# Challenge Problems

## Fizzbuzz style problems

# Return an array containing the numbers from 1 to N, where N is the parametered value. N will never be less than 1.

# Replace certain values however if any of the following conditions are met:

# If the value is a multiple of 3 & 5: use the value "FizzBuzz" instead
# If the value is a multiple of 3: use the value "Fizz" instead
# If the value is a multiple of 5: use the value "Buzz" instead

def fizzbuzz(n):
  # empty list to return
  fb_list = []
  # itereate up to n
  for i in range(1, n + 1):
  # use module to find if they are a multiple of 3 or 5
  # if i % 3 == 0 and i % 5 == 0:
  if i % 15 == 0: 
    fb_list.append('FizzBuzz')
  elif i % 3 == 0: 
    fb_list.append('Fizz')
  elif i % 5 == 0: 
    fb_list.append('Buzz')
  else: 
    fb_list.append(i)
  
  return fb_list

# If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

# Finish the solution so that it returns the sum of all the multiples of 3 or 5 below the number passed in.

def solution(number):
  # running total
  total = 0
  # 
  for i in range(1, number):
  if i % 3 == 0 or i % 5 == 0: 
    total += i

  return total

def recursive_solve(number, total = 0, i = 0):
  # base case
  if i >= number:
  return total
  
  if i % 3 == 0 or i % 5 == 0:
  total += i
  
  recursive_solve(number, total, i + 1)

# Given an array of integers, find the one that appears an odd number of times.

# There will always be only one integer that appears an odd number of times.

def find_it(seq):
  # hash table to store numbers
  hash = {}
  # count how many times each number appears
  for number in seq:
    if number not in hash: 
      hash[number] = 1 
    else:
      hash[number] += 1

  # find the one odd number
  for number in hash:
    if hash[number] % 2 != 0: return number

## Array Problems

# Sum Of Negatives And Count Of Positives

# Write a function that takes in a list of numbers and gives you back a tuple or list with two items: the sum of all negative numbers in the list, and the count of all positive numbers.

# For input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, -11, -12, -13, -14, -15], you should return [10, -65]. (There are 10 positive numbers, and the negative numbers add up to -65.)

# Using a loop.
def count_positives_sum_negatives(nums):
  count = 0
  total = 0
  for num in nums:
    if num > 0:
      count += 1
    else:
      total += num

  return [count, total]

# Using list comprehensions.
def count_positives_sum_negatives(nums):
  return [len([num for num in nums if num > 0]), sum([num for num in nums if num < 0])]

# Testing it!
print(count_positives_sum_negatives([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, -11, -12, -13, -14, -15]))

# Implement a function unique_in_order that will take in an argument either a string or an array and return a list of elements without any elements with the same # values next to each other, but preserving the original order.

# unique_in_order('AAAABBBCCDAABBB') == ['A', 'B', 'C', 'D', 'A', 'B']
# unique_in_order('ABBCcAD')         == ['A', 'B', 'C', 'c', 'A', 'D']
# unique_in_order([1,2,2,3,3])       == [1,2,3]


# Whenever your value is not what the last one is,
# update what the last one is,
# and append the new value to our result.
def unique_in_order(iterable):
    result = []
    prev = None
    for element in iterable:
        if element != prev:
            result.append(element)
            prev = element
    return result

# Or just check against the most recently thing added.
def unique_in_order(iterable):
    result = []
    for element in iterable:
        if len(result) == 0 or element != result[-1]:
            result.append(element)
    return result

## String Problems

# Sum Mixed Array: easiest
# Given an array of integers as strings and numbers, return the sum of the array values as if all were numbers.
# Return your answer as a number.

def sum_mix(arr):
  # # keep a running total
  total = 0

  # iterate over every item in the list
  for element in arr:
    # check the type if it is a string
    if type(element) == str:
      total += int(element)
    # add the item to our total
    else:
      total += element

  return total 
  # solution using list comprehension
  # https://www.w3schools.com/python/python_lists_comprehension.asp
  # return sum(int(element) for element in arr)

# Vowel Count: mediumest
# 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.

def get_count(input_str):
    num_vowels = 0
    # loop over input_str
    vowels = 'aeiou'
    for letter in input_str:
      # check if the letter is in vowels
      if letter in vowels: num_vowels += 1

    return num_vowels

# Convert string to camel case: hardest
# Complete the method/function so that it converts dash/underscore delimited words into camel casing. The first word within the output should be capitalized only if the original word was capitalized (known as Upper Camel Case, also often referred to as Pascal case).
# Examples
# "the-stealth-warrior" gets converted to "theStealthWarrior"
# "The_Stealth_Warrior" gets converted to "TheStealthWarrior"

def to_camel_case(text):
    # replace underscores with dashes so the strings can be split
    words = text.replace('_', '-')
    # split the string at dashes and make into a list 
    words = words.split('-')
    # this will be the return string -- start it off with the first word so it doesn't get capitalized
    string = words[0] 
    # loop over the rest of the words and and capitalize them (range 1 - list length)
    for i in range(1, len(words)):
        # add the capitalized words to the string 
        string += words[i].capitalize()
    
    # return the string
    return string