>When no more interesting kata can be resolved, I just choose to create the new kata, to solve their own, to enjoy the process  --myjinxin2015 said

# Description:
 Given two array of integers(`arr1`,`arr2`). Your task is going to find a pair of numbers(an element in arr1, and another element in arr2), their difference is as big as possible(absolute value); Again, you should to find a pair of numbers, their difference is as small as possible. Return the maximum and minimum difference values by an array: `[  max difference,  min difference  ]`

 For example:
 ```
 Given arr1 = [3,10,5], arr2 = [20,7,15,8]
 should return [17,2] because 20 - 3 = 17, 10 - 8 = 2
 ```
 
# Note:
 - arr1 and arr2 contains only integers(positive, negative or 0);
 - arr1 and arr2 may have different lengths, they always has at least one element;
 - All inputs are valid.
 - This is a simple version, if you want some challenges, [try the challenge version](https://www.codewars.com/kata/583c592928a0c0449d000099).
 
# Some Examples
 ```
 maxAndMin([3,10,5],[20,7,15,8]) === [17,2]
 maxAndMin([3],[20]) === [17,17]
 maxAndMin([3,10,5],[3,10,5]) === [7,0]
 maxAndMin([1,2,3,4,5],[6,7,8,9,10]) === [9,1]
 ```

Difficulty: introductory

Starter code:

def max_and_min(arr1,arr2):
	

Solutions:
def max_and_min(arr1,arr2):
    diffs = [abs(x-y) for x in arr1 for y in arr2]
    return [max(diffs), min(diffs)]


def max_and_min(arr1,arr2):
    max_min = [abs(y-x) for x in arr1 for y in arr2]
                        
    return [max(max_min), min(max_min)]

def max_and_min(arr1, arr2):
  return [op(abs(x-y) for x in arr1 for y in arr2) for op in [max, min]]

def max_and_min(arr1,arr2):
  arr1.sort()
  arr2.sort()
  i = 0
  difference = []

  while i < len(arr1):
    local = []
    for element in arr2 :
      local_dif = abs(arr1[i] - element)
      local.append(local_dif)
    local.sort()  
    difference.append(min(local))
    difference.append(max(local))

    i += 1
  difference.sort()
  maximum = max(difference)
  minimum = min(difference)
  return [maximum,minimum]

def max_and_min(lst1, lst2):
    diff = sorted(abs(b - a) for a in lst1 for b in lst2)
    return [diff[-1], diff[0]]

def max_and_min(arr1,arr2):
    def mini(i, j):
        if i == len(a1) or j == len(a2):
            return float('inf')
        val = a1[i] - a2[j]
        if val < 0:
            return min(-val, mini(i+1, j))
        return min(val, mini(i, j+1))    
    
    a1, a2 = sorted(arr1), sorted(arr2)
    return [max(a2[-1]-a1[0], a1[-1]-a2[0]), bool(set(a1)^set(a2)) and mini(0, 0)]

def max_and_min(arr1,arr2):
    return [max(abs(a-b) for a in arr1 for b in arr2), min(abs(a-b) for a in arr1 for b in arr2)]

def max_and_min(arr1, arr2):
    max_diff, min_diff = float('-inf'), float('inf')
    for a in arr1:
        for b in arr2:
            current_sum = abs(b - a)
            max_diff = max(max_diff, current_sum)
            min_diff = min(min_diff, current_sum)
    return [max_diff, min_diff]


def max_and_min(arr1,arr2):
    mi,ma=float("inf"), -float("inf")
    for x in arr1:
        for y in arr2:
            d = abs(x-y)
            if d < mi: mi = d
            if d > ma: ma = d
    return [ma,mi]

def max_and_min(arr1,arr2):
    l=[]
    for temp1 in arr1:
        for temp2 in arr2:
            l.append(abs(temp1-temp2))
            l.sort()
    return [l[-1],l[0]]