# Task
 You are given a string `s`. Every letter in `s` appears once. 
 
 Consider all strings formed by rearranging the letters in `s`. After ordering these strings in dictionary order, return the middle term. (If the sequence has a even length `n`, define its middle term to be the `(n/2)`th term.)

# Example

 For `s = "abc"`, the result should be `"bac"`.
```
The permutations in order are:
"abc", "acb", "bac", "bca", "cab", "cba"
So, The middle term is "bac".```

# Input/Output


 - `[input]` string `s`

  unique letters (`2 <= length <= 26`)

 - `[output]` a string

  middle permutation.

Source: codewars | Difficulty: MEDIUM
Tags: Ad-hoc

Starter code: def middle_permutation(string):
	

Solutions:
def middle_permutation(string):
	s = sorted(string)
	if len(s) % 2 == 0:
		return s.pop(len(s) // 2 - 1) + ''.join(s[::-1])
	else:
		return s.pop(len(s) // 2) + middle_permutation(s)


def middle_permutation(s):
	s = ''.join(sorted(s))
	return s[len(s) // 2 - 1:(len(s) + 1) // 2][::-1] + s[(len(s) + 1) // 2:][::-1] + s[:len(s) // 2 - 1][::-1]


from math import factorial

def nth_permutation(s, n):
	if not s:
		return ''
	m = factorial(len(s) - 1)
	(q, r) = divmod(n, m)
	return s[q] + nth_permutation(s[:q] + s[q + 1:], r)

def middle_permutation(s):
	return nth_permutation(sorted(s), factorial(len(s)) // 2 - 1)


def middle_permutation(s):
	s = ''.join(sorted(s, reverse=True))
	return s[len(s) // 2:(len(s) + 3) // 2] + s[:len(s) // 2] + s[(len(s) + 3) // 2:]


def middle_permutation(string):
	s = ''.join(sorted(string))
	mid = int(len(s) / 2) - 1
	if len(s) % 2 == 0:
		return s[mid] + (s[:mid] + s[mid + 1:])[::-1]
	else:
		return s[mid:mid + 2][::-1] + (s[:mid] + s[mid + 2:])[::-1]


def middle_permutation(string):
	letters = sorted(string, reverse=True)
	start = letters.pop(len(string) // 2)
	if len(string) % 2:
		start += letters.pop(len(string) // 2)
	return start + ''.join(letters)


from math import factorial

def factoradic_representation(n):
	res = []
	i = 1
	while n > 0:
		res.append(n % i)
		n = n // i
		i += 1
	return res[::-1]

def middle_permutation(string):
	if len(string) < 2:
		return string
	s = sorted(string)
	res = []
	fr = factoradic_representation(factorial(len(s)) // 2 - 1)
	fr = [0] * (len(string) - len(fr)) + fr
	for idx in fr:
		res.append(s.pop(idx))
	return ''.join(res)


def middle_permutation(string):
	(rev, l) = (sorted(string)[::-1], len(string))
	mid = rev[l // 2:l // 2 + l % 2 + 1]
	return ''.join(mid + [ch for ch in rev if ch not in mid])


def middle_permutation(s):
	s = ''.join(sorted(s))
	m = int(len(s) / 2)
	x = s[m - 1:m + 1] if len(s) % 2 else s[m - 1]
	return (s.replace(x, '') + x)[::-1]


def middle_permutation(string):
	sortedString = ''.join(sorted(string))
	reversedString = sortedString[::-1]
	result = ''
	result += reversedString[int(len(reversedString) / 2):int((len(reversedString) + 3) / 2)]
	result += reversedString[0:int(len(reversedString) / 2)]
	result += reversedString[int((len(reversedString) + 3) / 2):]
	return result