There is an array with some numbers. All numbers are equal except for one. Try to find it!

```python
find_uniq([ 1, 1, 1, 2, 1, 1 ]) == 2
find_uniq([ 0, 0, 0.55, 0, 0 ]) == 0.55
```

It’s guaranteed that array contains at least 3 numbers.

The tests contain some very huge arrays, so think about performance.

This is the first kata in series:

1. Find the unique number (this kata)
2. [Find the unique string](https://www.codewars.com/kata/585d8c8a28bc7403ea0000c3)
3. [Find The Unique](https://www.codewars.com/kata/5862e0db4f7ab47bed0000e5)

Difficulty: EASY
Starter code: def find_uniq(arr):
	

Solutions:
def find_uniq(arr):
	(a, b) = set(arr)
	return a if arr.count(a) == 1 else b


def find_uniq(arr):
	if arr[0] == arr[1]:
		return [x for x in arr[2:] if x != arr[0]][0]
	return arr[0] if arr[1] == arr[2] else arr[1]


def find_uniq(arr):
	s = set(arr)
	for e in s:
		if arr.count(e) == 1:
			return e


def find_uniq(arr):
	a = sorted(arr)
	return a[0] if a[0] != a[1] else a[-1]


from collections import Counter

def find_uniq(a):
	return Counter(a).most_common()[-1][0]


def find_uniq(arr):
	if arr[0] != arr[1]:
		if arr[0] != arr[2]:
			return arr[0]
		else:
			return arr[1]
	else:
		for i in arr[2:]:
			if i != arr[0]:
				return i


from collections import Counter

def find_uniq(arr):
	return next((k for (k, v) in Counter(arr).items() if v == 1))


def find_uniq(arr):
	return [x for x in set(arr) if arr.count(x) == 1][0]


def find_uniq(arr):
	arr.sort()
	return arr[0] if arr[0] != arr[1] else arr[-1]


def find_uniq(arr):
	res = sorted(arr)
	return res[0] if res[0] != res[1] else res[-1]


def find_uniq(arr):
	culprit = arr[1]
	if arr[0] == arr[2]:
		culprit = arr[0]
	for i in arr:
		if i != culprit:
			return i