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)

Difficulty: introductory

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