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Daily Challenge #201 - Complete the Pattern

dev.to staff on March 04, 2020

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 stri...

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La blatte • Mar 4 '20 • Edited

Well, here we go for some JS oneliner:

p=i=>[...Array(0|i)].reduce((a,_,l)=>[...a,Array(l+1).fill(l+1).join``],[]).join`\n`

It works like this :

First, build an array with a 0|i length (it will floor the number)
Use the reduce method to loop through the built array. Use the a parameter, which is the response array (initialized as [] and updated at each iteration), and the l parameter which is the current index.
For each iteration, add to the a array an item which is described as follow : Build a l+1-long array, fill it with l+1 (hence, if we are at l=1, the built item will be equal to [2, 2]). Then, use the join method to convert it to a string ('22'in this case)
Finally, once the array is built, use the join method one last time to add line breaks between lines

Lamonte • Mar 4 '20

Dart, didn't bother to deal with the float and use dynamic

void pattern(int n) {
  if(n < 1) {
    print(" ");
  }
  for(var x = 1; x < n+1; x++) {
    print(x.toString() * x);
  }
}

pattern(4);
pattern(8);
// pattern(0.5); // fails, not an int lol

// 1
// 22
// 333
// 4444
// 1
// 22
// 333
// 4444
// 55555
// 666666
// 7777777
// 88888888

Dharmen Shah • Mar 6 '20 • Edited

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'));
}

Lucas Frigo de Souza • Mar 4 '20

    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
//

Vidit Sarkar • Mar 4 '20

I think there is no new-line at the end.
Here is C++ solution:

string pattern(int number){
    string pat = "";

    // there is no new-line at the end so loop end is number-1
    for(int i=1;i<=number-1;i++){
        for(int j=0;j<i;j++)
            pat += to_string(i); // add the string version of number
        pat += "\n"; // add a new-line after adding the numbers each time 
    }

    // add the case of number at the end with no new-line
    for(int i=0;i<number;i++)
        pat += to_string(number);

    return pat;
}

Vidit Sarkar • Mar 4 '20 • Edited

Python one liner :

pattern = lambda number : '\n'.join([str(i)*i for i in range(1,int(number)+1)])

Julian Engel • Mar 5 '20

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

NecroTechno • Mar 9 '20

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));
}

Rohit Prasad • Mar 8 '20

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)

Avalander • Mar 6 '20

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.

kesprit • Mar 9 '20

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

Jesus Urias • Mar 4 '20

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]

Craig McIlwrath • Mar 4 '20

Haskell

import Data.List (intercalate)

pattern :: Int -> String
pattern n = intercalate "\n" $ map concat $ take n rows
  where rows = [ map show $ replicate i i | i <- [1..] ]