You are given an array of `n+1` integers `1` through `n`. In addition there is a single duplicate integer.

The array is unsorted.

An example valid array would be `[3, 2, 5, 1, 3, 4]`. It has the integers `1` through `5` and `3` is duplicated. `[1, 2, 4, 5, 5]` would not be valid as it is missing `3`.

You should return the duplicate value as a single integer.

Difficulty: EASY
Starter code: def find_dup(arr):
	

Solutions:
def find_dup(arr):
	return sum(arr) - len(arr) * (len(arr) - 1) / 2


def find_dup(arr):
	return max([i for i in arr if arr.count(i) > 1])


from collections import Counter
find_dup = lambda xs: Counter(xs).most_common(1)[0][0]


def find_dup(arr):
	return sum(arr) - sum(range(len(arr)))


find_dup = lambda l: [v for (i, v) in enumerate(l) if l.index(v) != i][0]


def find_dup(arr):
	for i in range(1, len(arr)):
		for j in range(i):
			if arr[i] == arr[j]:
				return arr[i]
	return None


def find_dup(arr):
	return next((n for n in arr if arr.count(n) == 2))