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.

Difficulty: introductory

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))
#    return next((a for a, b in zip(sorted(arr1), sorted(arr2)) if a != b), sorted(arr1)[-1])


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]