In this Kata your task will be to return the count of pairs that have consecutive numbers as follows:
```Haskell
pairs([1,2,5,8,-4,-3,7,6,5]) = 3
The pairs are selected as follows [(1,2),(5,8),(-4,-3),(7,6),5]
--the first pair is (1,2) and the numbers in the pair are consecutive; Count = 1
--the second pair is (5,8) and are not consecutive
--the third pair is (-4,-3), consecutive. Count = 2
--the fourth pair is (7,6), also consecutive. Count = 3. 
--the last digit has no pair, so we ignore.
```

More examples in the test cases. 

Good luck!

Please also try [Simple time difference](https://www.codewars.com/kata/5b76a34ff71e5de9db0000f2)

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Data structures

Starter code: def pairs(ar):
	

Solutions:
def pairs(ar):
	count = 0
	for i in range(0, len(ar), 2):
		try:
			(a, b) = (ar[i], ar[i + 1])
		except IndexError:
			return count
		if abs(a - b) == 1:
			count += 1
	return count


def pairs(arr):
	return sum((abs(a - b) == 1 for (a, b) in zip(arr[::2], arr[1::2])))


def pairs(ar):
	res = 0
	for i in range(1, len(ar), 2):
		if abs(ar[i] - ar[i - 1]) == 1:
			res += 1
	return res


pairs = lambda a: sum((abs(b - a) == 1 for (a, b) in zip(*[iter(a)] * 2)))


def pairs(ar):
	return sum((1 for (a, b) in zip(ar[::2], ar[1::2]) if min(a, b) + 1 == max(a, b)))


def pairs(ar):
	return sum((1 for n in range(0, len(ar) - 1, 2) if abs(ar[n] - ar[n + 1]) == 1))


def pairs(ar, count=0):
	for c in range(len(ar) - 1):
		if c % 2 == 0 and (ar[c + 1] == ar[c] + 1 or ar[c + 1] == ar[c] - 1):
			count += 1
	return count


pairs = lambda arr: sum((1 for i in range(len(arr) // 2) if abs(arr[i * 2] - arr[i * 2 + 1]) == 1))


def pairs(ar):
	a = []
	count = 0
	if len(ar) % 2 == 0:
		for i in range(0, len(ar), 2):
			if abs(ar[i] - ar[i + 1]) == 1:
				count += 1
	else:
		for i in range(0, len(ar) - 1, 2):
			if abs(ar[i] - ar[i + 1]) == 1:
				count += 1
	return count