# Task
Consider an array of integers `a`. Let `min(a)` be its minimal element, and let `avg(a)` be its mean.

Define the center of the array `a` as array `b` such that:

```
- b is formed from a by erasing some of its elements.
- For each i, |b[i] - avg(a)| < min(a).
- b has the maximum number of elements among all the arrays
  satisfying the above requirements.
```

Given an array of integers, return its center.



# Input/Output


`[input]` integer array `a`

 Unsorted non-empty array of integers.

`2 ≤ a.length ≤ 50,`

`1 ≤ a[i] ≤ 350.`

`[output]` an integer array


# Example

For `a = [8, 3, 4, 5, 2, 8]`, the output should be `[4, 5]`.

Here `min(a) = 2, avg(a) = 5`.

For `a = [1, 3, 2, 1]`, the output should be `[1, 2, 1]`.

Here `min(a) = 1, avg(a) = 1.75`.

Source: codewars | Difficulty: EASY

Starter code: def array_center(arr):
	

Solutions:
def array_center(lst):
	return [i for i in lst if abs(i - sum(lst) * 1.0 / len(lst)) < min(lst)]


from statistics import mean

def array_center(arr):
	(low, avg) = (min(arr), mean(arr))
	return [b for b in arr if abs(b - avg) < low]


from statistics import mean

def array_center(a):
	(n, m) = (min(a), mean(a))
	return [x for x in a if abs(x - m) < n]


def array_center(arr):
	(m, a) = (min(arr), sum(arr) / len(arr))
	return [n for n in arr if abs(n - a) < m]


def array_center(arr):
	avg = sum(arr) / len(arr)
	return [i for i in arr if abs(i - avg) < min(arr)]


def array_center(arr):
	a = min(arr)
	c = []
	b = 0
	for i in arr:
		b = b + i
	aver = b / len(arr)
	for i in arr:
		if abs(i - aver) < a:
			c.append(i)
	return c


def array_center(a):
	(mi, me) = (min(a), sum(a) / len(a))
	return [x for x in a if abs(x - me) < mi]


array_center = lambda a: [n for n in a if abs(n - sum(a) / len(a)) < min(a)]


from numpy import mean

def array_center(arr):
	(m, a) = (min(arr), mean(arr))
	return list(filter(lambda n: abs(n - a) < m, arr))


from statistics import mean

def array_center(a):
	(m, avg) = (min(a), mean(a))
	return [i for i in a if -m < i - avg < m]