You'll be given a list of two strings, and each will contain exactly one colon (`":"`) in the middle (but not at beginning or end). The length of the strings, before and after the colon, are random.

Your job is to return a list of two strings (in the same order as the original list), but with the characters after each colon swapped.


## Examples
```
["abc:123", "cde:456"]  -->  ["abc:456", "cde:123"]
["a:12345", "777:xyz"]  -->  ["a:xyz", "777:12345"]
```

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def tail_swap(strings):
	

Solutions:
def tail_swap(strings):
	(head0, tail0) = strings[0].split(':')
	(head1, tail1) = strings[1].split(':')
	return [head0 + ':' + tail1, head1 + ':' + tail0]


def tail_swap(arr):
	fmt = '{}:{}'.format
	((head, tail), (head_2, tail_2)) = (a.split(':') for a in arr)
	return [fmt(head, tail_2), fmt(head_2, tail)]


def tail_swap(strings):
	((h1, t1), (h2, t2)) = (s.split(':') for s in strings)
	return [f'{h1}:{t2}', f'{h2}:{t1}']


def tail_swap(ss):
	a = [s.split(':') for s in ss]
	(a[0][1], a[1][1]) = (a[1][1], a[0][1])
	return [':'.join(p) for p in a]


def tail_swap(lst):
	((a, b), (c, d)) = (s.split(':') for s in lst)
	return list(map(':'.join, ((a, d), (c, b))))


def tail_swap(s):
	(h1, t1) = s[0].split(':')
	(h2, t2) = s[1].split(':')
	return ['{}:{}'.format(h1, t2), '{}:{}'.format(h2, t1)]


def tail_swap(strings):
	s = [x.split(':') for x in strings]
	return [s[0][0] + ':' + s[1][1], s[1][0] + ':' + s[0][1]]


def tail_swap(strings):
	return '{0}:{3} {2}:{1}'.format(*[j for i in strings for j in i.split(':')]).split()


def tail_swap(strings):
	(a, b) = strings
	(a, b) = (a.split(':'), b.split(':'))
	return [f'{a[0]}:{b[1]}', f'{b[0]}:{a[1]}']


def tail_swap(strings):
	(a, b) = [ele.split(':') for ele in strings]
	return [f'{a[0]}:{b[1]}', f'{b[0]}:{a[1]}']