An onion array is an array that satisfies the following condition for all values of `j` and `k`:

If all of the following are `true`:

* `j >= 0`
* `k >= 0`
* `j + k = array.length - 1`
* `j != k`
  
then:

* `a[j] + a[k] <= 10`

### Examples:

```
[1, 2, 19, 4, 5]   =>  true  (as 1+5 <= 10 and 2+4 <= 10)
[1, 2, 19, 4, 10]  =>  false (as 1+10 > 10)
```

Write a function named `isOnionArray`/`IsOnionArray`/`is_onion_array()` that returns `true` if its argument is an onion array and returns `false` if it is not.

~~~if:php
Your solution should at least be moderately efficient.  Make sure you don't do any unnecessary looping ;)
~~~

Difficulty: introductory

Starter code:

def is_onion_array(a):
	

Solutions:
def is_onion_array(a):
    return all(a[i] + a[-i-1] <= 10 for i in range(len(a) // 2))

from typing import List


def is_onion_array(a: List[int]) -> bool:
    return all(aj + ak <= 10 for aj, ak in zip(a, a[:(len(a) - 1) // 2:-1]))


def is_onion_array(a):
    for i in range(0,len(a)//2):
        if a[i]+a[-i-1] > 10:
            return False
    return True

def is_onion_array(a):
    l = len(a)
    return all(a[i] + a[l - i - 1] <= 10 for i in range(l // 2))





def is_onion_array(a):
    for j in range(len(a)//2):
        if a[j] + a[-j-1] > 10:
            return False
    return True
    
    


def is_onion_array(a):
    if a == []:
        return True
    a = [sum(x) for x in zip(a, a[::-1])]
    a.pop(len(a)//2)
    for element in a:
        if element > 10:
            return False
    return True

def is_onion_array(a):
  i = 0; j = -1
  while i != len(a)//2:
    if a[i] + a[j] > 10:
      return False
    i += 1
    j -= 1
  return True

def is_onion_array(a):
    mid = len(a)//2
    
    for i in range(0,mid):
        if a[i] + a[-i-1] > 10:
            return False
    return True

def is_onion_array(a):
    print(a)
    for num in range(len(a)):
        if a[num] !=a[-(num+1)]:
            if not a[num] + a[-(num + 1)] <=10:
                return False
        
            
        
    
    return True

from itertools import combinations as cb
def is_onion_array(a):
    return not any([sum(c) == len(a)-1 and a[c[0]]+a[c[1]] > 10 for c in cb([*range(0,len(a))],2)])