✔ TECHNICAL ASSESSMENT

This project contains programming challenges that can be solved by using the language of your choice, preferred Java, Go, JavaScript or Ruby.

You´re welcome to look inside /src/ folder. You may find helpful the code already implemented in ApplicationSample and ApplicationSampleTest classes.

📑 Instructions

1. Create a fork of this repository into your own account.
2. There are 9 challenges described below. Read them and think for a solution.
3. Code your solution for each challenge. (Preferred using Java, Go, JavaScript or Ruby)
4. Test your solution. You are free to test the project as you wish.
5. Commit your code and push it to your own repository.
6. Create a pull request to the main repo(Skills Project).

🥋 Challenges

Across Years

Resume

Providing an input (years) and assuming it´s always an int, return the number of centuries for provided input. How many centuries have N years?

Testing cases

yearsToCenturies(1) ==> 1 why? 1st century

yearsToCenturies(99) ==> 1 why? 1st century

yearsToCenturies(101) ==> 2 why? 2nd century

yearsToCenturies(2020) ==> 21 why? 21st century

Scramble

Resume

This method search a text equality inside another text like SCRAMBLE. You can only use the amount of letters provided in the original word for your searching string. No more instructions needed, right?

Testing cases

scramble("hdsasdellsdasooawwadwdwawrl", "helloworld") ==> True

scramble("dsasdeldasooawwadwdwaw", helloworld) ==> False (There is no "h")

scramble("hd ell oo w rl", "helloworld") ==> True

Middle Char

Resume

Find a middle char(s) (max 2) in a text. Always expect a single word without spaces.

Testing cases

getMiddle("pair") ==> ai

getMiddle("odd") ==> d

getMiddle("finish") ==> ni

Duplicated Count

Resume

Find how many letters are repeated into a word. Note that case-sensitive is irrelevant and spaces don´t count.

Testing cases

duplicateCount("hey") ==> 0 No word is repeated

duplicateCount("hello") ==> 1 Letter 'l' is repeated

duplicateCount("AcCoMmOdatIOn") ==> 4 Letters 'a', 'c', 'o' and 'm' are repeated.

VIPs Vocals

Resume

Count the number of vocals that are contained in the provided text. Consider multiple words and upper/lower cases.

Testing cases

countVocals("n0n3") ==> 0

countVocals("aeiou AEIOU") ==> 10

countVocals("vowel counting is not an everyday thing") ==> 12

Recursive Multiplier

Resume

You should recursively multiply all numbers contained in certain number until you have a 1-digit number.

Testing cases

recursiveMultiply(20) ==> 0 why? [2 * 0] = 0

recursiveMultiply(236) ==> 8 why? [2 * 3 * 6] = 36 [3 * 6] = 18 [1 * 8] = 8

Recursive Sum

Resume

You must recursively sum all numbers contained in certain number until you have a 1-digit number.

Testing cases

recursiveSum(20) ==> 2 why? [2 + 0] = 2

recursiveSum(236) ==> 2 why? [2 + 3 + 6] = 11 [1 + 1] = 2

Array Couple

Resume

You must return an array of paired chars for the provided input phrase. Spaces and odd phrases are valid

Testing cases

coupleArray("hey") ==> ["he", "y"]

coupleArray("hello world") ==> ["he", "ll", "o ", "wo", "rl", "d"]

coupleArray("Give me some pairs please") ==> ["Gi","ve"," m","e ","so","me"," p","ai","rs"," p","le","as","e_"]

Give me this IP

Resume

This method convert a long into a valid IP address.

How?

1. Convert a long in binary (you will always get 32 bits or fewer, if there is less just apply the pad to the left with zero).
2. Get 4 bytes (4 groups of 8 bits will give you 4 bytes).
3. Converts each byte to a decimal number.
4. Apply the IP format to your 4 decimal numbers.

Testing cases

longToIP(2147483467L) ==> 127.255.255.75