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Daily Challenge #195 - No Zeroes for Heroes

dev.to staff on February 26, 2020

Setup It has officially been decided that numbers that end with zeroes are boring. They might be fun in your world, but definitely not h...

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Craig McIlwrath • Feb 27 '20 • Edited

Recursive Haskell solution

eradicateTrailingZeroes :: (Integral a) => a -> a
eradicateTrailingZeroes 0 = 0
eradicateTrailingZeroes n
  | n `rem` 10 == 0 = eradicateTrailingZeroes $ n `quot` 10
  | otherwise       = n

Avalander • Feb 27 '20

Does this work when passing 0 as argument?

Craig McIlwrath • Feb 27 '20

Good catch, guess I missed that case. Updated it.

zenmumbler • Feb 26 '20

I find it interesting that every solution here uses strings, here's one acting on the numbers itself. (tested against the above and other values).

function deathToZeroes(n) {
    while ((Math.abs(n) > 9) && (n % 10 === 0)) {
        n /= 10;
    }
    return n;
}

Cent • Feb 26 '20 • Edited

Javascript


const removeTrailingZeros = (number) => {
  if (number === 0) return number;
  if (number%10 === 0 ) removeTrailingZeros(number/10);
  else return number;
}

I just realized the function returns undefined when it's not 0 so let me try to fix that real quick lol


const removeTrailingZeros = (number) => {
  while(number % 10 === 0 && number !== 0) {
    number= number / 10;
  }
  return number;
}

codepen

Can someone tell my why the recursion I tried to do didn't work?

Vidit Sarkar • Feb 26 '20 • Edited

I guess you just need to add a 'return' in front of

removeTrailingZeros(number/10)

The code will be--

const removeTrailingZeros = (number) => {
  if (number === 0) return number;
  if (number%10 === 0 ) return removeTrailingZeros(number/10);
  else return number;
}

Cent • Feb 26 '20

Yep, that does it. Thanks!

SavagePixie • Feb 27 '20

Elixir

Elixir with recursion:

def dezerofy(0), do: 0
def dezerofy(n), do: _dezerofy(n, rem(n, 10))
defp _dezerofy(n, 0), do: _dezerofy(div(n, 10), rem(n, 100))
defp _dezerofy(n, _), do: n

Elixir without recursion, inspired by @avalander 's solution:

def dezerofy(0), do: 0
def dezerofy(n), do: n
  |> Integer.digits
  |> Enum.reverse
  |> Enum.drop_while(&(&1 == 0))
  |> Enum.reverse
  |> Integer.undigits

Avalander • Feb 26 '20 • Edited

Haskell

import Data.List (dropWhileEnd)

noZeroes :: Int -> Int
noZeroes 0 = 0
noZeroes n = (read . dropWhileEnd (== '0') . show) n

And the output:

main = do
  print (noZeroes 0)       -- 0
  print (noZeroes 9070)    -- 907
  print (noZeroes 210000)  -- 21
  print (noZeroes 10210)   -- 1021
  print (noZeroes 1450)    -- 145
  print (noZeroes 960000)  -- 96
  print (noZeroes 1050)    -- 105
  print (noZeroes (-1050)) -- -105

Vidit Sarkar • Feb 26 '20 • Edited

C++

int removeZeros(int number){
    // if number is 0 return 0
    if(number == 0)
        return 0;

    // if number is not 0 then divide the number by 10 until last digit of number is non-zero
    while(number%10 == 0){
        number /= 10;
    }
    return number;
}

Vidit Sarkar • Feb 26 '20 • Edited

Python one liner

eradicate = lambda number : 0 if number==0 else int(str(number).strip('0'))

print(eradicate(9070)) # 907
print(eradicate(210000)) # 21
print(eradicate(10210)) # 1021
print(eradicate(0)) # 0
print(eradicate(-1050)) # -105

Lucas Frigo de Souza • Feb 26 '20

        static void Main(string[] args)
        {
            System.Console.WriteLine("No Zero Challenge");
            Console.WriteLine(Eradicate(1450));   //145
            Console.WriteLine(Eradicate(960000)); //96
            Console.WriteLine(Eradicate(1050));   //105
            Console.WriteLine(Eradicate(-1050));  //-105
            Console.WriteLine(Eradicate(9070));   //907
            Console.WriteLine(Eradicate(210000)); //21
            Console.WriteLine(Eradicate(10210));  //1021
            Console.WriteLine(Eradicate(0));      //0
            Console.WriteLine(Eradicate(1234));   //1234
        }

        public static int Eradicate(int number)
        {
            if (number == 0) return number;
            while(number.ToString()[^1] == '0')
                number /= 10;
            return number;
        }

Lucas Frigo de Souza • Feb 26 '20

and Dart

void main() {
  print(eradicate(1450));   //145
  print(eradicate(960000)); //96
  print(eradicate(1050));   //105
  print(eradicate(-1050));  //-105
  print(eradicate(9070));   //907
  print(eradicate(210000)); //21
  print(eradicate(10210));  //1021
  print(eradicate(0));      //10
  print(eradicate(1234));   //1234
}

double eradicate (double number){
  if(number == 0) return number;

  while(number.remainder(10) == 0)
    number /= 10;

  return number;
}

Alexander Weleczka • Feb 26 '20

Felt like a bit of recursion is always fun to do.
And I couldn't remember when I used extensions in C# last...

public static int StripTrailingZeros(this int input)
{
    if(input != 0 && input % 10 == 0)
    {
        return (input / 10).StripTrailingZeros();
    }

    return input;
}

Project on GitHub

Alexander Weleczka • Feb 26 '20

Test Run Successful.
Total tests: 9
     Passed: 9
Total time: 1.6941 Seconds

Corentin Leffy • Mar 3 '20

A Dart solution using TDD :


void main() {
  group("Eradicate trailing zeroes :", () {
    final zeroesToHeroes = {
      1450: 145,
      960000: 96,
      1050: 105,
      -1050: -105,
      9070: 907,
      210000: 21,
      10210: 1021,
      0: 0
    };
    zeroesToHeroes.forEach((zero, hero) {
      test("$zero should become $hero", () {
        expect(eradicateTrailingZeroes(zero), equals(hero));
      });
    });
  });
}

num eradicateTrailingZeroes(num number) {
  for (; (number % 10 == 0) ^ (number == 0); number /= 10) {}
  return number;
}

Amin • Feb 26 '20 • Edited

Elm

eradicateZeros : Int -> Int
eradicateZeros number =
    if number == 0 then
        number

    else if remainderBy 10 number == 0 then
        eradicateZeros <| number // 10

    else
        number

Nijeesh Joshy • Feb 27 '20 • Edited

Ruby

Regex

# using strings
def remove_trailing_zeros(number)
  number
    .to_s
    .gsub(/0{1,}$/,"")
    .to_i || 0
end

Recursion

def remove_trailing_zeros(number)
  return 0 if number == 0
  return remove_trailing_zeros(number / 10) if number % 10 == 0
  return number
end

TESTS

require "test/unit"


class RemoveTrailingZerosTest < Test::Unit::TestCase
  def test_remove_trailing_zeros
    assert_equal 907, remove_trailing_zeros(9070)
    assert_equal 21, remove_trailing_zeros(210000)
    assert_equal 1021, remove_trailing_zeros(10210)
    assert_equal 0, remove_trailing_zeros(0)
  end
end

lormayna • Feb 26 '20

>>> def eradicate(n):
    n_as_str = str(n)
    if n_as_str == "0":
        return 0
    if n_as_str[-1] == "0":
        return eradicate(int(n_as_str[:-1]))
    else:
        return n


>>> eradicate(9070)
907
>>> eradicate(0)
0
>>> eradicate(96000)
96
>>> eradicate(21000)
21

Rafael Acioly • Feb 26 '20

Nice solution :)

Take a loot at strip method from str objects;
dev.to/rafaacioly/comment/m1ib

Empereol • Feb 26 '20 • Edited

TypeScript

function trimTrailingZeroes(input: number): number {
  return parseFloat(input.toString().replace(/(?<=\d)0+$/g, ""));
}

Ryan J • Feb 26 '20 • Edited

Javascript

function removeLast0(input) {
    if (input === 0) return 0;
    if (input % 10 === 0) return removeLast0(input / 10);
    return input;
}

Rafael Acioly • Feb 26 '20

Python solution 🐍

def eradicate(value: int) -> int:
    value_as_str = str(value)
    is_zero = value_as_str == "0"

    return 0 if is_zero else int(value_as_str.strip("0"))

matej • Feb 26 '20

Java with recursion

public int cutZeroes(int i) {
   return i == 0 || i%10 != 0 ? i : cutZeroes(i/10);
}

Rohit Prasad • Feb 27 '20

Python

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

while var != 0:
        if var%10 == 0:
                var = var/10
        else:
                break
print(int(var))