Daily Challenge #1 - String Peeler

#challenge

dev.to staff Jun 28, 2019 ・1 min read

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!

Posted on Jun 28 '19 by:

dev.to staff

@thepracticaldev

The hardworking team behind dev.to ❤️

Discussion

Alvaro Montoro

Jun 28 '19

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 '19

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 '19

Let people do something 😂

Jerome De Leon

Mar 30

HAHAHA damn. Let me finish my code LOL.

Clay Murray

Jun 29 '19

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?

Ibrahim Abdullahi Aliyu

Apr 15

This is indeed wild

Alvaro Montoro

Jun 28 '19

In JavaScript it could be 24 bytes:

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

Andrew Brown 🇨🇦

Jun 29 '19

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

Ben Halpern

Jun 28 '19

Ruby


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

MetaDave 🇪🇺

Aug 7 '19

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

Ceban Dumitru

Jun 30 '19

BASH

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

orenovadia

Sep 18 '19

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

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

Corey Alexander

Jun 28 '19

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 '19

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;
}

Andreas Jakof

Sep 24 '19

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
}

Bárbara Perdigão

Jul 17 '19

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.");
        }

}

Corey Alexander

Jun 28 '19

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

Jun 28 '19

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

Martin Himmel

Jun 28 '19

PHP

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

Claudio Scatoli

Jun 28 '19

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 '19

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 );

Tom Ekander

Jun 28 '19

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

Kasper Meyer

Jul 7 '19

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

Praneet Nadkar

Jul 16 '19

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 '19

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.

Michael Lang

Aug 9 '19

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

kespri

Oct 14 '19

Swift

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

jmc

Jun 30 '19

Clojure:

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

Adan ϟ

Nov 9 '19

I'll start doing this challenges in Dart :D

import 'dart:io';

main() {
  stdout.writeln('Phrase?');
  String phrase = stdin.readLineSync();
  phrase.length > 2 ? print(phrase.substring(1, phrase.length - 1)) : print('Invalid Phrase');
}

David Wickes

Oct 28 '19

Factor

USING: kernel math sequences ;
IN: dev.to-daily-challenge

: not-first-and-last ( str -- str ) dup length 2 >  [ rest but-last ] [ ] if ;

and some tests

USING: tools.test dev.to-daily-challenge ;
IN: dev.to-daily-challenge.tests

{ "ell" } [ "hello" not-first-and-last ] unit-test
{ "e" } [ "hel" not-first-and-last ] unit-test
{ "ha" } [ "ha" not-first-and-last ] unit-test
{ "" } [ "" not-first-and-last ] unit-test

Peter Bunting

Nov 23 '19

In F#. Not the most straightforward way to solve this, but I wanted to use the language Array splicing.

let stringToByteArray (s:string)= System.Text.Encoding.ASCII.GetBytes s
let byteArraySplice x y (b: byte[]) = b.[x..y]
let byteArrayToString (b: byte[]) = System.Text.Encoding.ASCII.GetString b

let stringPeeler s = 
    s
    |> stringToByteArray
    |> byteArraySplice 1 (s.Length - 2)
    |> byteArrayToString

printfn "%s" <| stringPeeler "abcdefgh"
printfn "%s" <| stringPeeler "ab"
printfn "%s" <| stringPeeler "a"

Peter Bunting

Nov 23 '19

And it turns out that you don't even need to explicitly convert it to an array. F# will let you do splicing on a string. So a much better solution is:

let stringPeeler (s: string) = s.[1..(s.Length - 2)]

Vivek

Dec 24 '19

Swift - 5

var str = "Hello, playground"
func stringPeeler(input: String) -> String? {
    if (input.count <= 2) {
        return nil;
    }

    return String(input.dropLast().dropFirst())
}

print(stringPeeler(input: str)!)

Rodrigo Sene

Jul 8 '19

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"));
    }
}

Shannon Crabill

Aug 2 '19

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"

Jarod Smith

Aug 16 '19


// "enterprise" solution
function peelString(str = '')
{
  //validate
  if(typeof(str) != 'string')
    throw new Error(`${str}, ${typeof(str)} is not a string`);
  if(str.length <= 2)
    throw new Error(`${str} does not have atleast 3 letters`);

  // process
  return str.substring(1,str.length-1);
}

My "enterprise" solution

David Figueroa

Jan 8

Powershell

function Remove-FirstAndLastCharacter {
    param(
        [parameter(Mandatory,ValueFromPipelineByName)]
        [string]$Input
    )
    if ($Input.length -ge 3)
    { 
        return $Input.SubString(1,$Input.Length - 2)
    }
}

Robin Victorino

Jul 9 '19

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]
}

flamangoes

Jul 26 '19

My approach in groovy...

Not sure if "ignore <2 chars" means "don't consider" or "ignore the trimming"

Don't consider =>

String peel(String toPeel) {
    toPeel[1..-2]
}

Ignore =>

String peel(String toPeel) {
    toPeel?.length() >2 ? toPeel[1..-2] : toPeel
}

I don't like returning null.

VIEW FULL DISCUSSION (112 COMMENTS) code of conduct - report abuse

Daily Challenge #1 - String Peeler

dev.to staff

Discussion

Ruby Language Version

With specs

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