1385. Find the Distance Value Between Two Arrays

1385. Find the Distance Value Between Two Arrays #

Problem #

Given two integer arrays arr1 and arr2, and the integer dreturn 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

Summary #

Given two integer arrays arr1 , arr2 and an integer d , return the distance value between the two arrays. The “distance value” is defined as the number of elements that satisfy this distance requirement: for an element arr1[i] , there does not exist any element arr2[j] such that |arr1[i]-arr2[j]| <= d .

Constraints:

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

Solution Approach #

  • Calculate the distance between the two arrays. The distance value is defined as the number of elements that satisfy the condition: for an element arr1[i], there does not exist any element arr2[j] such that |arr1[i]-arr2[j]| <= d.
  • This is an easy problem. According to the definition of the distance value, simply use nested loops to count.

Code #


package leetcode

func findTheDistanceValue(arr1 []int, arr2 []int, d int) int {
	res := 0
	for i := range arr1 {
		for j := range arr2 {
			if abs(arr1[i]-arr2[j]) <= d {
				break
			}
			if j == len(arr2)-1 {
				res++
			}
		}
	}
	return res
}

func abs(a int) int {
	if a < 0 {
		return -1 * a
	}
	return a
}


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