In this Kata, you will be given two strings `a` and `b` and your task will be to return the characters that are not common in the two strings. 

For example:
```Haskell
solve("xyab","xzca") = "ybzc" 
--The first string has 'yb' which is not in the second string. 
--The second string has 'zc' which is not in the first string. 
```
Notice also that you return the characters from the first string concatenated with those from the second string.

More examples in the tests cases. 

Good luck!

Please also try [Simple remove duplicates](https://www.codewars.com/kata/5ba38ba180824a86850000f7)

Source: codewars | Difficulty: EASY
Tags: String algorithms

Fundamentals

Starter code: def solve(a,b):
	

Solutions:
def solve(a, b):
	s = set(a) & set(b)
	return ''.join((c for c in a + b if c not in s))


def solve(a, b):
	return ''.join([c for c in a if not c in b] + [c for c in b if not c in a])


def solve(a, b):
	unique = []
	for i in a:
		if i not in b:
			unique.append(i)
	for i in b:
		if i not in a:
			unique.append(i)
	return ''.join(unique)


def solve(a, b):
	res = ''
	for char in a + b:
		if char in a and char in b:
			continue
		res += char
	return res


def solve(a, b):
	return ''.join((c for c in a + b if c in set(a) ^ set(b)))


def solve(left, right):
	result = ''
	for char in left:
		if char not in right:
			result += char
	for char in right:
		if char not in left:
			result += char
	return result


def solve(a, b):
	string = a + b
	return ''.join([i for i in a if i not in b] + [i for i in b if i not in a])


solve = lambda a, b: a.translate(str.maketrans('', '', b)) + b.translate(str.maketrans('', '', a))


def solve(a, b):
	e = [i for i in a if i not in b]
	f = [i for i in b if i not in a]
	return ''.join(e) + ''.join(f)