## Description

https://leetcode.com/problems/find-the-distance-value-between-two-arrays/

Given two integer arrays `arr1`

and `arr2`

, and the integer `d`

, *return the distance value between the two arrays*.

The distance value is defined as the number of elements `arr1[i]`

such that there is not any element `arr2[j]`

where `|arr1[i]-arr2[j]| <= d`

.

**Example 1:**

Input:arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2Output:2Explanation:For arr1[0]=4 we have: |4-10|=6 > d=2 |4-9|=5 > d=2 |4-1|=3 > d=2 |4-8|=4 > d=2 For arr1[1]=5 we have: |5-10|=5 > d=2 |5-9|=4 > d=2 |5-1|=4 > d=2 |5-8|=3 > d=2 For arr1[2]=8 we have:|8-10|=2 <= d=2|8-9|=1 <= d=2|8-1|=7 > d=2|8-8|=0 <= d=2

**Example 2:**

Input:arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3Output:2

**Example 3:**

Input:arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6Output:1

**Constraints:**

`1 <= arr1.length, arr2.length <= 500`

`-10^3 <= arr1[i], arr2[j] <= 10^3`

`0 <= d <= 100`

## Explanation

Iterate elements from the two lists and check how many distance between two elements are qualify.

## Python Solution

```
class Solution:
def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:
distance = 0
for num1 in arr1:
is_qualify = True
for num2 in arr2:
diff = abs(num1 - num2)
if diff <= d:
is_qualify = False
break
if is_qualify:
distance += 1
return distance
```

- Time Complexity: O(N^2).
- Space Complexity: O(1).