An array is called `centered-N` if some `consecutive sequence` of elements of the array sum to `N` and this sequence is preceded and followed by the same number of elements. 

Example:
```
[3,2,10,4,1,6,9] is centered-15
because the sequence 10,4,1 sums to 15 and the sequence 
is preceded by two elements [3,2] and followed by two elements [6,9]

```

Write a method called `isCenteredN` that returns :

- `true` if its array argument is `not empty` and `centered-N` or empty and centered-0
- otherwise returns `false`.

Difficulty: EASY
Starter code: def is_centered(arr,n):
	

Solutions:
def is_centered(arr, n):
	l = int(len(arr) / 2) if len(arr) % 2 == 0 else int((len(arr) - 1) / 2)
	return any((sum(arr[i:-i]) == n for i in range(1, l + 1))) or sum(arr) == n


from itertools import accumulate

def is_centered(arr, n):
	xs = [0] + list(accumulate(arr))
	return any((b - a == n for (a, b) in zip(xs[:len(arr) // 2 + 1], xs[::-1])))


def is_centered(arr, n):
	while True:
		if sum(arr) == n:
			return True
		try:
			(arr.pop(), arr.pop(0))
		except IndexError:
			return False


def is_centered(arr, n):
	if sum(arr) == n:
		return True
	for i in range(1, int(len(arr) / 2) + 1):
		if sum(arr[i:-i]) == n:
			return True
	return False


def is_centered(arr, n):
	x = sum(arr)
	if x == n:
		return True
	for i in range(len(arr) >> 1):
		x -= arr[i] + arr[-i - 1]
		if x == n:
			return True
	return False


def is_centered(a, n):
	if not n and (not len(a) % 2):
		return 1
	for i in range(sum(divmod(len(a), 2)) - 1, -1, -1):
		if sum(a[i:len(a) - i]) == n:
			return 1
	return 0


def is_centered(arr, n):
	return any((sum(arr[i:len(arr) - i]) == n for i in range(len(arr) // 2 + 1)))


def is_centered(arr, n):
	for i in range(len(arr)):
		for j in range(len(arr) + 1):
			if j > i:
				if sum(arr[i:j]) == n and i == len(arr) - j:
					return True
	return True if n == 0 and arr != arr[::-1] else False


def is_centered(arr, n):
	if len(arr) % 2 == 0 and n == 0:
		return True
	(left, right) = (len(arr) // 2, len(arr) // 2) if len(arr) % 2 else (len(arr) // 2 - 1, len(arr) // 2)
	while left >= 0 and right <= len(arr):
		if sum(arr[left:right + 1]) == n:
			return True
		left -= 1
		right += 1
	return False