = (1...Int(number.rounded())).reduce(into: []) { $0.append($1) }
    return numbers.reduce(into: "") { $0.append("\(String.init(repeating: "\($1)", count: $1))\n") }
}

swift

[Rohit Prasad]

Python

var = float(input("Enter the number: "))

def pattern(var):
        if var < 1:
                print("")
        else:
                var = int(var)
                for i in range(var+1):
                        for j in range(i):
                                print(i,end = '')
                        print('\n',end='')

pattern(var)

[Dharmen Shah]

JavaScript

function pattern(len) {
  if(isNaN(len) || len <1 ) {
    console.log('');
    return;
  }
  console.log( Array.from({ length: len }).map((item, index)=>{
    return Array.from({ length: index+1 }).fill(index+1).join('');
  }).join('\n'));
}

[Ganesh Raja]

Python one-line solution

[print(str(i)*i) for i in range(1,n+1)]

[Avalander]

Haskell

pattern :: Int -> String
pattern n
  | n > 0     = unlines [repeatNum x | x <- [1..n]]
  | otherwise = ""
  where
    repeatNum x = foldl (++) "" $ take x $ repeat $ show x

Output:

pattern 4
{-
1
22
333
4444
-}

pattern 8
{-
1
22
333
4444
55555
666666
7777777
88888888
-}

pattern 0.5
-- Compiler error, ha ha.

[Julian Engel]

Python ... slightly more verbose :)

def pattern(n):

pat = ''
if n < 1:
    return pat
else:
    for index,value in enumerate(range(n),start=1):
        if value:
            pat += '\n'
        for i in range(index):
            pat += str(index)
    return pat

[Amit Patel]

Shortest snippet among all! That's why I ❤️ Ruby.

[Lucas Frigo de Souza]

C#

    class Program
    {
        static void Main(string[] args)
        {
            Pattern(5);
            Pattern(11);
            Pattern(4);
            Pattern(8);
            Pattern(0.5);
        }

        public static void Pattern(double number)
        {
            for (var i = 1; i <= number; i++)
            {
                var str = string.Empty;
                for (var j = 0; j < i; j++)
                {
                    str += i;
                }
                Console.WriteLine(str);
            }
        }
    }

//1
//22
//333
//4444
//55555
//1
//22
//333
//4444
//55555
//666666
//7777777
//88888888
//999999999
//10101010101010101010
//1111111111111111111111
//1
//22
//333
//4444
//1
//22
//333
//4444
//55555
//666666
//7777777
//88888888
//

[Jesus Urias]

Here is my solution with Elixir

defmodule Meh do
  def pattern(number) when number < 1, do: " "
  def pattern(number) do
    Enum.map(1..number, fn x -> IO.puts(String.duplicate(Integer.to_string(x), x)) end)
  end
end

here is the output:

iex(1)> Meh.pattern(6)
1
22
333
4444
55555
666666
[:ok, :ok, :ok, :ok, :ok, :ok]