[LeetCode] 1385. Find the Distance Value Between Two Arrays

- Problem

1385. 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
• -1000 <= arr1[i], arr2[j] <= 1000
• 0 <= d <= 100

- 내 풀이

class Solution:
    def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:
        answer = 0
        
        for i in range(len(arr1)):
            for j in range(len(arr2)):
                if abs(arr1[i] - arr2[j]) <= d:
                    break
            else:
                answer += 1    
        
        return answer

- 결과

이해하는 데 한참은 걸린 문제이다. (결국엔 이해를 못하고 DIscuss 참고)
아니나 다를까 불만이 많던 릿코드 토론방

웩