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
Brute-force.
class Solution:
def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:
res = 0
for a in arr1:
exist = False
for b in arr2:
if abs(a - b) <= d:
exist = True
break
if not exist:
res += 1
return res
class Solution {
public int findTheDistanceValue(int[] arr1, int[] arr2, int d) {
int res = 0;
for (int a : arr1) {
boolean exist = false;
for (int b : arr2) {
if (Math.abs(a - b) <= d) {
exist = true;
break;
}
}
if (!exist) {
++res;
}
}
return res;
}
}
class Solution {
public:
int findTheDistanceValue(vector<int>& arr1, vector<int>& arr2, int d) {
int res = 0;
for (auto& a : arr1) {
bool exist = false;
for (auto& b : arr2) {
if (abs(a - b) <= d) {
exist = true;
break;
}
}
if (!exist) ++res;
}
return res;
}
};
func findTheDistanceValue(arr1 []int, arr2 []int, d int) int {
res := 0
for _, a := range arr1 {
exist := false
for _, b := range arr2 {
if math.Abs(float64(a-b)) <= float64(d) {
exist = true
break
}
}
if !exist {
res++
}
}
return res
}