In this Kata, you will be given a string and your task is to determine if that string can be a palindrome if we rotate one or more characters to the left.

```Haskell
solve("4455") = true, because after 1 rotation, we get "5445" which is a palindrome
solve("zazcbaabc") = true, because after 3 rotations, we get "abczazcba", a palindrome
```

More examples in test cases. Input will be strings of lowercase letters or numbers only.

Good luck!

Source: codewars | Difficulty: EASY

Starter code: def solve(s):
	

Solutions:
def solve(s):
	return any((s[i + 1:] + s[:i + 1] == s[i::-1] + s[:i:-1] for i in range(len(s))))


def solve(s):
	for i in range(len(s)):
		xs = s[i:] + s[:i]
		if xs == xs[::-1]:
			return True
	return False


from collections import deque

def solve(st):
	s = deque(st)
	return any((s == deque(reversed(s)) for _ in range(len(st)) if not s.rotate(1)))


is_pal = lambda s: all((s[i] == s[len(s) - i - 1] for i in range(len(s) // 2)))
solve = lambda s: any((is_pal(s[i:] + s[:i]) for i in range(len(s))))


def is_palindrome(s):
	for (i, (a, b)) in enumerate(zip(s, s[::-1])):
		if a != b:
			return False
		if i == len(s) // 2:
			return True

def solve(st):
	for i in range(len(st)):
		if is_palindrome(st[i:] + st[:i]):
			return True
	return False


def rotate(list_):
	return list_[1:] + list_[:1]

def is_palindrome(list_):
	return list_ == list_[::-1]

def solve(st):
	list_ = list(st)
	for _ in range(len(st)):
		if is_palindrome(list_):
			return True
		else:
			list_ = rotate(list_)
	return False


(q, solve) = (lambda p: p == p[::-1], lambda s: any((q(s[x:] + s[:x]) for x in range(len(s)))))


solve = lambda s, c=0: s == s[::-1] or solve(s[-1] + s[:-1], c=c + 1) if c != len(s) else 0


def solve(st):
	for x in range(len(st)):
		if st == st[::-1]:
			return True
		else:
			st = st[1:] + st[0]
	return False