Given two integer arrays where the second array is a shuffled duplicate of the first array with one element missing, find the missing element.

Please note, there may be duplicates in the arrays, so checking if a numerical value exists in one and not the other is not a valid solution.

```
find_missing([1, 2, 2, 3], [1, 2, 3]) => 2
```
```
find_missing([6, 1, 3, 6, 8, 2], [3, 6, 6, 1, 2]) => 8
```

The first array will always have at least one element.

Source: codewars | Difficulty: EASY
Tags: Data structures

Starter code: def find_missing(arr1, arr2):
	

Solutions:
def find_missing(arr1, arr2):
	return sum(arr1) - sum(arr2)


from functools import reduce

def find_missing(arr1, arr2):
	from itertools import chain
	from functools import reduce
	from operator import xor
	return reduce(xor, chain(arr1, arr2))


def find_missing(A1, A2):
	return sum(A1) - sum(A2)


def find_missing(a, b):
	for i in a:
		try:
			b.remove(i)
		except:
			return i


def find_missing(arr1, arr2):
	for x in set(arr1):
		if arr1.count(x) != arr2.count(x):
			return x


def find_missing(arr1, arr2):
	return next((a for a in set(arr1) if arr1.count(a) != arr2.count(a)))


from functools import reduce
from operator import xor

def find_missing(xs, ys):
	return reduce(xor, xs, 0) ^ reduce(xor, ys, 0)


find_missing = lambda *a, C=__import__('collections').Counter: C.__sub__(*map(C, a)).popitem()[0]


from collections import Counter

def find_missing(*args):
	(c1, c2) = map(Counter, args)
	return (c1 - c2).popitem()[0]


def find_missing(a, b):
	for x in b:
		a.remove(x)
	return a[0]