```
x = []
def find_outlier(integers):
count = 0
oddcount = 0
evencount = 0
#loop through length of integers
for n in integers:
#count should equal length of list by the end, because it goes up for every iteration through integers list
count += 1
#below counting number of odd and even integers with oddcount and evencount
if n % 2 != 0:
oddcount += 1
x[0] = n
if n % 2 == 0:
eventcount += 1
x[0] = n
#loop is over. If count memory is retained from loop, then it should go through the below conditional nest.
#Only one conditional should trigger, since the list is meant to contain only 1 even or one odd
if len(integers) == count:
if oddcount == 1:
return x[0]
if evencount == 1:
return x[0]
```

The instructions are

```
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.
```