# Codewars Practices

Welcome to [Codewars](https://www.codewars.com/) Archive Katas! This is a section specifically for MATB members to post their own exercises (katas) of code, here we will be using mainly practices from codewars, though other platforms or homework are welcomed!

| Kata | Short Analysis |
| - | - |
| Duplicate Encoder | applying `casefold()`
| Pyramid height | basic math and counter algorithms
| RGB To Hex Conversion | nested functions and list maniuplations
| URL Extraction | N/A
| Coefficients of the Quadratic Equation | basic math algorithms
| Factorial | nested functions
| Reverse Words | string manipulations
| Where My Anagrams at? | `set()` and nested functions
| Counting Characters in a String | nested loops and dictionary manipulations
| Counting Vowels | basic loop algorithms
| Categorize New Member | manipulating tuples in lists

### Made by:

## Duplicate Encoder

The goal of this exercise is to convert a string to a new string where each character in the new string is "(" if that character appears only once in the original string, or ")" if that character appears more than once in the original string. Ignore capitalization when determining if a character is a duplicate.

Example:

"din"      =>  "((("
"recede"   =>  "()()()"
"Success"  =>  ")())())"
"(( @"     =>  "))(("

### Code

def duplicate_encode(word):
    x = word.casefold()
    output = ''
    for char in x:
        if x.count(char) > 1:
            output = output + ")"
        else:
            output = output + "("
    return output

## Pyramid Height

Your task is to calculate the maximum possible height of a perfectly square pyramid (the number of complete layers) that we can build, given n number of cubes as the argument.

- The top layer is always only 1 cube and is always present.
- There are no hollow areas, meaning each layer must be fully populated with cubes.
- The layers are thus so built that the corner cube always covers the inner 25% of the corner cube below it and so each row has one more cube than the one below it.
- If you were given only 5 cubes, the lower layer would have 4 cubes and the top 1 cube would sit right in the middle of them, where the lower 4 cubes meet.

*If you were given 14 cubes, you could build a pyramid of 3 layers, but 13 wouldn't be enough as you would be missing one cube, so only 2 layers would be complete and some cubes left over!*

What is the tallest pyramid possible we can build from the given number of cubes? Simply return the number of complete layers.

Example:

4  -->  1
 5  -->  2
13  -->  2
14  -->  3

### Code

def pyramid_height(n):
    m = 0
    counter = 0
    deduction = n - m**2
    while deduction > 0:
        m += 1
        deduction = deduction - m**2
        counter += 1
    if deduction < 0:
        return counter - 1
    return counter

## RGB To Hex Conversion

The rgb function is incomplete. Complete it so that passing in RGB decimal values will result in a hexadecimal representation being returned. Valid decimal values for RGB are 0 - 255. Any values that fall out of that range must be rounded to the closest valid value.

Note: Your answer should always be 6 characters long, the shorthand with 3 will not work here.

The following are examples of expected output values:

rgb(255, 255, 255) # returns FFFFFF
rgb(255, 255, 300) # returns FFFFFF
rgb(0,0,0) # returns 000000
rgb(148, 0, 211) # returns 9400D3

### Code

def rgb(r, g, b):
    def rgb_out_of_range_values(a):
        if a < 0:
            a = 0
            return a
        elif a > 255:
            a = 255
            return a
        else:
            return(int(a/16))
    def rgb_out_of_range_values2(c):
        if c < 0:
            c = 0
            return c
        elif c > 255:
            c = 255
            return c
        else:
            return(int(c%16))
    Hex1 = rgb_out_of_range_values(r)
    Hex3 = rgb_out_of_range_values(g)
    Hex5 = rgb_out_of_range_values(b)
    Hex2 = rgb_out_of_range_values2(r)
    Hex4 = rgb_out_of_range_values2(g)
    Hex6 = rgb_out_of_range_values2(b)
    RGB = [Hex1,Hex2,Hex3,Hex4,Hex5,Hex6]
    Hex = []
    Hexdecimal_alphabets = "ABCDEF"
    def Hex_helper(y):
        if y >= 0 and y <10:
            return str(y)
        elif y >= 10 and y <= 16:
            y = y - 10
            x = Hexdecimal_alphabets[y]
            return x
        elif y < 0:
            x = min(Hexdecimal_alphabets)
            return x
        else:
            x = max(Hexdecimal_alphabets)
            return x
    for s in range(6):
        Hex.append(Hex_helper(RGB[s]))
    return("".join(Hex))

## URL Extraction

Write a function that when given a URL as a string, parses out just the domain name and returns it as a string. For example:

* url = "http://github.com/carbonfive/raygun" -> domain name = "github"
* url = "http://www.zombie-bites.com"         -> domain name = "zombie-bites"
* url = "https://www.cnet.com"                -> domain name = cnet"

### Code

def domain_name(url):
    #tried way too many times for this, can the author of this kata specify more on the test cases in the instructions section, the required 'domain' is not even a domain to begin with.
    #this is just a practice of trial and error tbh, it doesn't take subdomains into account, and i'm surprised the amount of test cases that are not included in instructions.
    print(url)
    domain = ""
    numbers = '1234567890'
    def domain_helper(loop):
        new_domain = ""
        for s in range(len(domain)):
            if domain[s] == '.':
                return new_domain
            else:
                new_domain = new_domain + domain[s]
                s += 1
    if url[4] == 's' and url[0] == 'h' and url[11] != '.':
        domain = url.split("https://")
        domain = domain[1]
        return domain_helper(domain)
    elif url[0] == 'w' and url[1] == 'w' and url[2] == 'w':
        domain = url.split("www.")
        domain = domain[1]
        return domain_helper(domain)
    elif url[4] == ':' and url[0] == 'h' and url[10] != '.':
        domain = url.split("http://")
        domain = domain[1]
        return domain_helper(domain)
    elif url[0] == 'h' and url[7] not in numbers and url[10] == '.':
        domain = url.split("http://www.")
        domain = domain[1]
        return domain_helper(domain)
    elif url[0] == 'h' and url[8] not in numbers and url[11] == '.':
        domain = url.split("https://www.")
        domain = domain[1]
        return domain_helper(domain)
    else:
        if len(url) <= 9:
            domain = url
            return domain_helper(domain)
        elif url[9] in numbers and len(url) <= 15:
            domain = url.split("https://")
            domain = domain[1]
            return domain_helper(domain)
        else:
            domain = url
            return domain_helper(domain)

## Coefficients of the Quadratic Equation

In this Kata you are expected to find the coefficients of quadratic equation of the given two roots (x1 and x2).

Equation will be the form of $ax^2 + bx + c = 0$

Return type is a Vector (tuple in Rust, Array in Ruby) containing coefficients of the equations in the order (a, b, c).

Since there are infinitely many solutions to this problem, we fix a = 1.

Remember, the roots can be written like $(x-x_1)(x-x_2) = 0$

- example:

quadratic(1,2) = (1, -3, 2)

This means $(x-1)(x-2) = 0$; when we do the multiplication this becomes $x^2 - 3x + 2 = 0$

Notes:

- Inputs will be integers
- When `x1 == x2`, this means the root has the multiplicity of two

### Code

def quadratic(x1, x2):
    return(1, -(x1 + x2), (x1 * x2))

## Factorial

In mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example:

$5! = 5 * 4 * 3 * 2 * 1 = 120$

By convention, $0! = 1$

Write a function to calculate factorial for a given input. If input is below 0 or above 12 throw an exception of type `ArgumentOutOfRangeException` (C#) or `IllegalArgumentException` (Java) or `RangeException` (PHP) or throw a `RangeError` (JavaScript) or `ValueError` (Python) or `return -1` (C).

### Code

def factorial(n):
    def fact_helper(k):
        if k == 1:
            return k
        else:
            return k * factorial(k-1)
    if n > 0 and n <= 12:
        return fact_helper(n)
    elif n == 0:
        return 1
    else:
        raise ValueError

## Reverse Words

Complete the solution so that it reverses all of the words within the string passed in.

Example:

"The greatest victory is that which requires no battle" --> "battle no requires which that is victory greatest The"

### Code

def reverse_words(s):
    s = s.split()[::-1]
    reverse = []
    for counter in s:
        reverse.append(counter)
    return " ".join(reverse)

## Where My Anagrams at?

What is an anagram? Well, two words are anagrams of each other if they both contain the same letters. For example:

'abba' & 'baab' == true

'abba' & 'bbaa' == true

'abba' & 'abbba' == false

'abba' & 'abca' == false

Write a function that will find all the anagrams of a word from a list. You will be given two inputs a word and an array with words. You should return an array of all the anagrams or an empty array if there are none. For example:

anagrams('abba', ['aabb', 'abcd', 'bbaa', 'dada']) => ['aabb', 'bbaa']

anagrams('racer', ['crazer', 'carer', 'racar', 'caers', 'racer']) => ['carer', 'racer']

anagrams('laser', ['lazing', 'lazy',  'lacer']) => []

### Code

def anagrams(word, words): 
    #random test cases are too little to judge the practicality of this code.
    #if u find there's errors or cases that don't work, message me.
    print(word, words)
    def words_helper(y):    
        unique_chars_words = set(y)
        unique_chars_words = list(unique_chars_words)
        if unique_chars_words == unique_chars and len(y) == len(word):
            anagrams.append(y)
            return anagrams
        else:
            false_anagrams.append(y)
            return false_anagrams
    unique_chars = set(word)
    unique_chars = list(unique_chars)
    anagrams = []
    false_anagrams = []
    for i in range(len(words)):
        words_helper(words[i])
    print(anagrams, false_anagrams)
    return anagrams

## Prime Numbers

- You can assume you will be given an integer input.
- You can not assume that the integer will be only positive. You may be given negative numbers as well ( or 0 ).
- NOTE on performance: There are no fancy optimizations required, but still the most trivial solutions might time out. Numbers go up to $2^{31}$ ( or similar, depending on language ). Looping all the way up to `n`, or `n/2`, will be too slow.

is_prime(1)  /* false */
is_prime(2)  /* true  */
is_prime(-1) /* false */

### Code

## Counting Characters in a String

The main idea is to count all the occurring characters in a string. If you have a string like `aba`, then the result should be `{'a': 2, 'b': 1}`.

What if the string is empty? Then the result should be empty object literal, `{}`.

### Code

def count(string):
    unique_chars = list(set(string))
    values = []
    test_dict = {}
    if len(string) == 0:
        return test_dict
    else:
        for counter in range(len(unique_chars)):
            value_counter = 0
            for i in range(len(string)):
                if string[i] ==  unique_chars[counter]:
                    value_counter = value_counter + 1
                else:
                    i += 1
            values.append(value_counter)
        test_dict = dict(zip(unique_chars, values))
        print(test_dict)
        return test_dict

## Permute a Palidrome

Write a function that will check whether ANY permutation of the characters of the input string is a palindrome. Bonus points for a solution that is efficient and/or that uses only built-in language functions. Deem yourself **brilliant** if you can come up with a version that does not use any function whatsoever.

madam -> True
adamm -> True
junk -> False

### Code

def permute_a_palindrome(input): #unfinished
    print(input)
    if input == input[::-1]:
        return True
    else:
        return False

## Counting Vowels

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.

### Code

def get_count(sentence):
    vowels = 'aeiou'
    counter = 0
    if len(sentence) == 0:
        return 0
    else:
        for i in range(len(sentence)):
            if sentence[i] in vowels:
                counter += 1
        return counter

## 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.**

Input:

Input will consist of a list of pairs. Each pair contains information for a single potential member. Information consists of an integer for the person's age and an integer for the person's handicap.

Output:

Output will consist of a list of string values (in Haskell and C: `Open` or `Senior`) stating whether the respective member is to be placed in the senior or open category.

Example:

input =  [[18, 20], [45, 2], [61, 12], [37, 6], [21, 21], [78, 9]]
output = ["Open", "Open", "Senior", "Open", "Open", "Senior"]

### Code

def open_or_senior(data):
    list = []
    def tuples_helper(tuple):
        if tuple[0] >= 55 and tuple[1] > 7:
            return "Senior"
        else:
            return 'Open'
    for i in range(len(data)):
        list.append(tuples_helper(data[i]))
    return list

## Complimentary DNA

Deoxyribonucleic acid (DNA) is a chemical found in the nucleus of cells and carries the "instructions" for the development and functioning of living organisms.

If you want to know more: http://en.wikipedia.org/wiki/DNA

In DNA strings, symbols "A" and "T" are complements of each other, as "C" and "G". Your function receives one side of the DNA (string, except for Haskell); you need to return the other complementary side. DNA strand is never empty or there is no DNA at all (again, except for Haskell).

More similar exercise are found here: http://rosalind.info/problems/list-view/ (source)

Example:

"ATTGC" --> "TAACG"
"GTAT" --> "CATA"

### Code

def DNA_strand(dna):
    dna_case = []
    compliment = {"A": "T", "T": "A", "G": "C", "C": "G"}
    for i in dna:
        dna_case.append(compliment.get(i,i))
    return "".join(dna_case)

Shortened:

def DNA_strand(dna):
    compliment = {"A": "T", "T": "A", "G": "C", "C": "G"}
    return "".join(compliment.get(i,i) for i in dna)

## Is This a Triangle?

Implement a function that accepts 3 integer values `a`, `b`, `c`. The function should return true if a triangle can be built with the sides of given length and false in any other case.

(In this case, all triangles must have surface greater than 0 to be accepted).

### Code

def is_triangle(a, b, c):
    if a <= 0 or b <= 0 or c <= 0:
        return False
    else:
        if a + b > c and b + c > a and a + c > b:
            return True
        else:
            return False

Shortened:

def is_triangle(a, b, c):
    return (a + b > c) and (b + c > a) and (a + c > b)

## Mumbling

This time no story, no theory. The examples below show you how to write function `accum`:

Example:

accum("abcd") -> "A-Bb-Ccc-Dddd"
accum("RqaEzty") -> "R-Qq-Aaa-Eeee-Zzzzz-Tttttt-Yyyyyyy"
accum("cwAt") -> "C-Ww-Aaa-Tttt"

Parameters: `a..z` and `A..Z`

### Code

def accum(s):
    mumble = ""
    for i in range(len(s)):
        mumble = mumble + s[i].upper() + i * s[i].lower() + "-"
    return mumble[:-1:]

## Speed Control

In John's car the GPS records every `s` seconds the distance travelled from an origin (distances are measured in an arbitrary but consistent unit). For example, below is part of a record with `s = 15`:

x = [0.0, 0.19, 0.5, 0.75, 1.0, 1.25, 1.5, 1.75, 2.0, 2.25]

The sections are:

0.0-0.19, 0.19-0.5, 0.5-0.75, 0.75-1.0, 1.0-1.25, 1.25-1.50, 1.5-1.75, 1.75-2.0, 2.0-2.25

We can calculate John's average hourly speed on every section and we get:

[45.6, 74.4, 60.0, 60.0, 60.0, 60.0, 60.0, 60.0, 60.0]

Given `s` and `x` the task is to return as an integer the `*floor*` of the maximum average speed per hour obtained on the sections of `x`. If `x` length is **less than or equal to 1 return 0 since the car didn't move.**

Example:

with the above data your function `gps(s, x)` should return `74`.

Note:

With floats it can happen that results depends on the operations order. To calculate hourly speed you can use:

(3600 * delta_distance) / s

### Code

def gps(s, x):
    if len(x) == 0 or sum(x) <= 1:
        return 0
    else:
        average_hour_speed = []
        for i in range(len(x)-1):
            hourly_speed = (3600 * (x[i+1] - x[i]))/s
            average_hour_speed.append(hourly_speed)
        return max(average_hour_speed)

Shortened:

def gps(s,x):
    if len(x) == 0 or sum(x) <= 1:
        return 0
    else:
        max_distance = max(x[i+1]-x[i] for i in range(len(x)-1))
        return (3600 * max_distance)/s