Daily Challenge #1 - String Peeler

dev.to staff Jun 28 ・1 min read

#challenge

Daily Challenge (98 Part Series)

Hey, everyone. We've decided to host a daily challenge series. We'll have a variety of challenges, ranging in difficulty and complexity. If you choose to accept the task, your goal is to write the most efficient function you can for your language. Bonus points for any features you add!

I’ll start you off with a simple, straightforward challenge. Our first one comes from user @SteadyX on CodeWars.

Your goal is to create a function that removes the first and last letters of a string. Strings with two characters or less are considered invalid. You can choose to have your function return null or simply ignore.

The fundamentals are important, it only gets more difficult from here.

Happy coding!

Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge for a future post? Email yo+challenge@dev.to with your suggestions!

Daily Challenge (98 Part Series)

DISCUSS (80)

dev.to staff

The hardworking team behind dev.to ❤️

Alvaro Montoro

Jun 28

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

Jun 29

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

Jun 29

Let people do something 😂

Clay Murray

Jun 29

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?

Alvaro Montoro

Jun 28

In JavaScript it could be 24 bytes:

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

Ben Halpern

Jun 28

Ruby


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

MetaDave 🇪🇺

Aug 7

I think you can get away with string[1..-2] there, Ben.

Andrew Brown 🇨🇦

Jun 29

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

Ceban Dumitru

Jun 30

BASH

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

orenovadia

Sep 18

I was surprised to find that -1 to work as well:

V=abcde
echo "${V:1:-1}"

Corey Alexander

Jun 28

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...

Exequiel Beker

Jun 29

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;
}

Vladislav Guleaev

Jun 28

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

Martin Himmel

Jun 28

PHP

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

Claudio Scatoli

Jun 28

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 :/

Rafi

Sep 18

solution in SQL (postgres)

\set str 'hello world'

select (
     case
       when (select length(:'str') > 2) 
        then (select substr(:'str', 2, length(:'str') - 2) ) 
else 'invalid string' end );

Andreas Jakof

Sep 24

in C# as Extension-Method

public static string Peel(this string toPeel)
{
    int residual = toPeel.Length -2;
    if (residual < 1) return null;

    return toPeel.SubString(1,residual); //skip the first and take all but the last char
}

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" ;

Andreas Jakof

Sep 24

But what if the input would be "aaajldflbbb"?

When using Trim(char[]), all appearances of the characters given to the method at the beginning and at the end, will be removed, leaving "jldfl".

Removes all leading and trailing occurrences of a set of characters specified in an array from the current String object.

Tom E

Jun 28

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

Corey Alexander

Jun 28

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

Michael Lang

Aug 9

Ruby Language Version

With specs

def trim value
  value.to_s[1...-1]
end

require "spec"

describe "#trim" do
  it { expect(trim "Foo Bar").to eq "oo Ba"}
  it { expect(trim "Foo").to eq "o"}
  it { expect(trim "Fo").to eq ""}
  it { expect(trim "F").to eq ""}
  it { expect(trim nil).to be_nil}
  it { expect(trim 777).to eq "7"}
end

output

>> rspec debook_ends.rb
......

Finished in 0.00696 seconds (files took 0.15187 seconds to load)
6 examples, 0 failures

Devendra Dhanal

Aug 12

const stringTrimmer = input => input.length <=2 ? null : input.substring(1, input.length - 1)



console.log(stringTrimmer("Lorem Ipsum")) // orem Ipsu
console.log(stringTrimmer("John Doe")) // ohn Do
console.log(stringTrimmer("Fox jumps over")) // ox jumps ove
console.log(stringTrimmer("Lorem")) // ore
console.log(stringTrimmer("Lo")) // null

ralph macchio net wo

Oct 10

free psn code generator is one of the most popular products designed by Sony. Players require PSN scratch codes if they desire to unleash a myriad of content from the PlayStation Store.

Vinicius Cavagnolli

Jun 28

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;

Iago Angelim Cavalcante

Sep 11

Elixir

def remove_primeiro_ultimo_caracteres(string, caracter_removido \\ 1) do
    unless String.length(string) > (caracter_removido * 2) do
      raise 'Tamanho da string é menor que 2'
    end

    inicio = caracter_removido
    fim = -1 - caracter_removido
    String.slice(string, inicio..fim)
  end

Ganesh Prasad

Aug 24


const test = require ('./tester');

const f = str => str.length > 2 ? str.slice(1, -1) : null;
test (f, [
    {
        in: ['Dev'],
        out: 'e',
    },
    {
        in: ['GP'],
        out: null,
    },
    {
        in: ['hello'],
        out: 'ell'
    }
]);

[PASSED]  Case #1: e
[PASSED]  Case #2: null
[PASSED]  Case #3: ell

kespri

Oct 14

Swift

func removeFirstAndLastLetter(value: String) -> String? {
    guard value.count > 3 else { return nil }
    var result = value.dropFirst()
    result.popLast()
    return String(result)
}

DoMiNeLa10

Jun 30

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'));

Peter

Sep 24

Solution in Go:

package main

import (
    "fmt"
)

func main() {
    p, err := peel("hello")
    if err != nil {
        panic(err)
    }
    fmt.Println(p)
}

func peel(s string) (string, error) {
    if len(s) < 3 {
        return "", fmt.Errorf("Input string too short")
    }
    return s[1 : len(s)-1], nil
}

Running example in Go Playground

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.");
        }

}

Werner Echezuría

Jun 28

Rust

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

Ilyes Chouia

Aug 23

PHP:


function removeFirstAndLastLetter($text){
    return substr($text, 1, -1);
}
echo removeFirstAndLastLetter("Oh no, I lost two letters");

// Outputs : h no, I lost two letter

Shannon Crabill

Aug 2

I liked today's challenge so I'm going back to do past ones

def string_peeler(string)
  if string.length > 2
    string.delete(string[0]).delete(string[-1])
  end
end

Returns

string_peeler("hi") => nil 
string_peeler("killer") => "ille" 
string_peeler("BOB") => "O"

Josh

Aug 2

Trivial in Ruby:

def double_chop(string)
  return string if string.length <= 2
  string[1..-2]
end

double_chop("hello world") # => "ello worl"
double_chop("book")        # => "oo"
double_chop("OOP")         # => "O"
double_chop("hi")          # => "hi"

VIEW FULL DISCUSSION (80 COMMENTS) code of conduct - report abuse

Daily Challenge #1 - String Peeler

dev.to staff Jun 28 ・1 min read

dev.to staff

Ruby Language Version

With specs

output

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