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Posted on Jun 28, 2019

Daily Challenge #1 - String Peeler

#challenge

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!

Top comments (114)

Alvaro Montoro • Jun 28 '19 • Edited

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.

Alvaro Montoro • Jun 29 '19

Thanks :)

Ryland G • Jun 28 '19 • Edited

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 '20

HAHAHA damn. Let me finish my code LOL.

Clay Murray • Jun 29 '19 • Edited

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 '20

This is indeed wild

Alvaro Montoro • Jun 28 '19 • Edited

In JavaScript it could be 24 bytes:

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

Andrew Brown 🇨🇦 • Jun 29 '19 • Edited

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 • Edited

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 • Edited

BASH

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

orenovadia • Sep 18 '19 • Edited

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

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

Corey Alexander • Jun 28 '19 • Edited

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;
}

Bárbara Perdigão • Jul 17 '19 • Edited

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.");
        }

}

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
}

Vladislav Guleaev • Jun 28 '19 • Edited

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 • Edited

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 • Jun 28 '19 • Edited

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

Praneet Nadkar • Jul 16 '19 • Edited

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.

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Daily Challenge #1 - String Peeler

Top comments (114)

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