Formatted question description: https://leetcode.ca/all/1385.html

# 1385. Find the Distance Value Between Two Arrays (Easy)

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=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=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=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

Related Topics:
Array

## Solution 1. Brute Force

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

// Time: O(N^2)
// Space: O(1)
class Solution {
public:
int findTheDistanceValue(vector<int>& A, vector<int>& B, int d) {
int ans = 0;
for (int i = 0; i < A.size(); ++i) {
bool found = false;
for (int j = 0; j < B.size() && !found; ++j) {
if (abs(A[i] - B[j]) <= d) found = true;
}
if (!found) ++ans;
}
return ans;
}
};


## Solution 2. Brute Force

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

// Time: O(N^2)
// Space: O(1)
class Solution {
public:
int findTheDistanceValue(vector<int>& A, vector<int>& B, int d) {
return count_if(begin(A), end(A), [&](const auto &a) {
return all_of(begin(B), end(B), [&](const auto &b) {
return abs(a - b) > d;
});
});
}
};


For each A[i], find the length of range [A[i] - d, A[i] + d] in array B. If the length is 0, then increment the answer.

The start point of the range (A[i] - d) can be found using lower_bound(begin(B), end(B), A[i] - d).

The next element after the end point of the range (the element after A[i] + d) can be found using upper_bound(begin(B), end(B), n + d).

If these two iterators are the same, it means the length of the range is 0.

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

// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int findTheDistanceValue(vector<int>& A, vector<int>& B, int d) {
sort(begin(B), end(B));
int ans = 0;
for (int n : A) {
if (lower_bound(begin(B), end(B), n - d) == upper_bound(begin(B), end(B), n + d)) ++ans;
}
return ans;
}
};


Java

class Solution {
public int findTheDistanceValue(int[] arr1, int[] arr2, int d) {
int count = 0;
int length1 = arr1.length, length2 = arr2.length;
for (int i = 0; i < length1; i++) {
int num1 = arr1[i];
boolean flag = true;
for (int j = 0; j < length2; j++) {
int num2 = arr2[j];
if (Math.abs(num1 - num2) <= d) {
flag = false;
break;
}
}
if (flag)
count++;
}
return count;
}
}