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 = 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
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).