Daily Challenge #1 - String Peeler

CSS

Just add the class removeFirstAndLastLetter to a tag, and see the first and last letter "removed" from the text 😋

.removeFirstAndLastLetter {
  background: #fff;
  font-family: monospace;
  white-space: nowrap;
  position: relative;
  display: inline-block;
}

.removeFirstAndLastLetter::before,
.removeFirstAndLastLetter::after {
  content: "\00a0";
  background: #fff;
  top: 0;
  display: block;
  position: absolute;
  height: 100%;
  width: auto;
}

.removeFirstAndLastLetter::after {
  right: 0;
}

And as an extra, it can be stylized and animated, so you can see the letters fade out:

Ben Halpern

Ha! This one is great.

Thanks :)

JavaScript

(someString) => someString.length > 2 ? someString.slice(1, -1) : undefined;

Python

someString[1:-1] if len(someString) > 2 else None

C++

someString.size() > 2 ? someString.substr(1, someString.size() - 1) : null;

int trimString(char someString[])
{
  if (strlen(someString) > 2) {
    char *copy = malloc((strlen(someString) - 2) * sizeof(char));
    memmove (copy, someString + 1, strlen(someString) - 2);
    printf("%s", copy);
    free(copy);
    return 0;
  }
  return 1;
}

Rafael Acioly

Let people do something 😂

Ben Halpern

Ruby


def remove_first_and_last(string)
  raise "Invalid" if string.size < 3
  string[1..string.size-2]
end

Alvaro Montoro

In JavaScript it could be 24 bytes:

f=s=>s.slice(1,-1)||null

Clay Murray

Let's just go wild

(someString) => {
   switch(someString.length){
   case 0:
   case 1:
   case 2:
       return null;
   default:
       const arr = someString.split("")
       arr.pop();
       arr.reverse();
       arr.pop();
       arr.reverse();
       return arr.join("")
   }
}

Why not?

Corey Alexander

Rust

fn remove_first_and_last(string: &str) -> &str {
    remove_first_and_last_n_chars(&string, 1)
}

fn remove_first_and_last_n_chars(string: &str, chars_to_remove: usize) -> &str {
    if string.len() <= (2 * chars_to_remove) {
        panic!("Input string too short")
    }
    let start = chars_to_remove;
    let end = string.len() - 1 - chars_to_remove;

    &string[start..end]
} 

fn main() {
    println!("Ans: {}", remove_first_and_last("Hello, world!"));
    println!("Ans: {}", remove_first_and_last_n_chars("Hello, world!", 2));
    println!("Ans: {}", remove_first_and_last_n_chars("aa", 1));
    println!("Ans: {}", remove_first_and_last_n_chars("aa", 2));
}

View it in the Rust Playground here: play.rust-lang.org/?version=stable...

Andrew Brown 🇨🇦

I am surprised no one wrote test code. Sometimes in interviews with challenge this simple and you have access to run ruby they are expecting to see test code. Check out my ruby test code tutorials buds.

I notice lots of people are raising an error instead of ignoring or returning null so they have failed the challenge's instructions.

require "test/unit"

# remove the first and last letter of a string
# if there is less than 2 characters return zero.
def quartered value
  raise ArgumentError, 'Argument is not a string' unless value.is_a? String
  return value unless value.size > 2
  value[0] = ''
  value.chop
end


class QaurteredTest < Test::Unit::TestCase
  def test_quartered
    assert_equal 'orl', quartered('world'), "quartered('world') should return a string called 'orl'"
  end

  def test_quartered_ignore
    assert_equal 'hi', quartered('hi'), "quartered('hi') should return 'hi'"
  end

  def test_quartered_invalid
    assert_raise_message('Argument is not a string', "quartered(2) should raise exception") do
      quartered(2)
    end
  end
end

Martin Himmel

PHP

function remove_string_ends($str) {
    if (strlen($str) <= 2) {
        return null;
    }
    return substr($str, 1, -1);
}

Claudio Scatoli

I like that substr trick with the -1, didn't think of that!

How about this one liner?

<?php

function trimThis(string $str)
{
    return (mb_strlen($str) <= 2) ? null : substr($str,1,-1);
}

Use mb_string to support multibyte chars, such as Chinese

Also typehinted the argument, you never know...

This one is even shorter, possible only if the arg is typehinted tho

<?php
function trimThis(string $str) {
    return substr($str,1,-1) ?: null;
}

When the string is shorter than 2, substr returns false;
when the length is 2, substr returns "".

In both cases it's false, the the return is null.

Edit: many typos, I'm on mobile :/

Corey Alexander

Ruby

Basic

def remove_first_and_last(some_string)
  raise 'Not long enough' unless some_string.length > 2

  some_string[1..-2]
end

Extra

def remove_first_and_last_n_characters(some_string, chars_to_remove: 1)
  raise 'Not long enough' unless some_string.length > (chars_to_remove * 2)

  start_index = chars_to_remove
  end_index = -1 - chars_to_remove
  some_string[start_index..end_index]
end

Vladislav Guleaev

function removeChar(str){
 return str.substr(1, str.length - 2)
};

Praneet Nadkar

Jul 16

In C#, I would use in built string function Trim()
In my opinion we don't need Substring() here.
Just call :

readedInput.Trim(readedInput[0], readedInput[readedInput.Length - 1])

With 2 length validation:

var trimmed = readedInput.Length > 2 ? readedInput.Trim(readedInput[0], readedInput[readedInput.Length - 1]) : "Invalid" ;

Werner Echezuría

Rust

fn trim_string(string: &str) -> &str {
    if string.len() <= 2 {
        panic!("Input string too short");
    }
    &string[1..string.len() - 1]
}

Robin Victorino

Jul 9

Hi!

A bit late to the party, but here's a Groovy one:

String peel(String toPeel) {
    if(!(toPeel?.length() > 2)) {
        throw new Exception('Can\'t peel a string shorter than 3 characters')
    }
    return toPeel[1..-2]
}

Exequiel Beker

I tried to not use any str function.

char* trimFirstAndLastLetters(char* str)
{
    int index = 1;

    if(str[0] == '\0' || str[1] == '\0') {
        return NULL;
    }

    while(*(str + index) != '\0') {
        *(str + (index - 1)) = *(str + index);
        index++;
    }

    /* Remove last one */
    *(str + (index - 2)) = '\0';
    return str;
}

Kasper Meyer

Jul 7

Ruby solution

require "minitest/autorun"

class WordTrimmer
  def initialize word
    @word = word
  end

  def trim
    @word.length <= 2 ? @word : @word[1..-2]
  end
end

class WordTrimmerTest < MiniTest::Test
  def test_removing_first_and_last_letter
    assert_equal "aspe", WordTrimmer.new("kasper").trim
  end

  def test_ignoring_two_letter_word
    assert_equal "hi", WordTrimmer.new("hi").trim
  end

  def test_ignoring_one_letter_word
    assert_equal "I", WordTrimmer.new("I").trim
  end
end

Vinicius Cavagnolli

VB5 / VB6

Function Challenge1(input As String) As String
    If len(input) <= 2 Then
        Challenge1 = Nothing
    Else
        Challenge1 = Mid$(input, 2, len(input) - 2) 
    End If
End Function

Visual Basic .NET

Function Challenge1(input as String)
    Return If(input.Length > 2, Mid(input, 2, len(input) - 2), Nothing)
End Function

string Challenge1(string input) => input.Length > 2 ? input.Substring(1, input.Length - 2) : null;

C# using LINQ (because yes)

string Challenge1(string input) => input.Length > 2 ? string.Concat(input.Skip(1).Take(input.Length - 2)) : null;

Tom E

ReasonML

let strip = text =>
  String.(
    length(text) > 2
      ? switch (sub(text, 1, length(text) - 2)) {
        | s => Some(s)
        | exception _ => None
        }
      : None
  );

strip("hello"); // Some("ell")
strip("ab"); // None
strip(""); // None

Bárbara Perdigão

Jul 17

I wanted to take my chances with these challenges, but I decided to start from the beginning, so here's my (super) late entry, done in Java :

public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        System.out.println("Type the word:");
        String word = scan.nextLine();

        if (word.length() > 2) {
            System.out.printf("Result: %s%n", word.substring(1, word.length() - 1));
        } else {
            System.out.println("Invalid word.");
        }

}

DoMiNeLa10

const strip = (input) => {
  const str = String(input);
  const { length } = str;

  if (length < 3) {
    return null;
  }

  return str.substr(1, length - 2);
};

console.log(strip(''));
console.log(strip('12'));
console.log(strip('test'));

Rafael Acioly

func strip(content string) (string, error){
    if len(content) <= 2 {
        return "", error.New("invalid length")
    }

    return content[1:len(content)-1], nil
}

Rodrigo Sene

Jul 8

JAVA

public class Solution {

    public static String getStringCore(String input){
        if(input.length() > 2){
            return input.substring(1, input.length() -1);
        }
        return "invalid input";
    }

    public static void main(String[] arg){
        System.out.println(Solution.getStringCore("let's do it agin"));
    }
}

Ceban Dumitru

BASH

if [[ ${#1} > 2 ]]; then
    echo "${1:1:$(($#1-2))}";
else echo "not validated";
fi

jmc

Clojure:

(defn trim [s]
  (apply str (drop-last (rest s))))

Victoria Fluharty

Ruby

def remove_char(s)
  raise "invalid string" if s.size < 3

  return s[1..-1].chop
end

Jordon Replogle

Perl?

$testing = "Happy coding!";
print &peeler($testing) . "\n";

sub peeler {
    my $str = shift;
    if(length($str) > 2) {
        return substr($str, 1, -1);
    }
}

E. Choroba

Jul 3

Or maybe just

sub peeler { shift =~ s/^.|.$//gr }

Ben Lovy

Jul 9

I'm late to the party, catching up one by one!

Haskell:

trimAtBothEnds :: String -> Maybe String
trimAtBothEnds s = if ((length s) <= 2) then Nothing else Just (init.tail $ s)

Steve Moon

Jul 1

Erlang

fltrim (Input) when length(Input) > 2 ->
    string:slice(Input, 1, length(Input) - 2).

Akash Deep

Java:

someString.length()<=2?someString:someString.substring(1,someString.length()-1)

willsmart

Seat of your pants GNU C

printf(“String: %s”, s);
printf(“ trin : %s”, ({s[strlen(s)-1]=0;s+1;}));

(untested, also a bad idea in multiple ways)

Harshal Raverkar