You are given an array (which will have a length of at least 3, but could be very large) containing integers. The array is either entirely comprised of odd integers or entirely comprised of even integers except for a single integer `N`. Write a method that takes the array as an argument and returns this "outlier" `N`.

## Examples

```python
[2, 4, 0, 100, 4, 11, 2602, 36]
Should return: 11 (the only odd number)

[160, 3, 1719, 19, 11, 13, -21]
Should return: 160 (the only even number)
```

Source: codewars | Difficulty: EASY

Starter code: def find_outlier(integers):
	

Solutions:
def find_outlier(int):
	odds = [x for x in int if x % 2 != 0]
	evens = [x for x in int if x % 2 == 0]
	return odds[0] if len(odds) < len(evens) else evens[0]


def find_outlier(integers):
	(even, odd) = (0, 0)
	for i in integers:
		if i % 2:
			even += 1
			last_even = i
		else:
			odd += 1
			last_odd = i
	return last_even if even == 1 else last_odd


def find_outlier(integers):
	parity = [n % 2 for n in integers]
	return integers[parity.index(1)] if sum(parity) == 1 else integers[parity.index(0)]


def find_outlier(integers):
	parity = [n % 2 for n in integers]
	return integers[parity.index(sum(parity) == 1)]


def find_outlier(integers):
	listEven = []
	listOdd = []
	for n in integers:
		if n % 2 == 0:
			listEven.append(n)
		else:
			listOdd.append(n)
	if len(listEven) == 1:
		return listEven[0]
	else:
		return listOdd[0]


def find_outlier(nums):
	base_parity = sum((x % 2 for x in nums[:3])) // 2
	for i in range(len(nums)):
		if nums[i] % 2 != base_parity:
			return nums[i]


def find_outlier(integers):
	assert len(integers) >= 3
	bit = (integers[0] & 1) + (integers[1] & 1) + (integers[2] & 1) >> 1
	for n in integers:
		if n & 1 ^ bit:
			return n
	assert False


def find_outlier(integers):
	evens = []
	odds = []
	for each in integers:
		if each % 2 == 0:
			evens.append(each)
		else:
			odds.append(each)
		if len(evens) > 1 and len(odds) == 1:
			return odds[0]
		elif len(odds) > 1 and len(evens) == 1:
			return evens[0]


def find_outlier(integers):
	determinant = [x for x in integers[:3] if x % 2 == 0]
	if len(determinant) > 1:
		mod = 1
	else:
		mod = 0
	for i in integers:
		if i % 2 == mod:
			return i


def find_outlier(num):
	odd = 0 if sum((n % 2 for n in num[:3])) > 1 else 1
	return next((n for n in num if n % 2 == odd))