Simple coding challenges with Ruby - 01

Doing coding challenges is an excellent way to imporve your programming language & problem-solvings skills.

_{I'am learning and trying to improve my English. So I'm really sorry if my grammar mistakes annoy you. Thank you for your understanding.}

The Hacking Project (THP)#

You may be asking: "What is The Hacking Project?"

The Hacking Project (THP) is a type coding bootcamps within 12 weeks for the full-stack web development on Ruby & Ruby on Rails. The Hacking Project has been created in France since 2017 to help the people who want to learn a new coding skills, who want to create a startup or even for the people who want to make their career change into web development.

In 12 weeks, you will learn how to create a web application with the front-end skills (HMLT, CSS, JavaScript), the back-end skills with Ruby, Ruby on Rails and many others skills in IT developement.

This bootcamps is now free for every one. It's very impressive, isn't it?

If you interessed in THP, you can check out the website The Hacking Project for more details.

Why this serie "Ruby Coding Challenges with THP"?#

I'm a student of THP since 29/06/2020 (we are now in the 13e promo). In the second week, we started learning to code with Ruby language. After do the basic exercies, in the last day of that week, we have been suggested to complete some coding challenges on Codewars. If you don't know Codewars yet, it is a excelent plateform for software developers where contains a lot of programming puzzles with the different levels of difficulty. Those challenges have known also as "Kata Challenges". We can solve them with any programming language.

Practing katas suggested by THP helped me a lot to understand the fundamentals of Ruby and take some best practices in this language. It also help me to improve my problem-solving skills that is very important in the IT developement. I really love the Codewars now.

So, I write this serie to teach me again, more deep learning and also to share my though, my idea on each katas.

Let's talk about it!

Kata challenges#

As the beginer in Ruby, THP given us some fundamental challenges (7kyu level)

(Level of difficulty on Codewars: easy to difficult => 7kyu - 1kyu).

If you have already familiarized with Ruby, this first chapter may be not interesting for you. You can skip it and take the next one.

1. Vowel count (7kyu)#

For the first challenge, we will count the number of vowel in a given string.

:bell: Task: Vowel count

Return the number (count) of vowels in the given string. We will consider a, e, i, o, u as vowels for this Kata (but not y). The input string will only consist of lower case letters and/or spaces.

:arrow_right: My solution:

I use the method scan of String. It allows us to interate through a string, and matching with a given pattern of regular expression. If you don't know regular expression yet, you can take a look on Regular Expression Ruby.

Regular expression give us a lot of incredibles methods to work with string and it is also used in all programming languages.

def get_count(inputStr)
    inputStr.scan(/[aeiou]/).size
    # [aeiou]: search all volwel caracter
    # .scan: return new array which matching with the pattern
end

:heavy_check_mark: Best solution on codewars

def getCount(inputStr)
    inputStr.count("aeiou")
end

Another big advantage of codewars is learning to solve the problem from each other. When you submit your solution successfully, you can see the solution of others. There are a lot of developers who give the best practice and the very clever ideas to solve problems.

2. Remove the minimum (7kyu)#

:bell: Task: Remove the minimum

Given an array of integers, remove the smallest value. Do not mutate the original array/list. If there are multiple elements with the same value, remove the one with a lower index. If you get an empty array/list, return an empty array/list. ex: remove_smallest([1,2,3,4,5]) = [2,3,4,5]

:arrow_right: My solution:

• If the input array is empty, I will return an empty array
• We can't mutate the original array/list, so I create a clone array with the method ".dup" to keep the original.
• To remove the smallest values, I use the method ".delete"_at and give an index of the min values in array

def remove_smallest(numbers)
  if numbers.size <1
    return []
  end
  clone = numbers.dup
  clone.delete_at(numbers.index(numbers.min))
  clone
end

:heavy_check_mark: Best solution on codewars

def remove_smallest(numbers)
    numbers.reject.with_index { |_,i| i == numbers.index(numbers.min) }
end

Method "reject" is a reverse method of "select", we remove the items of array by condition. This method does not change the original array.

3. Get the middle caracter (7kyu)#

:bell: Task: Get the middle caracter

You are going to be given a word. Your job is to return the middle character of the word. If the word's length is odd, return the middle character. If the word's length is even, return the middle 2 characters. ex: Kata.getMiddle("test") should return "es" ex:Kata.getMiddle("testing") should return "t"

:arrow_right: My solution:

def get_middle(s)
  len = s.length
  if len <=1
    return s
  end
  if len % 2 == 0
    return s[len/2 -1] + s[len/2]
  else
    return s[len/2]
  end
end

:heavy_check_mark: Best solution on codewars

def get_middle(s)
  s[(s.size-1)/2..s.size/2]
end

def get_middle(s)
 mid = (s.length - 1) / 2
  s.length.odd? ? s[mid] : s[mid..mid+1]
end

Very impressive, is'nt it?

4. Jade casing strings (7kyu)#

:bell: Task: Jade casing strings

Your task is to convert strings to how they would be written by Jaden Smith. The strings are actual quotes from Jaden Smith, but they are not capitalized in the same way he originally typed them. ex: Not Jaden-Cased: "How can mirrors be real if our eyes aren't real" ex: Jaden-Cased: "How Can Mirrors Be Real If Our Eyes Aren't Real"

:arrow_right: My solution:

This kata is quite simple, I just use the ".map" method and convert each word with "capitalize" method.

def to_jade_case(str)
  return str.split.map(&:capitalize).join(' ')
end

Finally, I got a first success on codewars, because I have the same solution with the best solution on codewars :blush:

5. Disemvowel trolls (7kyu)#

:bell: Task: Disemvowel trolls

Your task is to write a function that takes a string and return a new string with all vowels removed.

For example, the string "This website is for losers LOL!" would become "Ths wbst s fr lsrs LL!".

:arrow_right: My solution:

• The regular expression is very strong with string. So I will use it for this solution.
• gsub: is a method to replace all caracters in string which match with the given text or pattern.

def disemvowel(str)
  str.gsub(/[aeiouAEIOU]/,'')
end

:heavy_check_mark: Best solution on codewars

def disemvowel(str)
  str.delete('aeiouAEIOU')
end

So they use a method "delete" more simple that I don't know yet. So cool!

6. Square every digit (7kyu)#

:bell: Task: Square every digit

YWelcome. In this kata, you are asked to square every digit of a number.

For example, if we run 9119 through the function, 811181 will come out, because 92 is 81 and 12 is 1.

:arrow_right: My solution:

• Using ".map" method to solve this problem

def square_digits num
  num.to_s.split('').map{|x| (x.to_i)**2}.join
end

:heavy_check_mark: Best solution on codewars

def square_digits num
  num.to_s.chars.map{|x| x.to_i**2}.join.to_i
end

def square_digits num
  num.digits.map { |d| d*d } .reverse.join.to_i
end

I see that there are many ways to solve this problem. With Codewars, I can learn more some tricks from others.

7. Shortest word (7kyu)#

:bell: Task: Shortest word

Welcome. In this kata, you are asked to square every digit of a number.

For example, if we run 9119 through the function, 811181 will come out, because 92 is 81 and 12 is 1.

:arrow_right: My solution:

• Using ".min_by" method

def find_short(s)
  return s.split.min_by{|word| word.length}.length
end

:heavy_check_mark: Best solution on codewars

def find_short(s)
  s.split.map(&:size).min
end

An alternative better way to solve it!

8. List filtering (7kyu)#

:bell: Task: List filtering

In this kata you will create a function that takes a list of non-negative integers and strings and returns a new list with the strings filtered out.

For example

filter_list([1,2,'a','b']) == [1,2]
filter_list([1,'a','b',0,15]) == [1,0,15]
filter_list([1,2,'aasf','1','123',123]) == [1,2,123]

:arrow_right: My solution:

def filter_list(l)
  return l.select{|item| item.is_a? Numeric}
end

:heavy_check_mark: Best solution on codewars

def filter_list(l)
  l.reject { |x| x.is_a? String }
end

9. Credit card mark (7kyu)#

:bell: Task: Credit card mark

Usually when you buy something, you're asked whether your credit card number, phone number or answer to your most secret question is still correct. However, since someone could look over your shoulder, you don't want that shown on your screen. Instead, we mask it. Your task is to write a function maskify, which changes all but the last four characters into '#'.

For example

maskify('4556364607935616') # should return '############5616'
maskify('64607935616')      # should return '#######5616'
maskify('1')                # should return '1'
maskify('')                 # should return ''

# "What was the name of your first pet?"
maskify('Skippy')                                   # should return '##ippy'
maskify('Nananananananananananananananana Batman!') # should return '####################################man!'

:arrow_right: My solution:

• I keep only 4 last caracters of string, and add the "#" (string.length-4) times in the first.

def maskify(cc)
  return cc.length <= 4 ? cc : ('#' * (cc.length-4)) + cc[cc.length-4..cc.length-1]
end

:heavy_check_mark: Best solution on codewars

def maskify(cc)
   cc.gsub(/.(?=....)/, '#')
end

def maskify(cc)
  cc.size <= 4 ? cc : "#" * (cc.length-4) + cc[-4..-1]
end

Very easy to read the code with the best solutions on codewars.

10. Categorize new member (7kyu)#

:bell: Task: Categorize new member

The Western Suburbs Croquet Club has two categories of membership, Senior and Open. They would like your help with an application form that will tell prospective members which category they will be placed.

To be a senior, a member must be at least 55 years old and have a handicap greater than 7. In this croquet club, handicaps range from -2 to +26; the better the player the lower the handicap.

Example input

[[18, 20],[45, 2],[61, 12],[37, 6],[21, 21],[78, 9]]

Example output

["Open", "Open", "Senior", "Open", "Open", "Senior"]

:arrow_right: My solution:

• The method ".map" is again take the job done

def open_or_senior(data)
  return data.map{|person| (person[0]>=55 && person[1]>7) ? "Senior": "Open"}
end

:heavy_check_mark: Best solution on codewars

def openOrSenior1(data)
  data.map { |age, handicap| age >= 55 && handicap > 7 ? "Senior" : "Open" }
end

11. Find The Parity Outlier (6kyu)#

For the last challenge, i will take an kata of 6kyu. It will be a little more complex of the previous challenges.

:bell: Task: Find The Parity Outlier

You are given an array (which will have a length of at least 3, but could be very large) containing integers. The array is either entirely comprised of odd integers or entirely comprised of even integers except for a single integer N. Write a method that takes the array as an argument and returns this "outlier" N..

For example

[2, 4, 0, 100, 4, 11, 2602, 36]
Should return: 11 (the only odd number)

[160, 3, 1719, 19, 11, 13, -21]
Should return: 160 (the only even number)

:arrow_right: My solution:

• Because our array contains an unique even number or odd number. So I will check 3 first items in array. If it contains 2 and more even number, so the number that we need to find is a odd number, and vice versa.
• After checking, we can easily find the result with the method ".select"

def maskify(cc)
  is_all_even = integers[0..2].select{|x| x%2 ==0}.size > 1
  return is_all_even ? integers.select{|x| x%2 != 0}[0] : integers.select{|x| x%2 == 0}[0]
end

:heavy_check_mark: Best solution on codewars

def find_outlier(integers)
  integers.partition(&:odd?).find(&:one?).first
end

def find_outlier(integers)
  outlier = integers.first(3).count(&:even?) < 2 ? :even? : :odd?
  integers.find(&outlier)
end

Thanks to the developers on the codewars, I could learn about the "partition" method. That return two arrays by the given condition. I could not never know this method if I just learn on the basic document.

Reference#

You can find the code in my github repository Codewars Ruby

You can also check out my another repository that contains all the programs day by day I worked with The Hacking project thehackingproject

The Hacking Project website thehackingproject.org

Codeware website coderwars.com

Series Ruby Coding Challenges#

1. Ruby Coding Challenges with TheHackingProject - 01
2. Ruby Coding Challenges with TheHackingProject - 02
3. Ruby Coding Challenges with TheHackingProject - 03

I hope you like this serie. The next challenges will be so much more instesting.