Given an array of integers, find the one that appears an odd number of times.

There will always be only one integer that appears an odd number of times.

Difficulty: EASY
Starter code: def find_it(seq):
	

Solutions:
def find_it(seq):
	for i in seq:
		if seq.count(i) % 2 != 0:
			return i


def find_it(seq):
	counts = dict()
	for n in seq:
		if n not in counts:
			counts[n] = 0
		counts[n] += 1
	for (n, n_count) in list(counts.items()):
		if n_count % 2 != 0:
			return n


def find_it(seq):
	for n in set(seq):
		if seq.count(n) % 2 != 0:
			return n


def find_it(seq):
	return [x for x in seq if seq.count(x) % 2][0]


from collections import Counter

def find_it(l):
	return [k for (k, v) in Counter(l).items() if v % 2 != 0][0]


def find_it(seq):
	nums = set()
	for num in seq:
		if num in nums:
			nums.remove(num)
		else:
			nums.add(num)
	return nums.pop()


def find_it(seq):
	for elem in set(seq):
		if seq.count(elem) % 2 == 1:
			return elem


def find_it(seq):
	result = 0
	for x in seq:
		result ^= x
	return result


from itertools import groupby

def find_it(seq):
	return next((k for (k, g) in groupby(sorted(seq)) if len(list(g)) % 2))