In this kata, you've to count lowercase letters in a given string and return the letter count in a hash with 'letter' as key and count as 'value'. The key must be 'symbol' instead of string in Ruby and 'char' instead of string in Crystal.

Example: 

```python
letter_count('arithmetics') #=> {"a": 1, "c": 1, "e": 1, "h": 1, "i": 2, "m": 1, "r": 1, "s": 1, "t": 2}
```

Source: codewars | Difficulty: EASY

Starter code: def letter_count(s):
	

Solutions:
from collections import Counter

def letter_count(s):
	return Counter(s)


def letter_count(s):
	return {i: s.count(i) for i in s}


from collections import Counter as letter_count


def letter_count(s):
	letters = {}
	for i in list(s):
		if i in letters:
			letters[i] += 1
		else:
			letters[i] = 1
	return letters


import collections

def letter_count(s):
	res = collections.defaultdict(int)
	for _char in s:
		res[_char] += 1
	return res


def letter_count(s):
	return {c: s.count(c) for c in set(s)}


def letter_count(s):
	return dict(((letter, s.count(letter)) for letter in sorted(set(s))))


def letter_count(s):
	count = {}
	for c in s:
		count[c] = count.get(c, 0) + 1
	return count


import collections

def letter_count(s):
	result = {}
	for (i, d) in enumerate(s):
		count = 1
		if d not in result:
			for j in range(i + 1, len(s)):
				if d == s[j]:
					count += 1
			result[d] = count
	return collections.OrderedDict(sorted(result.items()))


def letter_count(s):
	count = {}
	for i in s:
		if i in count:
			count[i] += 1
		else:
			count[i] = 1
	return count