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 = 2 Output: 2 Explanation: 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 = 3 Output: 2 

Example 3:

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

Constraints:

• 1 <= arr1.length, arr2.length <= 500
• -10^3 <= arr1[i], arr2[j] <= 10^3
• 0 <= d <= 100

给你两个整数数组 arr1 ， arr2 和一个整数 d ，请你返回两个数组之间的 距离值 。「距离值」 定义为符合此距离要求的元素数目：对于元素 arr1[i] ，不存在任何元素 arr2[j] 满足 |arr1[i]-arr2[j]| <= d 。

提示：

• 1 <= arr1.length, arr2.length <= 500
• -10^3 <= arr1[i], arr2[j] <= 10^3
• 0 <= d <= 100

• 计算两个数组之间的距离，距离值的定义：满足对于元素 arr1[i] ，不存在任何元素 arr2[j] 满足 |arr1[i]-arr2[j]| <= d 这一条件的元素数目。
• 简单题，按照距离值的定义，双层循环计数即可。

funcfindTheDistanceValue(arr1 []int, arr2 []int, dint) int { res:=0fori:=rangearr1 { forj:=rangearr2 { ifabs(arr1[i]-arr2[j]) <=d { break } ifj==len(arr2)-1 { res++ } } } returnres } funcabs(aint) int { ifa<0 { return-1*a } returna }