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

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

Description

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 = 2Output: 2Explanation: 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=2For 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 = 3Output: 2

Example 3:

Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6Output: 1

Constraints:

1 <= arr1.length, arr2.length <= 500-10^3 <= arr1[i], arr2[j] <= 10^30 <= d <= 100

分析

题目的意思是：给定两个数组，找到arr1的数与arr2所有的数的绝对值差都大于d的所有的数，我仔细思索了一下，还是两个循环暴力破解一下算了。然后我看了其他人的解法好像跟这个双循环差不多。

代码

class Solution: def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int: m=len(arr1) n=len(arr2) cnt=0 for i in range(m): flag=True for j in range(n): dis=abs(arr1[i]-arr2[j]) if(dis<=d): flag=False break if(flag): cnt+=1 return cnt

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